We have had a lot of seriousness over the last year. Covid Madness, possibly wonky elections, riots of the Left and riots of the Right, petty criminals become heroes and heroes are characterized as criminals. “Enough!” I say. Time for some intellectual amusement.

When I was a youth, and continuing into my late teens and early 20s, I was a semi-professional stage magician. That is, I was occasionally paid to perform on stage doing magic – it was only a hobby and I never earned anything like a living at it. But in the run up to my semi-professional career, I studied every book I could find — buy, beg or borrow — on the performance of stage magic, illusions, card tricks and the lives of the great magicians of the 19^{th} and early 20^{th} century. Sadly, I did not have the patience to learn coin manipulation tricks (see prestidigitation).

My real strength was the art of misdirection – shifting the attention of the audience away from my manipulation of the objects necessary to carry off the trick. I used an example of one of the techniques in an essay on climate science many years ago: The Magician’s ‘Red Scarf Trick’.

Stage magic, and stage illusions, all rely on the same basic underlying approach in order to make something impossible occur – to make magic. The Magician tells the audience he is doing one thing, the Magician lets the audience see that one thing, but all the while the Magician actually does something else altogether and the results, as far as the audience is concerned, are magical. That requires misdirection, both verbal and physical, and a great deal of *sleight of hand*.

Today I write about something that is very akin to “sleight of hand” – which we can think of as *sleight of mind* in which the Magician makes the audience (even if just an audience of one) think one thing when something else is the reality.

Our kind-hearted magician, *The Magnificent Sam*, is a little short of cash, so, sort of like the famous Penn and Teller, sets up a act in which he can earn some money*. It goes like this:

Sam invites a couple from the audience, Mr. and Mrs. Dupe, to the stage. Sam shows us all three cards, two Jokers and one Ace, then lays them on the table, face down and swirls them around a lot, mixing them up.

Sam then tells the Dupes that he will pay them $10 if they can “find the Ace”, and if they fail, they pay him $10. (Sounds a lot like street-side Three Card Monte doesn’t it?) The Dupes pick the middle card and the Magician puts a little Yellow Star sticker on it, but the Magician doesn’t turn it over yet.

You see, the newlywed Dupes are so young, so cute and needy looking, that the Magician takes pity on them and says, “Look, this card is a Joker”, and turns over the card on the left, leaving two cards – the one the Dupes picked and one other. Sam asks “Now, do you want to stick to your choice – card B? Or switch to the other one, card C?

Remember Mr. and Mrs. Dupe are playing only this once.

**Questions:**

Now, our Kind-hearted Magician, The Magnificent Sam, says “I can’t bring myself to cheat such a nice young couple. When you first picked, you had only a one-in-three chance of being right and winning $10. I’d like to improve your odds to make it fair. Here’s what I’ll do, using my Magic Wand as an Event Eraser, I’ll wave it over the cards like this . . . (waving wand slowly over the three cards now on the table) . . . and erase the last few minutes of time, when you were picking from three cards.”

Sam waves the Magic Wand and lo-and-behold the Joker on the left, in position A, begins to fade from sight leaving just a little shadow. “That’s better,” Sam says, “now, would you like to move your Yellow Star sticker to card C or leave it on card B?” If your star is on the Ace, then you win $10, and if on the Joker, you pay me $10.”

Remember Mr. and Mrs. Dupe are playing only this once.

**Question:**

Take your time and answer the questions in the comments below. I think it will be more interesting if the questions are answered before we read on.

Many of you will have recognized the problem posed here as the classic Monty Hall Problem – alternately named the Monty Hall Dilemma. Marilyn vos Savant famously answered the question in her column in *Parade* in 1990 [ wiki ] resulting in a wide-ranging controversy in the mathematics and statistics worlds. [Here is Briggs’s classic take on the MH Problem.]

Marilyn vos Savant answers, quite correctly, that in our Q2, the Dupes will double their chances of winning if they switch their choice. The number of proofs of this is nearly endless and can be found in the Wiki link.

However, Sam the Kind-hearted Magician gives us another look at the problem. He uses magic to erase the original circumstances of their choice, which was to pick one of three cards (or doors in the original TV show). Having “erased” that event left the Dupes with exactly the same situation as they found themselves when we asked Q1 and Q2.

Exactly? Yes, precisely that.

Let’s check carefully, comparing the scenarios presented by vos Savant and the scenario presented by Sam, the Kind-hearted Magician:

First Point:

Savant – Two remaining cards are in play, one each in positions B and C.

Magic – Two remaining cards are in play, one each in positions B and C.

Second Point:

Savant – One of the two cards still in play is the Ace and one is the Joker.

Magic — One of the two cards still in play is the Ace and one is the Joker.

Point Three:

Savant – No one changes the positions of cards B and C from the moment they are laid on the table.

Magic — No one changes the positions of cards B and C from the moment they are laid on the table.

Point 4:

Savant – The Dupes can choose either card B or pick card C. Presented as “sticking with their original choice” or “switching their choice”.

Magic — The Dupes can choose either card B or pick card C. Presented as leaving the Yellow Star on card B or moving it to card C.

Now, think carefully, are there *any differences* in the real physical universe situations presented above? All three points above are apparently identical in the physical world.

Marilyn vos Savant has been declared, without doubt, to be correct in her classical presentation and answer: The Dupes should switch their choice – doing so doubles their chances of winning – if they switch they win 2/3 of the time and if the stick they win 1/3 of the time.

However, Sam, our Kind-hearted Magician, shows us that if he magically erases the past in which the Dupes chose one card out of three, he creates the exactly the same two-card situation as in Savant’s presentation, but the Dupes now magically have an equal chance of winning or losing when forced to make the exact same choice. There is no longer any advantage in switching or not switching.

The ONLY thing that has changed are the *words of the presentations* of the scenarios – or some idea about it – but *nothing* in the physical world changed at all. Same series of steps, same cards in the same positions, same choices to be made. But, when the Dupes get down to making the fateful choice to “stick or switch”, the odds of winning in the two scenarios end up being quite different.

What has happened here?

**Questions:**

Q4: Was Marilyn vos Savant wrong after all?

Q5: Is Sam’s wand really magical and, after it has been waved over the card table, we find it has actually changed the probabilities?

One Last Question:

Q6: Has there been some *Sleight of Mind*? If so, what was it and who performed it?

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Categories: Statistics

Probability is not in the physical world.

Odds of winning – 0 and their odds won’t change.

Stop thinking about the cards. Imagine that they had picked the left-most card (call it A). Now draw a line between A and the other two. There is only 1/3 chance the ace is on their side of the line. So the chance that it’s NOT on the side they picked is 2/3.

Ah but indeed it is; some Physicists contend that subatomic matter is simply a collection of probability functions. Niels Bohr was reported to have believed that a complete understanding of reality lies forever beyond the capabilities of rational thought.

Thank you Kip Hansen for this beautifully explained example of the conundrums that lie within probability theory.

The easiest way to reason about the Monty Hall problem is to scale it to more than just 3 doors (or 3 cards). Imagine there are 1000 cards, 999 of which are jokers and one of which is an ace. You select a card, then 998 jokers are revealed. Would you switch? Of course you would. The only reason it is so obvious is that your original chance was 1/1000 instead of 1/3, but the fundamentals have not changed.

The whole thing is fixed! Clearly, Sam *knew* that if he turned over the first card, it would be a Joker. He wouldn’t have turned it over otherwise. I think he switched out the Ace for a third Joker, so it doesn’t matter what card the couple picks, he’ll always win. His next move will be to turn over whatever card has a star on it, to show that it’s also a Joker, which means the game is over. Everyone will assume the third card is the Ace, even if it’s no longer on the table.

Sheri ==> An interesting answer to the over all dilemma. But if probability is not in the physical world, and is only an intellectual concept, why do we bother with it? If it doesn’t translate into the physical world, then there is no point to it at all. By this I mean, if understanding the probabilities doesn’t help us make more useful choices then it is no better than sudoku.

DAV ==> Absolutely true — then The Magnificent Sam’s wand is truly magical, as it changes the probabilities when it erases the original choice step.

DAV ==> And, this must tell us something important about the whole concept, the idea, of probability.

Kip,

then The Magnificent Sam’s wand is truly magical, as it changes the probabilities when it erases the original choice step.Maybe I misunderstood something. Originally there were three cards: one chosen and the other two not chosen. Then the magician removes a not-chosen card (which apparently is not an ace). Because the chosen card was one of three the probability that it is an ace is 1/3. There’s a 2/3 chance it is not. It remains so unless one of the other two is shown to be an ace. It’s irrelevant what was done to them: turned over, removed, set on fire, whatever. We already knew one of the not-chosen was a joker. Knowing which one changes nothing.

You seem to be saying the probability has changed. Or am I getting that wrong?

If their odds of winning increase if he removes the joker and he needs the money, why doesn’t he just flip their card over the first time? why would he increase their odds of winning?

Phil R,

by allowing them to switch, he increases their odds of winning. Monty Hall’s motive was to extend the suspense making it more interesting plus he likely didn’t care if they won or not. Not at all clear what the magician gains.

Q1. Does the Magician’s revealing of one of the two Jokers improve the couple’s chance of winning? Does the Magician’s allowing them to pick again improve the couple’s chance of winning?

No, because it’s a con from the start. They don’t have *any* chance of winning. The con man does not play the game with you if he will lose. However, if he’s running many of these scams he’ll let a few winners through once in awhile, so perhaps they have a small chance. Picking again is useless and it’s all on the magician to decide if they win or not.

Q2. Regardless of your answer to Q1, what would you advise our young couple to do? Stick with their first choice, switch their choice, or flip a coin to pick between the two remaining cards?

All these options are immaterial.

Q3. What are the odds – the probabilities – of the Dupes winning now? Do the odds change if they change their pick?

If it’s a con, then *very low*. And no.

