MONTY HALL LOGIC - Not adequate for real life situations.

[Ghost Post]

Yeah, there may be truly random events. The coin toss, for instance, may not be predetermined - if it's tossed by a human, and there is some non-deterministic thing called "free will", for instance, that affects the outcome. But in either case, we don't know enough to determine the outcome, so we use probability.

I've always had a problem with the argument that you can't talk about the probability of something that already happened - that it's either 0 or 1. Whether it's in the past or the future seems irrelevant to me. It's a matter of what you know about it.

And I certainly have a problem with the argument that there is some distinction between "the probability that X occurred" and "the probability that you would be correct if you geussed that X occurred". They can't be any different.

[Ghost Post]

Extro,
I agree. I think that you and I currently have a much different philosophical concept of what probability is. But you raised a very good point with your checkered room example. I'm surprised that I have never really been forced to consider the question of exactly where collection of data on the outcome begins. Now I can think of several examples that never occurred to me as being unclear before - like a horse race.

Does collection of data on the outcome begin after the event starts and you start receiving info or does it begin after the event is complete and you know the *actual* outcome? Additionally, is probability really the potential of something to occur or is it really a starting point constantly moving until it actually becomes the outcome (E.G. is the probability of a coin landing tails ever 1)? I had always understood a probability as not influencing the actual outcome and the actual outcome not influencing the probability of that outcome - each existing independently of each other and had always held strong to that concept. It looks like you have prodded me into climbing into the attic tonight to find my old philosophy books and seeking enlightenment.

[Wonko the Sane]

Well, technically speaking, since we, as mere humans, are pretty much planted at one point in the fifth dimension for our entire lives, there is no probability. If we could move in more than one direction across the fourth dimension, we wouldn't be having this discussion because we could know ahead of time what would happen. Let me explain.
The fifth dimension is what we call first fundamental probability. It contains every possible timeline and in other places in the 6th dimension it contains a fair number of impossible ones too (okay, infinite numbers of both). The 5th dimension is the first level where probability matters. If you're moving in the 5th dimension, then you're constantly shifting timelines. The problem is, you can control that movement. So here comes that all-important 6th dimension. The 6th dimension basically contains all possible 5th dimensions. Essentially, depending on where you are in the 6th dimensions, there will be different 5th dimensions that have the 4-dimensional time frames in different orders. Basically, things are pretty random. However, if you mapped out every 4th and 5th dimension as they correspond with each other and with the 6th, you could go around and hop on timelines with the best outcome.
What I'm saying, is that somewhere out there, it already happened in the future. The only reason we argue probability is because we haven't seen it yet because we're stuck in the 3rd dimension being pushed in the same direction along the 4th dimension for our entire lives. We can say that a coin will land heads roughly 1/2 the time when we through it. But the probability of it landing heads on a specific toss is already 1 or 0. It's outcome will be decided on the forces your hands generate on the coin, the air pressure and wind patterns, and the texture of the ground. Whether or not you think you can trick reality, you can't, you're a slave to your own timeline. You already did it, you just don't know that you already did it because at the point in time that you're conscious of, you don't necessarily know what you will in a later point in time.
I realize this is rather confusing and drags in concepts of dimensions a lot higher than some people credit the existance of. However, I should point out that modern superstring theory says that there are 11 dimensions (old string theory said 18). However, I have never seen the practical applications of anything above the 6th dimension, so let's not worry about dimensions 7-11, alright? Feel free to ignore this if it makes no sense.

On to something a little more relavent. Let's look at what probability means, alright? Probability doesn't mean that something has an x/y chance of happening. It means that if the same situation happens y times, that something will happen approximately x times.
So, you don't have a 2/3 chance of surviving. But if you ask the warden who will live and he says B 3 (or more realistically, 30000) times, you will live 2 (or in the more realistic case, approximately 20000) times. Make sense? Because I've gotta get back to doing real school work...or not.

[Tom]

Well, Wonko, there are actually a couple of defnitions of what probability "means" .. Bayesian is one description, there is at least one other ; if anyone has a description of these, I'd love to read it, it's something I've always wondered about.

I'm not trying to be dismissive, but the dimension of probability, etc, are all just useful thought constructs .. I don't know how much they tell us what is actually happening. But I'm not saying you don't have a point, just that I get a bit shy of any argument that starts "since we, as mere humans, are pretty much planted at one point in the fifth dimension" ..

[Wonko the Sane]

Well Tom, you bring up an interesting point. For all we know, we jump randomly about the higher dimensions. And then maybe there is such a thing as probability. However I'd prefer not to think that I'm jumping dimensions left and right as I move through space (relative to a point) at hundreds of thousands of meters per second. I was just trying to raise another viewpoint, though.

[jdlfkjhkajhfkljsdhflksjdf]

hjkghk

[HyToFry]

[Bump] Fixed the thread

[Wonko the Sane]

w00p. Thanks HyToFry, I was getting really irritating with having to read the source code to get the responses

[HyToFry]

Edwarding for ?Tink?. I'm not sure if that was your name.. sorry if not [= if so.

[Ghost Post]

I am relatively new here and I have been reading this particular discussion with some interest. I am neither particularly good at puzzles nor maths (especially probability), so perhaps you could clarify a couple of points for me?