Is “sets up a act in which he can earn some money*” supposed to have a footnote?

Q4: Was Marilyn vos Savant wrong after all?

No, if the situation is not run by a con man.

Q5: Is Sam’s wand really magical and, after it has been waved over the card table, we find it has actually changed the probabilities?

Of course not.

One Last Question:

Q6: Has there been some Sleight of Mind? If so, what was it and who performed it?

You performed it on your readers, by convincing them that the magician is on the up-and-up.

Don’t forget the “Mexican Turnover”. He’s a magician after all.

DAV ==> (In response to your “Maybe I misunderstood something . . .) The classic answer to the Monty Hall Problem, as given by Marilyn vos Savant and many many others, including our ever-striving host, is: “Thus, conditional on all the evidence, the probability of winning is 2/3 if you switch and 1/3 if you stay.” (This after a goat or a joker is revealed and a second chance to choose is offered).

So, in the classic solution, the contestant (in our case, the Dupes), after the one Joker is revealed but before any intervention by the Magic Wand, have a 1/3 chance of winning if they stick to their original choice, but a 2/3 chance of winning if they switch. If this is unclear, read the link to Marilyn vos Savant’s original column, the Wiki link, and/or Brigg’s classic blog piece on Monty Hall — all linked in the essay above.

Despite the classic solutions, when The Magnificient Sam waves his wand and “erases the past event” (in which the Dupes chose one card out of three) and they are given the identical-in-every-way chance to stick or switch, their chances of winning, either way, are now 1/2 or fity-fifty as we Americans say.

Magic or Sleight-of-Mind?

Phil R and DAV ==> The original problem is derived from the TV game show, Let’s Make A Deal.

It is usually formulated and solved as follows (from the Wiki):

” Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice?

Vos Savant’s response was that the contestant should switch to the other door. Under the standard assumptions, contestants who switch have a 2/3 chance of winning the car, while contestants who stick to their initial choice have only a 1/3 chance. ”

Monty, the game show host, didn’t care if the contestant won the car or not — the car’s were promotional and provided by auto dealers or manufacturers. The purpose of giving a second chance to choice was, as DAV says, to increase the suspense and excitement for the in studio and home viewing audiences.

Our fictional Kind-heaterd Magician stated his motive, he wanted to make it fairer for those nice newlyweds, the Dupes. But did he?

Nate ==> Actually, in both the original TV game show and in our fictional Magic act, the contestants do have a chance to win if their final choice reveals the prize. Many housewives won brand-new cars on “Let’s Make a Deal” and our fictional The Magnificent Sam would have paid off (he’s no scoundrel!)

Before the Dupes arrive, Sam puts 51 cards face down on his table (the cards consist of a standard desk with jokers, minus 3 aces.

Sam removes 48 cards in turn, leaving the three cards above remaining.

The Dupes then arrive.

Is the initial probability of picking the ace 1/51 or 1/50 or … 1/3?

Kip,

before any intervention by the Magic Wand, have a 1/3 chance of winning if they stick to their original choice, but a 2/3 chance of winning if they switch. If this is unclear, read the link to Marilyn vos Savant’s original columnI thought I said the same thing so, no, it’s not unclear.

when The [Magnificent] Sam waves his wand and “erases the past event” (in which the Dupes chose one card out of three) and they are given the identical-in-every-way chance to stick or switch, their chances of winning, either way, are now 1/2 or fity-fifty as we Americans say.No because nothing really changed. In the final scenario, the remaining un-chosen card is still the same card. The probability that it is an ace is still 2/3 (assuming the ace was ever available for selection — he is a con man). If he had reshuffled using only an ace and a joker and had them pick from only those two then, yes, the probability would be 1/2. But he didn’t do that. He removed a card known to be a joker. It was already known that one of them was a joker. Knowing which doesn’t change your knowledge in any useful way.

The only sleight-of-mind here is the obfuscation created by removing a known joker instead of just leaving it there which is actually irrelevant.

Incidentally, this is the probability from the Dupes perspective. The magician seems to know the position of the ace with certainty.

If I understood the original background to the monty hall solution, the probability was affected by the presenter’s prior knowledge. He revealed a door that wasn’t the right door, because the stage hands controlling the curtains know which is the right spot.

While I like the line of thought on this question, I am walking away from Sam the Benevolent because i know just how susceptible I am to being manipulated by these guys. After the fact, I have a chance of seeing how I was conned. Before the fact, I just know I should walk away or had my cash to him for the lesson.

Our author hints that the star might get moved rather than the nice couple changing their card choice.

Nate ==> Yes, in the one form of the manuscript, there was a footnote linking to the wiki entry for the Penn and Teller show: https://en.wikipedia.org/wiki/Penn_%26_Teller:_Fool_Us somehow lost in multiple edits. Penn and Teller made their money by hosting a popular TV show — not performing a version of Three-card Monty. Thanks for catching that!

As for Monty Hall and our fictional The Magnificent Sam, both are on the “up-and-up” and their is no con involved — we are looking at a serious question about probability through a well-known rather frivolous example.

I should point out that the probability of the remaining unchosen card being and ace changed from 1/3 to 2/3 once it was revealed which of the two was not an ace. Again, think of the choice as a line drawn between chosen and not chosen. The probability that one side has the ace remains unless the unknown cards are changed or the actual position of the ace is revealed.

brad,

If I understood the original background to the monty hall solution, the probability was affected by the presenter’s prior knowledge.No. The presenter’s knowledge is irrelevant outside of allowing him to show which of the remaining choices is not the target. All of the probability that we are considering is from the perspective of the contestants. It’s only THEIR knowledge which matters here.

Or, what about this …

Sam lays out 999,999,999,999 jokers and one ace face down. The Dupes pick a card, Sam then turns over all but the one joker and one ace. Is it now almost certain that switching is a win?

Sam lays out 999,999,999,999 jokers and one ace face down … Is it now almost certain that switching is a win?Yes.

Jim Fedako ==> An interesting similar type of problem. Why not remove 50 cards? Then the probability of the remaining card being that only Ace is 1/51, as it was in the original modified deck. What changes about the probability of the Ace being a viable choice when the dealer leaves three cards on the table? I’m not sure at what point your contestant is picking.

One view is that the contestant is still picking one card out of 51, the original proposition. The other view is, if she is picking when there are three cards left on the table, then the probability depends on the odds of the Ace being one of those three cards. You see, it depends on from which point in time one calculates the probability. If the Ace is not still on the table, then the probability of picking it at this point, is zero. If the Ace is one of the three, then at this point, the probability of picking is is one in three. If you want to know the probability of picking the Ace from the outset, when all 51 cards are first laid on the table…….

Alternately, you could lay it out for the readers here.

DAV ==> Yes, I agree, in your “June 14, 2021 at 11:49 am” you state the standard solution precisely.

But that precisely stated solution must needs ignore the physical fact that after The Mag Sam erases the revealed Joker and gives the loving newlyweds, the Dupes, a chance to pick again, their probability of picking the Ace and winning $10 — at this exact moment in time and under the then existing circumstances — really is 1/2.

Our question today is: How can both of these viewpoints be equally true? (IF THEY ARE . . . )

Probability is conditional. they’re both true, but not at the same time.

DAV ==> (re: June 14, 2021 at 12:02 pm)

You are quite right about the two Unchosen Cards having a 2/3 chance of being the Ace — but only because there are two of them. We do not yet know if the Chosen Card is the Ace or not. That one of the Unchosen Cards is a Joker has a probability of 100% and revealing this fact is just that — we already knew that but still do not know if the OTHER Unchosen Card is a Joker or the Ace — but we have a 100% probability that it is one or the other. Is it not true that *now* the Chosen Card also has a 100% probability if being either an Ace or a Joker?

(re: June 14, 2021 at 12:06 pm)

Your answer to Brad is spot on — “The presenter’s knowledge is irrelevant outside of allowing him to show which of the remaining choices is not the target” (in our case, is a JOKER). All of the probability that we are considering is from the perspective of the contestants. It’s only THEIR knowledge which matters here.”

I’m tempted to write a small program to test the theory out empirically over large numbers of iterations. It seems like someone would have already done that. Anyone aware of an example?

Ann

Ugh, I’ll try to post once more, I enjoyed this brain-teaser.

Q1. Does the Magician’s revealing of one of the two Jokers improve the couple’s chance of winning? Does the Magician’s allowing them to pick again improve the couple’s chance of winning?

My Answer is, Yes and No. The Dupes’ chances were 1 out of 3, so 1/3. when the Magician revealed one of the two Jokers, leaving one Joker and one Ace overturned, their chances became 1 out of 2, or 1/2. So yes, revealing a Joker improved their chances. However, allowing them to pick again from those two cards, still only gives them a 1 out of 2, or 50/50 chance of winning.

Whether you turn over the Joker, or turn it over and then make it disappear, the odds between the two remaining cards is still 1 out of 2 or 50/50.

Q2. Regardless of your answer to Q1, what would you advise our young couple to do? Stick with their first choice, switch their choice, or flip a coin to pick between the two remaining cards?

My Ans: Normally I’d advise them to stick with their first answer. If they’d been wrong, why didn’t the magician just take his $10 then? But we’re told he “felt sorry” for the Dupes, and so he says, “this one’s a Joker” and then turns over a card they didn’t pick, to reveal a Joker.

Now, if he randomly shuffled the three cards as the story claims, how did he know it was a Joker before turning it over? Because he’s cheating. All the more reason to stick with their first choice, since their odds are still 50/50 “on paper.”

Q3. What are the odds – the probabilities – of the Dupes winning now? Do the odds change if they change their pick?

For the above-stated reasons, I’d say the odds of the Dupes winning if they go with their original choice, is better than 50/50, because the Magician already revealed that he’s cheating, and the “probability” is that he’s trying to trick them into changing their pick.

Jim Fedako ==> (re: June 14, 2021 at 12:08 pm)

On the “999,999,999,999 jokers and one ace face down” — the contestant picks when ALL the cards are face down — and then the Host turns over Unchosen cards, all of the revealed cards being Jokers, leaving only our two face-down cards, each of which is either the Ace or a Joker?

I am not quite sure if that’s what you mean.

If so, you are right, we have an identical problem as demonstrated by The Magnificent Sam. There now remains choice between two cards, one an Ace and one a Joker. The question posed in this essay is: Is there really a better chance that the one remaining Unchosen Card is an Ace than that the Chosen Card is the Ace?

And, in fact, someone has done that very thing. The internet provides:

https://www.mathwarehouse.com/monty-hall-simulation-online/

DAV ==> (re: June 14, 2021 at 12:12 pm)

You have answered Jim’s question with an unequivocal “Yes”. From that view we can increase the probability of the remaining Unchosen Card being the Ace to higher and higher degrees of certainty simply by increasing the number of Jokers in the original deck ad infinitum? and then turning all but two cards — one the Chosen Card and one an Unchosen Card — back over to reveal them as Jokers?

Don’t we still just end up with two cards, one the Ace and one a Joker — their identities still unknown?

@Kip –

Again, there are two probabilities being conflated: the probability that a given card is the ace ; and the probability of getting the ace by switching.

Me picking a card at any point in the sequence has no affect on the probability someone else faces when confronted with the remaining cards.

Phil R ==> (re: June 14, 2021 at 12:32 pm) “Probability is conditional. they’re both true, but not at the same time.”

Now I think you’re getting somewhere.

So, if both are true and “not at he same time” — if we were betting our lives and not $10 (or vying for a car and not a goat), how do we know what point in time to select as the basis for making our life-and-death decision to increase our hopes of staying alive?

Remember, in all the examples in the essay, the contestant is only playing once. There are no do-overs.

Kip,

But that precisely stated solution must needs ignore the physical fact that after The Mag Sam erases the revealed Joker and gives the loving newlyweds, the Dupes, a chance to pick again, their probability of picking the Ace and winning $10 — at this exact moment in time and under the then existing circumstances — really is 1/2.If one didn’t know the original scenario then yes the probability is 1/2 because two choices would be the only thing known. Erasing the known joker from the original scenario is no different than showing it was a joker. It doesn’t matter at all what is done with it. It doesn’t help to know where the ace isn’t. It’s why I used thinking about one side of the line vs the other side.

Using knowledge from the original scenario, the probability of the original chosen card being an ace is 1/3 unless it shown that one of the unchosen cards is actually the ace. That means there is a probability of 2/3 the ace is on the unchosen side of the line.

Is it not true that *now* the Chosen Card also has a 100% probability if being either an Ace or a Joker?Yes, assuming the choices are only Ace and Joker. It’s called a tautology.

Cloudbuster ==> (June 14, 2021 at 12:38 pm)

In actual fact, thousands of such tests have been performed, even using the super-computer at Los Alamos National Laboratory. All confirm, using the problem exactly as laid out to Marilyn vos Savant and her conditionals, that switching double your chance of winning.

I have also confirmed using a handy Excel spreadsheet to confirm the same, out to a thousand tries.

That leaves us with the problem that The Magnificent Sam has shown us — that the odds are ALSO 50-50.

Ann Cherry ==> (re: June 14, 2021 at 12:43 pm)

An interesting analysis but hundreds of PhDs and professional statisticians finally agreed with Marilyn vos Savant that switching after the reveal doubles one’s chances of winning.

That’s the classical — it does not, however, explain the fact that The Magnificent Sam’s Magic Wand changes the odds to your 50-50.

I assure you, neither Monty Hall nor Sam are scamming anyone — it’s all on the up-and-up. (Even the magic!)

that leaves us with the problem that The Magnificent Sam has shown us — that the odds are ALSO 50-50.Probability is conditional on given knowledge. Sam didn’t change that knowledge. We still know the choice was to pick one out of three cards. You seem to think that Sam somehow changed that when he discarded the joker.

Jim Fedako ==> (re: June 14, 2021 at 1:00 pm)

Hmmmm — You may be right. When we are presented with three cards (three doors), only one of which is the Winner, with no prior insider knowledge and given one choice to pick one of three, our odds are certainly 1 in 3 of being correct.

We are then presented with a follow-on choice: one of the two unchosen unidentified cards (doors) is shown to be a Loser (a goat or a Joker). In the classical Monty Hall Problem, this is presented as the choice to Stick with the original choice or to Switch to the other remaining door. Sam the Magician does exactly the same things and offers the exactly same choice: stay with your first pick or pick the other card. The ONLY difference is the Magic Wand bit in which Sam claims to have magically erased the past choice of 1 in 3. The classical presentation could be said to simply ignore that the final situation is “two cards — choose one” — just like the Magician’s case.

In fact, in classical presentation (Marilyn vos Savant et al.) depends on the belief that the probability depends only on the original choice being the Ace (or the car)– odds 1 of 3. It is only in the classical presentation that The probability of WINNING is presented as “winning by switching”.

Sam, our Kind-hearted Magician, just says chose one of the two remaining cards — by switching or sticking.

There are no differences between the two situations in the physical world — each ends with two cards, both unknown, each randomly either the Ace or a Joker. The contestant/Dupe is given a free choice to pick either card.

You are right — the Dupes chances to not depend on the when they chose the card — do they?

Read Briggs’ Monty Hall piece first so I’ll take a stab at it:

“Q1. Does the Magician’s revealing of one of the two Jokers improve the couple’s chance of winning?”No, remains 1/3.

“Does the Magician’s allowing them to pick again improve the couple’s chance of winning?”Yes, introduces new rule, now 2 out of 3 scenarios switch wins while only 1/3 scenarios not switching wins.

“Q2. Regardless of your answer to Q1, what would you advise our young couple to do? Stick with their first choice, switch their choice, or flip a coin to pick between the two remaining cards?”Switch.

“Q3. What are the odds – the probabilities – of the Dupes winning now? Do the odds change if they change their pick?”[After Sam’s magic card vanishing act]Nothing substantial changed, so outcomes remain: 2 out of 3 scenarios switch wins while only 1/3 scenarios not switching wins.

Sam did not make that card vanish, he only made it

seemto vanish. It’s like that Lincoln joke; if you call a tail a leg then how many legs does a dog have? Four, ’cause a tail ain’t a leg.A good lesson for this Golden Age of Deception.

How about a one minute video of you doing a magic trick?

That would be interesting.

And skip the article.

This story was boring when it surfaced about 30 years ago.

I feel like Mr. Dupe after wasting my time reading this article.

You’ve written lots of good articles, Hanson,

with an unusually high “batting average”,

but this was not one of them.

Richard Greene

Bingham Farms, Michigan

Proud Lifetime Member of the

American “Knit Pickers” Association

DAV ==> (re: June 14, 2021 at 1:03 pm)

We can easily see that something is going on that makes the Monty Hall Problem no so cut and dried — no so “one right answer”.

You manage to construct a scenario in which the probability of Winning or Losing depends on whether or not you are “in the know” about the original scenario (“If one didn’t know the original scenario then yes the probability is 1/2”). But if you know the original scenario, you have a 2/3 chances of winning if you switch your original choice.

Let’s see how this would look: Monty (or Sam) says: Pick B or C. One is an Ace and one is a Joker. You pick B. Your chance of winning is 1/2. But but but — then someone sneaks up and whispers in your ear “There were originally three cards, A, B and C, and A was a Joker!” Now your chances are only 1/3 if you stick with B and magically become 2/3 if you now pick C?

Kip, I hate to harp, but if this “kind-hearted” magician is “on the up and up” as you say, how did he know the card on the left was a Joker before turning it over?

Do you tell us he’s “kind-hearted” so we’ll think he’s trying to help the Dupes win money? If he knew ahead of time that card A was a Joker, wouldn’t he also know where the Ace is located? If he were “cruel-hearted”, we’d assume he’s trying to get them to change their answer from B to C, so he can win the $10. If he’s so kind, and they have the right answer, why is he leading them on?

It would seem, that changing their answer does not depend upon “the odds” but rather on whether they believe the cheating Magician is cruel, or kind. But either way, once you’ve revealed that third card is a Joker, and taken it off the table literally or figuratively, the Dupes actual chances ought to be 50/50, right?

Then, all those other things besides “chance” come into play, such as observing that the magician is a demonstrable cheater who knew at least one card ahead of time

.

Dean Ericson ==> (re: June 14, 2021 at 1:56 pm)

An interesting view So, your answer is that Sam the Kind-hearted Magician is trying to pull some “Sleight of Mind” by pretending to erase the original fact of “a choice of 1 out of 3” (“There is no magic!”). That Sam’s just saying “the odds are now fifty-fifty” isn’t true, isn’t reality, isn’t the correct view of the probability of our Dupes picking the Ace with a free unforced, random choice between two cards, one of which is an Ace and one of which is a Joker?

How does this sit with you after taking into account what I respond to Dav here: https://wmbriggs.com/post/36178/#comment-198815 ?

@Kip –

I guess it all comes down to probability is not a thing, per se. It is simply has to do with lack of info.

Ann Cherry ==> (re: June 14, 2021 at 2:35 pm)

Oh, that’s quite alright. I don’t mind at all — I accept that I may have not been as clear as I could have been. Those familiar with this old nugget — The Monty Hall Problem — have an advantage and most of those who have ever toyed with probability and mind teasers are familiar with the classical Monty Hall scenario and the rules and conditionals. Folks my age watched the original TV show . . . .

You can still watch whole episodes of “Let’s Make a Deal” via YouTube. I had trouble finding an episode that shows the exact classic scenario as used in the original Monty Hall Problem: Here is the classical scenario:

“Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice?

Vos Savant’s response was that the contestant should switch to the other door.[3] Under the standard assumptions, contestants who switch have a 2/3 chance of winning the car, while contestants who stick to their initial choice have only a 1/3 chance.” https://en.wikipedia.org/wiki/Monty_Hall_problem

I start with a standard presentation, but use cards instead of doors (the rules are the same). Monty Hall and Sam play the same roles. It is true that both Monty and Sam know where the Ace is, but they make only one choice in the game — revealing one Joker (or Goat).

That choice does not affect the outcome, as there is always a Joker (goat) to be revealed between the two unchosen cards (doors).

In the TV show, Monty doesn’t care if the contestant wins or loses — it is all showmanship. The cars are supplied by the show’s sponsors. (I never did think that the contestants actually took the goats home if they “won” them — nor the camels either.)

You’ll have to take Sam at face-value — I made him up and he is a kind-hearted guy.

The question here is whether or not Sam’s “magic” actually CHANGES the odds of the Dupes winning as he claims? Some say Yes, some say No, some say “Depends”.

Given the above, what do you say?

Richard Greene ==> (re: June 14, 2021 at 2:04 pm)

Thanks for checking in but I only perform now for my grandchildren — who can still believe in magic and are easily amazed.

Not everyone is fascinated by mind-teasers and statistical/probability problems. More here than at Anthony’s I suppose.

My wife was reading Marilyn vos Savant’s book “Power of Logical Thinking” (1997) in which vos Savant tells her story about the Monty Hall Problem. My wife asked me to walk her through the logic and probability issues.

Since it is not cut-and-dried, not a “one answer” issue, this essay resulted.

You manage to construct a scenario in which the probability of Winning or Losing depends on whether or not you are “in the know” about the original scenario … someone sneaks up and whispers in your ear “There were originally three cards, A, B and C, and A was a Joker!”

Someone didn’t have to tell us. You told us what the original scenario was and the Dupes are also aware of it. Sam did nothing to change that awareness. He just made the joker he turned over vanish. Vanished/unvanished still the same thing.

There may be those who walked in after the joker vanished. As far as they know, there were only two cards with equal probability of being the ace. But they can’t be the Dupes who were present the whole time. They know there were originally three options.

You are making the classic error of changing the proposition. In a way, it’s like those who add conditions when asked the probability of a particular outcome given there are only N outcomes. Instead you are dropping known facts. You’re still changing what was given. You can’t disregard what you already been given when it’s convenient.

I’m with Ann

[Ann Said : Kip, I hate to harp, but if this “kind-hearted” magician is “on the up and up” as you say, how did he know the card on the left was a Joker before turning it over?]

If he’d picked the card on the left and it was an ACE – whoops – Game over – gonna happen a third of the time so it becomes a variation on the sexing the dog problem

Maybe the Magician KNOWS ONLY about the leftmost card, if it IS in fact the ACE then the magician knows the other two are NOT the Joker and you still have the Monty Hall problem

NOT the ACE – sorry – or was that “NOT the Mama”

Jim Fedako ==> (re: June 14, 2021 at 3:02 pm)

“I guess it all comes down to probability is not a thing, per se. It is simply has to do with lack of info.”

Our host has written an entire book — THE book — on Uncertainty. I think he would agree that probability is just a way of looking into the uncertainties of various situations.

The comment section here today — if read as as a whole — is leading somewhere.

Since it is not cut-and-dried, not a “one answer” issue, this essay resulted..There really is only one answer.

Call the ace 1 and the joker 0.

E is the original scenario description.

The probability of A=1 is 1/3 making Pr(A=0 | E) = 2/3.

That means there is a 2/3 probability of (A=0 & B=1 & C=0 OR A=0 & B=0 & C=1).

Knowing that C=0 just eliminates one term from the OR leaving the other. Knowing that B=0 would do the same thing.

(I’ll have to ponder what DAV just wrote.) Kip, I went back and read Briggs 2018 post that you linked, in honor of Monty Hall. It would seem that everything depends upon Monty knowing behind which door lies the prize.

In your illustration, when you said the Magician mixed up the cards, I assumed the Dupes weren’t being duped right from the start. But he is a magician….

Do the odds change to 50/50 after one card’s been removed, if all other things are equal, but the card dealer isn’t a magician or Monte Hall, and doesn’t know in advance where the prize (card) is located?

Let’s say that I, a clueless person (so true for math and magic) lay out three cards. You pick “B”. I give your card its marker, and then randomly turn over one of the other two, and I choose card A. Lucky for you, it’s not the Ace.

If neither of us knows which card is the Ace, how does this change your odds either way? Aren’t they now exactly 50/50, or 1 out of 2?

When you remove one Joker, or one Goat, you still have a Joker and Ace, or Goat and Car. When you remove a factor, you change the equation or odds, re-figure them, right? You don’t use the original “model” of three cards or three prizes, your model is now two, so your “new” odds 50/50.

I guess I’m still missing something here? 🙂

DAV ==> You said, far above, “Probability is conditional on given knowledge. Sam didn’t change that knowledge. We still know the choice was to pick one out of three cards. You seem to think that Sam somehow changed that when he discarded the joker.”

My point about the Whisperer is simply to test the proposition that the probabilities change (or not) depending on whether the Dupes know or don’t know that there were originally three cards.

My Whisperer hypothesis is just that, if the Dupes arrive at the table when there are only the two showing there is no argument that their odds of randomly picking the winner out of those two cards is 50-50.

Does their chance of picking he correct one-card-out-of-two change when the Whisperer runs out and gives them the news that there was once, before the arrived at the table, a third card and it had been revealed to be one of two Jokers (in the original stack of two Jokers and one Ace). ??

I am exploring here, not giving answers.

Game 1:

There are three facedown cards, two are jokers and one is Ace. The game is played in four phases.

In the first phase, Mr. Dupe will select a card. In the second phase, Sam will reveal a card (not the Ace).

In the third phase, Sam will pretend he didn’t reveal a card. In the fourth phase, Mr. Dupe will decide to switch or not switch, and either win or lose.

Evidence:

At the start of the game, the Ace is either (L)eft, (M)iddle, (R)ight. The others are jokers.

Scenario A:

1) Assume the Ace is the Middle Card.

2) Mr. Dupe selects the middle card.

3) Sam chooses to reveal either the Left or the Right card (either one)

4) Sam pretends he didn’t reveal card.

5a) Mr. Dupe doesn’t switch: Win

5b) Mr. Dupe switches: Lose

Scenario B:

1) Assume the Ace is the Left Card.

2) Mr. Dupe selects the middle card.

3) Sam is forced to reveal either the right card, a joker.

4) Sam pretends he didn’t reveal the right card. (‘erasing’ evidence #3 in the mind of Mr. Dupe)

5a) Mr. Dupe doesn’t switch: Lose

5b) Mr. Dupe switches: Win

Scenario C:

1) Assume the Ace is the Right Card.

2) Mr. Dupe selects the middle card.

3) Sam is forced to reveal either the left card, a joker.

4) Sam pretends he didn’t reveal the left card. (‘erasing’ evidence #3 in the mind of Mr. Dupe)

5a) Mr. Dupe doesn’t switch: Lose

5b) Mr. Dupe switches: Win

In two of the three scenarios, switching means Mr. Dupe wins. Sam’s pretending doesn’t matter at all.

Now, in one of your comments, you have changed the rules of the game:

Game 2:

There are two facedown cards, Left card is a Joker and Right One is Ace. The game is played in three phases.

In the first phase, Mr. Dupe will select a card. In the second phase, someone will whisper to Mr. Dupe “there was originally another card that was a joker, and sam took it away earlier.” In the third phase, Mr. Dupe will decide to switch or not switch, and either win or lose.

Scenario A:

1) Assume the Ace is the First Card.

2) Mr. Dupe selects the Second Card.

3) Whisper

4a) Mr. Dupe doesn’t switch: Lose

4b) Mr. Dupe switches: Win

Scenario B:

1) Assume the Ace is the First Card.

2) Mr. Dupe selects the First Card.

3) Whisper

4a) Mr. Dupe doesn’t switch: Win

4b) Mr. Dupe switches: Lose

Note that the key here is that Mr. Dupe selected the card under different conditions in Game 2, and Sam was forced to reveal a card in Game 1. To *truly* erase the past few minutes, Mr. Dupe would have had to not select any card at all, because then Sam could have revealed either joker, he would not be forced, and you would only have two scenarios.

The end of my comment matches DAV’s above: You are making the classic error of changing the proposition.

Ann,

There are only three possible outcomes given there cards with a single ace. Ace=1 and Joker=0

x) A=1, B=0, C=0

y) A=0, B=1, C=0

z) A=0, B=0, C=1

Each has a probability of 1/3.

Assume x is the desired outcome given the original selection. That means the probability of the outcome NOT being x is 2/3 = Pr(y)+Pr(z) = 1 – Pr(x).

Knowing that z is 0 makes Pr(y) = 2/3 = Pr(y)+Pr(z) = Pr(y)+0

It really doesn’t matter what Sam knows. The only way he can affect the probabilities for anyone who doesn’t know the location of the ace is to reveal the ace making the probability of that outcome (x, y, or z) one making the probability of the other two zero.

If what Dr. Briggs says in the doc referred to: “If you did not pick the prize, Monty opens a door; not the one you picked, and obviously not the one with the prize behind it” then switching is guaranteed to win – because if your choice had been “right” then Monty would have opened that door.

If that is a mis-statement of the game – i.e. if Monty opens a door he knows is not a winner whether or not you picked a winner, then the classic logic that it’s one chance in 3 etc etc applies.

In the game put forward here, however, the dupes get to play two separate games: in the first one they have one chance in three, in the latter one in two – assuming an honest magician who really does reset the game.

So my advice to the dupes would be to switch because if the game is not rigged it’s 1 in 2 no matter which choice they make, and if it is rigged it’s 2 out of 3 with the switch and 1 in 3 without.

Nate ==> (re: June 14, 2021 at 3:58 pm)

You are not wrong, of course. Your Game 1 is correct. That is the classical solution to the Monty Hall Problem. Nearly everyone agrees and I can demonstrate your position with a handy Excel spreadsheet that runs a thousand random combinations of prize-locations, contestant selection and the door Monty opens to reveal a goat.

Your Game 2 is equally correct. Except when you try to explain it. You are not right when you say Sam was not forced to reveal a Joker. He did so, but the Dupes just didn’t know it until someone, The Whisperer, snuck (sneaked, for you Brits) on stage and told them. Once they had been told that there were once three cards, one of which was removed in the past, their chances of winning still did not change. Thus, we “prove” that it is not the knowledge or lack of knowledge about the original conditions that change that final probability. In Game 2, the Dupes have a 50-50 chance — whether they have a knowledge of “what went before”.

However, “You are making the classic error of changing the proposition.” It is not an error, it is a choice in the exploration of the probabilities involved in the Monty Hall Problem.

In the Magician version, we discover that there is another view of the problem that changes the odds.

In The whisperer version we see that it is not the merely the knowledge of the three-card-ness of the original scenario that forces the probabilities.

So, the question remains:

Q6: Has there been some Sleight of Mind? If so, what was it and who performed it?

The dupes selecting a card forces the choice. Nothing else does, regardless of *how many cards* were on the table prior, there *is no choice* being made. The Dupes did not ‘select the card’ out of three, they selected out of the two presented.

Extend it to absurd leves. The magician could have placed 500 cards on the table, a rabbit, and three magic wands, and removed them all, before the dupes could pick two cards, one an ace, one a joker. It’s *the pick* and the removal that adds information to the problem. The whisper does *not* add information, because it that ‘removal’ was not in response to the pick.

The point is, the odds of the specific games, as stated, are 2/3 if you switch in the original game, and 2/3 if you switch in the ‘Sam’ game, and 1/2 in the ‘lots of whispered information but irrelevant because the pick came later’ game.

Paul Murphy ==> (re: June 14, 2021 at 4:42 pm)

Before this essay was published, I raised the issue with Briggs — the original version of the MHP does not contain the caveat “If you did not pick the prize, Monty opens a door…”

The original formulation is:

“Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice?”

So that is a valid point, but both Briggs and I do not think it changes either of our presentations.

“if Monty opens a door he knows is not a winner whether or not you picked a winner, then the classic logic that it’s one chance in 3 etc etc applies.”

The original version insists that Monty opens a goat door (there is always one of the two goats available regardless of the contestant’s pick).

“So my advice to the dupes would be to switch because if the game is not rigged it’s 1 in 2 no matter which choice they make, and if it is rigged it’s 2 out of 3 with the switch and 1 in 3 without.”

That is a novel viewpoint. I’m not convinced that the vos Savant version can be considered “rigged” or that the Magician version is any different.

Nate ==> (re: June 14, 2021 at 5:04 pm)

The Dupes did in fact, in the Magician’s version, after the Magic Wand trick, pick one card out of two.

But equally true, in the vos Savant version (the original), they did in fact pick one card out of three originally, but after the reveal of one Joker, were then given the choice to pick one out of two — one an Ace and one a Joker — with the knowledge that one of the not chosen cards, the revealed card, was a Joker.

If we add in The Whisperer to the Magician’s version, the Dupes have a choice of one out of the two remaining, but know that their choice is “really one of of three” oddly augmented by the knowledge of the location of one of the losers.

So, our final question is: What is the true difference between the three final choices and why do we consider their probabilities to be different?

This isn’t quite how you described the magic wand trick as above, there is a little faint joker there. Nobody ‘forgot’ anything.

Now, if this is real magic and you are somehow able to completely make past ‘events’ disappear, including all knowledge of what happened and what the initial conditions were, including the rules of the game:

Magician’s Game:

There are three facedown cards, two are jokers and one is Ace. The game is played in four phases.

In the first phase, Mr. Dupe will select a card. In the second phase, Sam will reveal a card (not the Ace).

In the third phase, everyone (including us) will forget everything that happened except knowing that there is an ace and a joker on the table. We will even forget the the rules of the game that there *were three cards*.

*ignore the past three lines, we have all magically forgot them*.

In this game, Mr. Dupe will discover a mark placed somehow on a card, and Sam will ask if Mr. Dupe wants to look at that card or the other card. If he guesses an Ace, he wins.

Scenario A:

1. Evidence A_E1: There are three cards, the ace = middle. Mr. Dupe Picks middle, Sam reveals one of the cards.

2. Everyone forgets the rules of the game or that a game was being played. This scenario no longer applies.

Scenario B:

1. Evidence B_E1: There are three cards, the ace = left. Mr. Dupe Picks middle, Sam reveals one of the cards.

2. Everyone forgets the rules of the game or that a game was being played. This scenario no longer applies.

Scenario C:

1. Evidence B_E1: There are three cards, the ace = right. Mr. Dupe Picks middle, Sam reveals one of the cards.

2. Everyone forgets the rules of the game or that a game was being played. This scenario no longer applies.

Scenario D:

1. Evidence D_E1: There are two cards, ace = left. The left card is marked. Mr. Dupe is offered the choice and selects the marked card. He wins.

Scenario E:

1. Evidence E_E1: There are two cards, ace = right. The left card is marked. Mr. Dupe is offered the choice and selects the marked card. He loses.

The result is that switching makes no difference, because nobody knew that there were three cards originally. Probability always is in your head. It is not a physical thing. It *all hinges* on who knows the rules of the game, and what they know when. If you *change the rules* halfway through, then of course there will be different odds.

This is where the whisperer might make a difference – if not everyone knows ALL the rules up front, and they *add* to the rules, very explicity.

I walk up to a table and one card is marked and I am offered to switch. If it’s an ace, I win $10.

If someone says, “after someone with no knowledge of the cards selected one, there only was one ace in the original set, and the magician flipped one of the others over” then the same information is added that the original monty hall problem adds, which changes my odds.

If, however, they say “there were once three cards on the table and now there are two” they haven’t added anything new.

I think they’re two separate problems (actually three) conjoined by a sleight-of-mind. the first problem is two Jokers and an Ace. their odds (all else being equal, as they say) are 1/3 of picking the Ace.

When the first Joker is removed, as long as they are given the choice of selecting one of two cards, their odds are 1/2. the magic removal is a sleight of mind that gets people focused on the non-existent Joker.

in the third problem, their odds are still 1/2 if there are only two cards. The whisperer is another sleight of mind that shifts the focus from a problem that is (two cards) to a problem that isn’t. the whisperer could have told them anything, but doesn’t change the fact that the current choice involves only two cards.

That’s my story and I’m sticking to it…

Phil R ==> (re: June 14, 2021 at 6:40 pm)

Good man! You stick to your story.

The only issue is that almost every statistician in the world disagrees.

I can demonstrate conclusively that Marilyn vos Savant is absolutely correct.

Of course, I wrote this essay and also firmly believe that the Magician’s version shows that you are correct ALSO.

Nate ==> (re: June 14, 2021 at 6:29 pm)

I left a faint shadow of the Joker so you, the reader, wouldn’t forget about it totally, even though the choice made when there was three cards and the existence of that now-oh-so-faint Joker have been “magically erased”. We forget those two things to afford us the luxury of making the scenarios equal — same cards and same choice.

I’m not sure that there needs to be another mark on one of the cards. In both of the scenarios in the essay, there is a Yellow Star sticker placed on the originally “selected” card. In the Magician scenario, the Dupes, having had recent events erased by the Magic Wand, are confronted with a table that has two cards on it, B and C. B has a Yellow Star sticker, which they may leave to select card B or move to select card C. They can choose either card. This is the same choice as in the vos Savant scenario (the classical scenario), except in the classical, the revealed Joker is still on the table. The Dupes can select either card B or card C.

The Whisperer version is the last stage of the Magician’s scenario, two cards, one Ace and one Joker, one card with a Yellow Star, free choice between cards, and the Dupes unaware that there had been three cards originally. The Whisperer comes and tells them “there was once, before you arrived at the table, a third card and it had been revealed to be one of two Jokers (in the original stack of two Jokers and one Ace).” So, this is really new information, at least to the Dupes, and leaves the Dupes in the exact physical and knowledge state as in the classical vos Savant scenario.

The question is: Are the odds of winning really different between the classical and the Magician scenarios? and does the Whisperer can that?

Great post. 72 comments. Briggs, this Hansen dude – he seems to be one of my people – is a real troublemaker. He should post here more often.

Dean Ericson ==> (re: June 14, 2021 at 8:04 pm)

Thank you, sir — very kind.

Here’s where it goes wrong: “This is the same choice as in the vos Savant scenario (the classical scenario), except in the classical, the revealed Joker is still on the table. The Dupes can select either card B or card C.” It is not the ‘same choice’ in the knowledge sense.

Probability is always and everywhere about what does one know. If we know that Mr. Dupe made a choice and the magician revealed a card, the probability, from our evidence, is 2/3. Mr. Dupe does not have our evidence, if it’s been erased from his brain. So to him, the probability of choosing the correct card is 1/2. He has no knowledge of the previous game, the joker existing at all.

Imagine this:

Scenario 1: A card shuffling machine spits out a single card from a standard 52 card deck. What is the probability that it is the ace of spades, to you, me, and Mr. Briggs?

Scenario 2: A card shuffling machine spits out a single card from a standard 52 card deck. I look at it, and then I secretly tell Mr. Briggs that is is a spade. I don’t tell you anything about the card. What is the probability that it is the ace of spades, to you, me, and Mr. Briggs (it is the same physical situation, but *three* different probabilities).

This is the same as these scenarios. It’s always about who knows what when.

Nate ==> (re: June 14, 2021 at 8:49 pm)

“Here’s where it goes wrong: “This is the same choice as in the vos Savant scenario (the classical scenario), except in the classical, the revealed Joker is still on the table. The Dupes can select either card B or card C.” It is not the ‘same choice’ in the knowledge sense.

It is ONLY the “same choice” and “same knowledge state” in the Magician scenario with The Whisperer — who informs the Dupes of the original three-card choice and the revealed Joker.

Now, let’s be a bit more pragmatic. We’re betting our first born child — one chance, win or lose. We really really want to make the best possible choice — to get the best possbile odds — and save that darling litte brat. If we accept the classical view, then we must switch — the odds are doubled in our favor. If we take the other view,The Magician scenario, it matters not — we have only a 50-50 random chance. If The Magician scenario has been “enhanced” by The Whisperer — hmmmm….not sure.

I do get your point — but the question is in the reality of the probabilties.

When your card shuffling and card spitting machine kicks out that one card, it has a real physical world probability of being the Ace of Spades — even if we are not present and don’t even know the game is on.

This situation shifts if we change and begin to talk about “The chances of Briggs, or Kip, or Nate making a winning bet, after chosing their calculated odds based on their knowledge about the deck and the card.”

This is the very aspect of probability this essay is meant to explore.

re: Robin June 14, 2021 at 9:28 am ” some Physicists contend that subatomic matter is simply a collection of probability functions. Niels Bohr was reported to have believed that a complete understanding of reality lies forever beyond the capabilities of rational thought. ”

We are, I think, at a stage to challenge that. Hear me out, before decrying that it is impossible, or that those that contend so are insane! I also ask the forbearance of the host of a longish post (I can’t really do this topic in a paragraph, but CAN come close with half a dozen!)

HERE’S what I want to get at, and it begins with an INDICTMENT of “QM” (Quantum Mechanics) which is now on the order of 100 years old as contrasted with *new* theories built on the observations over the last +50 some years and which QM inadequately “explains”/provides the “first principles” in physics.

—————————-

Typically freshman students are introduced to classical laws that they apply to physical problems that can be understood intuitively and solved in closed form. As they advance to the second year, they are introduced to a contradictory view—that the atomic-scale world is nonphysical, counter intuitive, and incapableof being understood in physical, intuitive terms.

In addition, they are asked to take for granted many fantastical concepts such as electrons being probability waves having an infinite number of energies and positions simultaneously, until measured, spooky actions at a distance, and virtual particles which occupy every point in space but can not be detected. With the introduction of quantum mechanics, which is not a theory of physical reality, students are taught to abandon all that they initially learned for laboratory scale systems and to accept that these laws do not apply to atomic systems; even though, they learned by direct experimental observation that these laws worked perfectly well and that laboratory scale objects are made up of atoms.

…

QM has never dealt with the nature of fundamental particles … it postulates the impossible situation that they occupy no volume; yet are everywhere at once. In contrast, CP (GUToCP) solves the structure of the electron using the constraint of nonradiation based on Maxwell’s equations. CP gives closed-form physical solutions for the electron in atoms, the free electron, and excited states that match the observations. With these solutions, conjugate parameters can be solved for the first time, and atomic theory is at last made predictive and intuitive.

Application of Maxwell’s equations precisely predicts hundreds of fundamental spectral

observations and atomic and molecular solutions in exact equations with no adjustable parameters (fundamental constants only). Moreover, unification of atomic and large scale physics [this gives] … a natural relationship between Maxwell’s equations, special relativity, and general relativity. CP holds over a scale of spacetime of 85 orders of magnitude …

The Maxwellian approach allows the solution of previously intractable problems such as the equations of the masses of fundamental particles.

…

GUToCP (the new theory):

o Successfully predicted the mass of the top quark before it was reported and

o Correctly predicted the acceleration of the expansion of the universe before it was observed.

o It correctly predicts the behavior of free electrons in superfluid helium and

o Further predicts the existence of new states of hydrogen that are lower in energy than the n=1 state that represents a new energy source and a new field of chemistry that has far reaching technological implications in power generation, materials, lighting, and lasers.

The existence of such states has been confirmed by the data presented in over 100 published journal articles and over 50 independent test reports and articles.

——————————-

If the “card shuffling and card spitting machine kicks out …the Ace of Spades” and “we are not present and don’t even know the game is on,” does a probability exist?

Oh man. Are we getting into Uncertainty, Heisenberg?

Robin ==> (re: June 14, 2021 at 9:28 am)

(Somehow I missed this comment earlier in the day — so sorry).

Thank you for the kind comment.

Dr. Mabuse ==> (re: June 14, 2021 at 10:07 am)

(Somehow I missed this comment earlier in the day — so sorry).

I assure you the game is not rigged in any way — no card tricks, no Mexican Turnovers.

Monty and Sam do know the positions of the cards (Monty knows what’s behind the doors). Sam turns over a card he knows is a Joker. (Monty opens a door with a goat).

brad.tittle ==> (re: June 14, 2021 at 11:50 am)

Sam (or Monty) does know the location of the Ace/Car. I mention moving the star as in the Magician scenario, the star is still one one of the cards. In reality, the Dupes are just given a chance to choose card B or card C.

Phil R ==> (re: June 14, 2021 at 9:31 pm)

Good question. If a card has been spit out, then it must be some card and if some card then it might, with some probability, be the the Ace of Spades.

Kip: This is where we fundamentally disagree: “it has a real physical world probability of being the Ace of Spades”.

No, it does not. It has no physical real world ‘probability’ at all. Some set of causes made the the machine to do this. The initial state of the cards, the friction between the cards, even, if you want to go there, the likely unmeasurable “quantum” initial conditions. But all of those things are the *causes* for the ace of spades to be spit out. The more of those causes we know, the better our guess at what the card will be. But if we knew ever condition, had every single initial condition identified, then we could say, with certainty, what the outcome would be. Probability literally just quantifies our uncertainty.

@Kip –

Consider this …

Sam lays out two aces and one joker, flips them over and mixes them, has the Dupes pick a card, turns one card over revealing an ace, magically erases the selection event, and has the Dupes decide on their best choice.

Again, the Dupes are faced with an ace and a joker. Applying the standard rules, switching now loses 2/3 of the time.

Wait, the choice is between the two same cards (one ace and one joker) yet the probabilities have switched. Hmmm.

Is there something magical about the past that influences the probability of the situation at hand?

Nate ==> (re: June 14, 2021 at 10:29 pm)

Yes, Probability literally just quantifies our uncertainty. That is certainly true.

But, as the card must be some card. There are 52 (53, 54 with Joker[s]) cards. One of them is the Ace of Spades. Therefore the dealt card might be the Ace of Spades with some assignable probability.

It may be true that there is no truly random shuffle and dealing of the card, due to a lot of factors — a million deals may favor some one particular card with some tiny increase in its probability and some other particular card with some decrease in probability.

But probability can also be defined as an understanding that we are willing to bet on if we get to choose the odds (that we think favor us). In that case, I would want better than 1 in 54 — maybe my one dollar against your 1000 dollars, and be allowed to play as many times as I want…..that might satisfy.

READERS ==> I am out of here for the day. I will take up responding to your comments in the morning, Eastern Time.

Thanks to all who participated in the discussion.

All,

I’ll respond with a full post, I hope next week.

“But, as the card must be some card. There are 52 (53, 54 with Joker[s]) cards. One of them is the Ace of Spades. Therefore the dealt card might be the Ace of Spades with some assignable probability.”

In this statement, you have created a set of evidence to which a probability can be assigned. There is no probability without the evidence – P(X) does not exist, P(X|E) does. You can only every have a probability when you have an E (this includes the definition of the words and shared understanding of the evidence).

That is my point – probability is a function of the mind, never a feature of reality.

See Briggs post:

https://wmbriggs.com/post/22534/

“Probability, absolutely all of it all of the time, is conditional. You walk to a corner and desire to cross. At this point you must form premises on which to act. You might say, “I might get hit”, which adds nothing to your ability to form a probability of “I will get hit”. (This, and everything else, is proved in Uncertainty.)”

My advice to the dupes (and to “win” or not lose):

do not play the game!

Kip Hansen: “That leaves us with the problem that The Magnificent Sam has shown us — that the odds are ALSO 50-50.”

No, the odds are not also 50-50. Saying you’re erasing information doesn’t actually mean you’re erasing information. Pretending information doesn’t exist is not the same as it not existing.

The two remaining cards arrived at their current configuration through a different path than they would have if Sam had initially dealt two cards, one being a Joker and one being an Ace. The fact that they were part of a group of three cards — two Jokers and one Ace, and that one of the Jokers has been revealed — is information that cannot be destroyed simply by waving a wand.

Now, if Sam dismissed the couple, and brought in a different couple from outside the room who did not possess that information and had them pick from between those two cards, then that couple would have a 50/50 chance of picking the correct card. Because they lack the information that the first couple possesses.

Information affects probability. It is not unlike Briggs’ frequent statement that “Random means unknown cause, nothing more.”

I made a video on this: https://youtu.be/BiKXpQDopDk

Jim Fedako ==> (RE: June 14, 2021 at 10:49 pm)

“Wait, the choice is between the two same cards (one ace and one joker) yet the probabilities have switched. Hmmm….. Is there something magical about the past that influences the probability of the situation at hand?”

A consensus is forming that (as always has been taught) Probability depends on what you know about the situation. Some say that probability doesn’t exist outside of a set of known information.

So, maybe what we know about the cards and previous choices does change the odds? But, the Whisperer entering the stage at the last minute to reveal previous information doesn’t seem to change the odds? or does it?

HanSolo (re: June 14, 2021 at 11:38 pm)

Come on Hansy – be a sport. Watch an episode of “Let’s Make a Deal!” and see how absolutely over-the-moon those housewives are about the chance to win a brand new car — with a fortune in those days to the average Los Angeles housewife.

Cloudbuster ==> (re: June 15, 2021 at 4:42 am)

“Information affects probability.” Yes, that appears to be the case – maybe.

If there is really no such thing as “physical probability” — Nate’s card shuffling and card spitting machine chucking out a single card example — the machine selects one card, is it the Ace of Spades? Some say that if we weren’t there and looking, there would be no probability that the card was the Ace of Spades at all — probability would not exist because there was no one to have knowledge about the situation and without knowledge there can be no probability.

Don’t ask me — what do I know?

HansSolo ==> “worth a fortune” … I have lazy fingers this morning.

Kip Hansen: ““Information affects probability.” Yes, that appears to be the case – maybe.”

Not “maybe.” Absolutely. This only appears uncertain because in the scenario, the amount of information revealed allowed for some remaining uncertainty. Instead, imagine that Sam showed the couple where the Ace was, then asked them whether they wanted to stick or switch. There is zero remaining uncertainty — full knowledge — so the couple will choose the Ace 100% of the time. Yet, if you then dismissed them, removed one of the Jokers, brought in a couple without the Dupes’ knowledge, their odds would still be 1/2. Or, if you left the three cards unchanged for the new couple, the odds would be 1/3 on their initial guess (and so on if you recursively led them through the same dilemma). The only thing that changes is knowledge.

The Monty Haul Dilemma is mental sleight of hand. The “grift” that is being played is that since in the second stage there are two cards, one of which is the Ace, the probability of a particular card being the Ace is 1/2, but the mental sleight of hand is getting you to focus on that rather than your knowledge of how the cards came to be in that configuration.

’ve read, and re-read the article and comments, and set aside the (alleged) red herrings as to Sam’s knowledge and motivation.

When the Dupes chose Card B as Ace, their odds were 1 in 3. Cards “A” and “C” also had 1 in 3 odds of being the Ace. When card A was “taken off the table” as a Joker, using your (apparently correct) logic, the odds of Card B being an Ace remained 1/3, because the original odds didn’t change (I’m with you so far), whilst Card C was given all of Card “A’s” 1/3….likelihood of being Ace? That’s where you lose me.

Card “C” is given (entirely unmerited, in my opinion) a whopping 2/3 chance of being the Ace. Why? What changed? Why did Card C, like Card B, start at 1/3 chance, and then move to 2/3 chance, while Card B stayed at 1/3 chance?

Using the logic, why not give all of Joker Card “As” helpful 1/3 chance to “B” instead, meaning the Dupes have 2/3 chance of winning? What is it about the unchosen card, because it was unchosen, that makes it more likely to be the Ace? See how confused I am?

And yet, you tell us that PhDs, super-computers, excel spreadsheets, and Marilyn Vos Savant, say that Card C gets A’s 1/3 of the cheese. I, for one, believe you, not because I understand, but because you do.

Maybe this is why we need these math equations, to sort things out, when all our brains see is unfairness and deceit. Thanks a lot, Magician Sam. I’ll change to Card C after all. Either way, my chances are 50/50! (See, I learned nothing.)

Ann – think of it this way. I put down 1,000,000 cards, all Jokers, and one ace. You pick one of them. Now I remove 999,998 cards, leaving just two – a joker and an ace. Do you switch? Do you *really* think you picked the right card the first time in that situation?

Q1. Does the Magician’s revealing of one of the two Jokers improve the couple’s chance of winning? Does the Magician’s allowing them to pick again improve the couple’s chance of winning?

A1 – It helps reinforce the idea that they still have a chance of having guessed right from a 1/3 chance. Picking again does not improve upon that, that’s an illusion that they now have a 1/2 chance of winning, but that’s also a 1/2 chance of losing. So it’s a wash. You either doubt you picked right from a 1/3 chance or you are confident.

Q2. Regardless of your answer to Q1, what would you advise our young couple to do? Stick with their first choice, switch their choice, or flip a coin to pick between the two remaining cards?

A2- Go with whatever their gut instinct suggests which, like the coin has a 1/2 chance of being right or wrong.

Q3. What are the odds – the probabilities – of the Dupes winning now? Do the odds change if they change their pick?

A3- No, because remaining on the same one is also a choice.

Q4: Was Marilyn vos Savant wrong after all?

A4- Abstractly, she is not wrong. But you are still making a decision from what was originally 1/3. That you might change it does not affect the reality of what’s on the table.

Q5: Is Sam’s wand really magical and, after it has been waved over the card table, we find it has actually changed the probabilities?

A5- No.

One Last Question:

Q6: Has there been some Sleight of Mind? If so, what was it and who performed it?

A6 – Yes. In the couple’s gut the feeling of being confident in their 1/3 choice or the new risk of a 1/2 choice. The magician produced it by not turning over the chosen card in the first place, instead delaying the reveal producing time that would lead to a 1/2 chance of either doubt or confidence in the couple’s mind.

The truth is that the more time one takes to make a decision based solely on a 1/2 probability the more pressure comes to bear on them. Psychologically speaking, it was easier to make a decision from a statistically higher probability of failure for one would either choose not to play, or would consider the $10 a suitable sum for a momentary amusement having comfortably settled on the probability that they are likely wrong. But the 1/2 probability is more annoying as the subject would take the loss more personally as there was perfect equality presented to them and yet they failed to grasp it.

The world is an easier place to live in with the odds stacked against you to begin with, rather than to have been given fairer opportunities and yet to have come out a loser. Such a result can be crippling.

An idea for the gamble would be interesting if the wager involved had a direct influence on the winnings of risk/reward.

Picking right from a 1/3 chance earns you $30 for your $10 investment. Choosing to change on a 1/2 chance reduces your winnings to $20 if correct. Do you stay or do you change?

Nate, who would take that one-in-a-million bet? I would, if I knew in advance you were going to even my odds by removing 999,998 cards, leaving me a joker and an ace. I didn’t have to pick the ace, and very likely, I didn’t. But you did pick an ace, if, as you say, I’m left with a joker and ace. So at that point, if you shuffle the two cards, my chances are 50/50. If you don’t shuffle and it’s an honest game, I’d be crazy not to switch cards. Am I missing something here?

Cloudbuster ==> (re: June 15, 2021 at 11:13 am)

Well, another vote for the “probability depends on knowledge” hypothesis.

“knowledge of how the cards came to be in that configuration.” In the Magician scenario, where The Magnificient Sam erases the recent past with his Magic Wand — the odds seems to change to 50-50 instead of 1/3 nd 2/3.

But if we add in The Whisperer who runs out and informs the Dupes(who have forgotten that they originally made a choice of one out of three cards as a result of the Magic Wand) that they actually previously chose one out of three, and that one card of the three had been revealed to be a Joker — the odds then change for them from 50-50 to 1/3 and 2/3 just because they now know about the original choice? Really?

Ann Cherry ==> (re: June 15, 2021 at 3:07 pm)

“When the Dupes chose Card B as Ace, their odds were 1 in 3. Cards “A” and “C” also had 1 in 3 odds of being the Ace. When card A was “taken off the table” as a Joker, using your (apparently correct) logic, the odds of Card B being an Ace remained 1/3, because the original odds didn’t change (I’m with you so far), whilst Card C was given all of Card “A’s” 1/3…likelihood of being Ace? That’s where you lose me.”

I sympathize — but that is the classical answer — explained over and over in the literature on the Monty Hall Problem. Marilyn vos Savant explains it, the Wiki explains it, Briggs explains it and there are computer programs that simulate it and show that it is absolutely correct.

In this essay, I am acting as intellectual provocateur — intentionally stirring the pot on this problem — showing that the 50-50 is not entirely wrong as many claim — because — *it depends*.

I am sorry if I confused you — I got into this trying to lay it out for my wife, who was also confused by it all. She is still uncertain about Monty Hall, even after editing my essay several times.

Briggs promises to do a follow-up post to expose my nonsense — please have patience and wait for him to make it crystal clear to all. (or not 🙂

Ann – “But you did pick an ace, if, as you say, I’m left with a joker and ace. So at that point, if you shuffle the two cards, my chances are 50/50. If you don’t shuffle and it’s an honest game, I’d be crazy not to switch cards. Am I missing something here?”

You’re correct. My point was that the same rules apply to the 3 cards as to the 1 million. So long as the original choice is *still marked* after removing the cards, you are better off switching, because you probably picked the wrong card the first time (2/3 chance of being wrong, instead of 1/1M). If the original choice is lost, then that information (that you *probably picked wrong*) is lost too, and thus you’re back to a 50/50 shot.

Johnno ==> (re: June 15, 2021 at 5:41 pm)

I’ll put you in the 50-50 crowd — with a psychological overlay.

Stop back next week to see what Briggs has to say about this.

Thanks for making your point so clearly.

Kip Hansen: “the odds then change for them from 50-50 to 1/3 and 2/3 just because they now know about the original choice? Really?”

(Aside: it wasn’t clear to me in the original example that their memory is erased. If so, that is the equivalent of what I did in my examples — replacing them with a couple that lacks their knowledge)

As for your above quote, yes, really. Take my previous example where Sam shows them the Ace, thus giving them a 100% chance of picking the Ace. If he wiped that knowledge from their minds, their chance becomes 1/2, but if he or a Whisperer restored that knowledge to their minds, nobody would find it confusing that they once again had a 100% chance of picking the Ace.

The entire point of Marilyn vos Savant’s solution to the Monty Haul Dilemma is that the original choice and the exposure of the joker provide information about the probable location of the Ace. Take away the knowledge and you take away the change in probability as certainly as knowledge, or not, of exactly where the Ace is would affect the probability of picking the correct location. I don’t see how this is debatable.

Oh, an addendum to my previous comment. It is also useful to consider Sam’s perspective. At every stage of the game, Sam knows with 100% certainty where the Ace is, otherwise he wouldn’t be able to successfully perform the reveal of the first Joker. But he’s looking at the same cards as the Dupes. Why is his chance of picking the Ace always 100% while the Dupes’ chances vary? Knowledge.

Ann Cherry ==> (re: June 15, 2021 at 8:30 pm)

When you are left with two cards out of a million — knowing that one is an Ace and one is a Joker — regardless of the fact there there had once been 999,999,998 other Jokers in the pack (assuming that you were shown each Joker and the Ace has not been removed) — what are the odds that *one* of the two cards on the table is an Ace? There isonly one answer — 50-50.

Is it really a different question to ask: “What are the odds that the card you picked (once one of a milion and now one of two cards on the table) is the Ace?”

Nate ==> (re: June 15, 2021 at 10:19 pm)

Ah, now you are maintaining that the odds depend entirely on knowing what the “original choice” was. That means the odds (the probabilities) change based on knowledge of the past alone! and are not related to the pysical present.

Kip

But the Physical Present HAS the Star Marking the card! So being honest, Sam? related the present with the past. This is something we do everyday relating the present with the past which is why people with different experience and knowledge have different probability assignments

Cloudbuster ==> (re: June 16, 2021 at 12:43 am)

“Take my previous example where Sam shows them the Ace, thus giving them a 100% chance of picking the Ace. If he wiped that knowledge from their minds, their chance becomes 1/2, but if he or a Whisperer restored that knowledge to their minds, nobody would find it confusing that they once again had a 100% chance of picking the Ace.”

Now, *knowing where the Ace is* the point of the probabilities — so that’s not quite fair.

I am still trying to get someone to tell us how they are so sure that our 50-50 odds (two cards, both unknown, one Ace, one Joker) *change* when we know how many cards there were before we were left in our current state — as long as our current state is as given.

“the original choice and the exposure of the joker provide information about the probable location of the Ace.” Just checking here — but *What probable location of the Ace?* We know for certain, 100%, that it is one of the two remaining cards. We do not know which one, and, can not know which one. (The host, Monty or Sam, know but the contestant doesn’t).

This leads, of course, to the conclusion that Probability is about something other than the actual cards in play, their original states, numbers of cards, whatever shuffling, moving revealing, magic wanding, etc happens to them.

Cloudbuster ==> (re: June 16, 2021 at 12:50 am)

If there is no uncertainty, there is no probability — so, there can be no probability involved with Sam (or Monty) . 100% is not a *Probability* — it is simply Knowledge.

For probability to exist — there must needs be uncertainty.

“Ah, now you are maintaining that the odds depend entirely on knowing what the “original choice” was.”

Kip, *nothing that I have said has changed at all*. This was and is always the case, because that original choice was selecting one out of X cards. When all the other cards except for the ‘target’ card are removed, that original choice matters because the question is – did I guess right at the start or not. This is why extending this to 1 million cards makes the problem simple.

You guessed this card, I removed 999,998 cards that *aren’t* it, would you like to switch? Of course you would. Do you really think you picked the ace on the first try? Every removal makes it more likely you didn’t pick right!

“I am still trying to get someone to tell us how they are so sure that our 50-50 odds (two cards, both unknown, one Ace, one Joker) *change* when we know how many cards there were before we were left in our current state — as long as our current state is as given.”

It’s not and never was *only* that we know how many cards were left. It’s that we also *had made a selection* before, then all the non-correct cards were removed (except for our card and the final card, either of which could be the ace), and now we have the option to change that selection or not.

If we *had not* made a selection, then our odds of picking the right card from two is 50/50. But we constrained the problem by picking BEFORE removal.

john b() ==> (re: June 16, 2021 at 10:56 am)

“the Physical Present HAS the Star Marking the card!” — Yes — quite right. The Yellow Star Sticker is on one of the cards the Dupes see after the Magic Wand erases from their minds the recent past (or maybe erases the recent past altogether). They are given a chance to leave it where it is thus marking their Pick or move it.

You are quite right in this sense: If every Tuesday morning for the past year the milkman has left a pint of cream with your daily quart of milk (as is your present order) — your view would be that the probability is high, nearly certain, certainly expected, that today, which is Tuesday, there will be (yet again) a pint of cream with your quart of milk. This IS how we live our lives.

And when we are confronted with two cards, one reliably declared to be an Ace and the other to be a Joker, and we have no knowledge of which is which — no clue — no secretly marked cards — we expect that (this will seem a tautology but must be said) that one really is the Ace and the other is really the Joker. We would also accept that we can not know which is which — and come to the conclusion that our probability for each card being the Ace is 50% — 1 out of 2.

The presence or absence of a yellow Star sticker on one of the cards might make us curious — but tells us NOTHING about which card is the Ace. Thus, we are stuck at 50-50.

Aren’t we?

One lase comment.

Imagine it this way. 50 cards. You pick one. The magician says, sorry, that’s not it, but I’ll give you a second chance, and he removes all the cards *including* your card. Now there are two completely unknown cards. It’s the act of picking that changes how the game works. The ‘select from two cards’ game is not the same as the ‘select a card, remove all cards but it and one other, then offer a switch’ game.

Confusion of the two games is why we confuse the probabilities. *one* probability is the probability of selecting from two unknowns.

The other is selecting from many unknowns, keeping the selection while removing most of the unknowns, then computing the probability of switching being the right choice. They are completely different problems and sets of evidence. That is why they are different probabilities.

I did not realize quite what you meant by the “event” eraser

“Now, think carefully, are there any differences in the real physical universe situations presented above? All three points above are apparently identical in the physical world.”

IF time is part and parcel of the physical world, then how ever much time you’ve “erased” totally creates a whole ‘nother physical world … be it a several seconds, a couple of minutes or twenty minute, you’ve got a different universe

Hezekiah had a 100% chance of dying … SOON … get your house in order NOW

This event (the original prediction) was erased

Hezekiah now had 100% chance of dying in fifteen years … get your house in order LATER

New prediction and a whole new reality

Nate ==> (re: June 16, 2021 at 11:35 am)

First, please understand that I am being intentionally provacative — I’m “messing with you” in the modern parlance.

I am forcing a discussion — like your professor in University might have done in a smaller group discussion with his students (not that I am attempting to *teach” you anything).

So, let’s consider that it is a fact that the DEALER, who laid out the million cards, must know which card is the Ace and which 999,999,999 are Jokers. He chooses which 999,999,998 of the 999,999,999 Jokers to turn over and MAY NOT turn over the single Ace. So, what knowledge drives the selection of cards that leaves two of them on the table?

If the contestant has “picked” the Ace (made a correct pick) then the dealer may turn over ANY of the other Jokers, leaving only one. If the contestant has picked Joker (picked wrong) then the dealer has no choice of which Joker to leave — he must leave the chosen one.

If the contestant hasn’t been asked to pick at the beginning, then the odds clearly, for the contestant, are 50-50. For the dealer, there is no probability involved, he knows where the Ace is.

The contestants knowledge of the original 1 million cards make his original odds of picking the right card, the Ace, 1 in 1 million. After 999,999,998 Jokers are removed . . . .

There are just two cards — the contestant still has a 1 in 1 million chance of “having picked” the right card, the one Ace in a million. But, when if he is allowed to pick again, if he blinks his eyes and looks at the problem anew, his chances of picking the Ace out of two cards, in present time, are 1 out of 2.

If the magician’s Card A vanishing trick has gone back in time and erased even the possibility that the Dupes could have chosen that card in the first place, then the probability that the starred card is the ace has changed from 1/3 to 1/2; otherwise, it has not. Likewise with the whisperer; because he is whispering after the fact of the choice having been made, his information does not create the possibility of having chosen the card whereof he whispers, so the probability that the star is on the ace remains 1/2.

But in any case, it is meaningless to speak of probability in reference to events of which there is only one iteration. It is like asking, “What is the probability that you were born on your birthday?”

Nate ==> (re: June 16, 2021 at 11:41 am)

Last one for me too! (Really folks, I have a life outside of this blog and the fascinating world of probability.)

This has been fun — and thanks for your dogged participation, Nate.

I have not been fair — intentionally I confess. I never intended to be fair, I intended to push and prod and force readers to think more and think deeper and not just give “the answer I learned in school”.

If probability and uncertainty were simple issues, Briggs would not have had to write a 277 page book trying to explain it from his point of view.

Briggs will expose me as a fraud next week when he writes in response to my essay and all the comments — well, because the *me* you read here is “having you on”.

The next time someone tells you “the odds are in you favor” — look a little closer.

READERS:

I am out of time to follow the comments here any longer.

Readers who still want to ask questions or make comments without my participation should still do so.

If you want to address me and me alone — you can email me at my first name at i4.net.

Thank you all for playing along.

Jerome Wildeberg ==> (re: June 16, 2021 at 12:44 pm)

(Well, OK, one last response — I missed Jerome’s comment earlier)

“…it is meaningless to speak of probability in reference to events of which there is only one iteration. What is the probability that you were born on your birthday?”

I like this idea – but does this mean that if there is to be a “you get to play one time only” game, that probability doesn’t apply?

This thing has been bugging me. In a good way, since it got me to thinking, which is something I don’t naturally do. Not about quantifying uncertainties, anyway, like some people around here, who nonchalantly construct half-a-dozen formulas for choosing a tie in the morning. But I can see in this magician scenario that while the rules of the game determine probabilities for uncertain outcomes, defining the rules — what info is relevant and what isn’t — is a tricky business.

Dean Ericson ==> (re: June 18, 2021 at 8:56 am)

“…defining the rules — what info is relevant and what isn’t — is a tricky business.:

That’s why there is a Magician!

Seriously though, you have arrived at a Truth . . . . which is a good thing!

Probability isn’t for the timid.