I have not had a chance to read all three pages of this discussion, but I have read the first page and the prisoner problem intrigues me. Firstly, it would appear to me that this is not anagulous to the Monty Hall problem for the following reason:

In the Monty Hall problem (this point has been made before with a deck of cards), one can intuitively see that if there were 100 doors instead of 3, then once you chose a door and the host opened the other 98 then the probability of winning by switching goes up from 1/100 to 99/100. However, this is because the host hase revealed _all but one_ of the other doors.

However, in the prisoner problem the warden does not reveal _all but one_ of the prisoners that will live. If there were 100 prisoners and prisoner "1" asked who would live, the warden would still give him only one answer, not 98. Therefore, I do not see how this may then increase prisoner "1"'s chances of survival. I think this is problem is also easier to illustrate with 100 prisoners (99 go free), rather than 3, on an intuitive level.

The point I cannot understand is that surely prisoner "1"'s chances of survival cannot change after the warden told him who would go free, as this information was known to prisoner "1" already. Therefore, if the warden said "Prisoner 52 will go free", that does not, in my opinion, change Prisoner "1"'s chances of being executed.

However, as this approach does not work out mathematically (1/3 + 1/3 <> 1 or 1/100*99 <> 1 (depending on the number of prisoners)) and someone _has_ to be executed, this leads to somewhat of a conundrum. I cannot spot the flaw in the logic, yet I know it must be there.

Perhaps someone can paraphrase what I am trying to say a bit more clearly and/or explain my mistake.

[Quailman]

I see the parallel to Monty Hall as being:

Execution, although clearly bad, = Monty's winning door

Prisoner A = the door first chosen.

The Prisoner who goes free = Monty's now-open door which held the Rice-a-roni.

[Ghost Post]

What I am saying is this:

If there were more than 3 prisoners (let's say 100 prisoners). Would the problem be phrased like this:

1). Prisoner "A" asks the guard to name 98 names of the remaining prisoners that will go free.

2). Prisoner "A" asks the guard to name 1 name of the other prisoners that will go free.

In case 1) the problem _is_ analogous to the Monty hall problem. In case 2) I don't think it is.

And although with only 3 prisoners the guard will always reply with only one name regardless in which of the two ways the question is phrased, I think this may have an impact on the solution?

Thoughts?

[This message has been edited by Leo (edited 08-18-2000).]

[This message has been edited by Leo (edited 08-18-2000).]

[Ghost Post]

Yes, case 1 is analogous, case 2 is not. But with three prisoners, the question will be the same - "name 1 prisoner, of the other 2, who will go free".

[BigDaddy]

Actually you had a 33 percent chance of winning, by giving the unseen door up to take one of the remainingg two, you gave yourself a 50 percent chance of losing.
Cauce either you had the big one in your hand or not

[tv snake]

Looks like I get a chance to explain this to people before I go

OK. If you pick the right door to start with (a 33% chance), Monty will open a wrong door. If you then switch, you will swith to the other wrong door and then lose.

If you pick a wrong door to start with (a 67% chance), Monty will open a wrong door. If you then switch, you will pick the right door (the only door left) and win.

So if you switch, you always win if you pick the wrong door to start with. So you have a 67% chance of winning if you switch.

[Ghost Post]

You had a 1/3 chance of either dying or going free.
once knowing that b goes free
the chance of either for a or c was 1/2
give me a 33 percent chance of losing
compared to a 50 percent chance of losing

now apply monty
he says door 2 is the soso prize
and you have door one
by then taking door 3
you give yourself a 50 percent chance of losing, not a better chance of winning

[tv snake]

OK... let's assume that the guard has the name of the person who will be executed written down in his office.

Person A asks the guard which prisoner out of B and C will go free.

The guard goes into his office and looks at the sheet that says who he has to kill.

There are three equally likely possibilities.

A will die.
B will die.
C will die.

If A will die, the guard can say that either one of B or C can go free (each of these has prob. 17%)

If B will die, the guard will say that C will go free. (prob. 33%)

If C will die, the guard says that B will go free (prob. 33%)

Since the guard says that B goes free, two of the cases (1a and 2) will be eliminated, leaving one case where C dies (33%) and one case where A dies (17%).

If we double these so they are out of 100 again we get:

C dies 67%
A dies 33%

------------------
"Teach a man to make fire, and he will be warm for a day. Set a man on fire, and he will be warm for the rest of his life." -- John A. Hrastar

[Ghost Post]

Unbeknownst to A, C also secretly approaches the guard.

The guard tells him that B will go free. Thinking about the statistics of it, C walks away happy in the knowledge that he ALSO has a 67% chance of going free!

[mathgrant]

HEY! I thought this problem was over with by now! I thought that you were supposed to be working on the new potpourri! Finish that first, and then you can get to the trivial things like Monty Hall.

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[This message is being edited by mathgrant (being edited right now).]

[Ghost Post]

Well how is it possible for them both to have prob. 67? From the original GL puzzle, my brother thinks that once an option is eliminated, the remaining 2 possibilities actually change their probability to 50%. I don't know how to answer this. Any help?

[knightofni]

Mathgrant: would you kindly stop ordering people not to discuss other puzzles? It's not taking anything away from the efforts to figure out the potpourri puzzles, because quite frankly, I'm not sure if I can do any of the remaining ones, and I'm sure there are others in the same situation. So, I could spend 15 hours working on it and get nowhere. And you telling people that they are not allowed to discuss things at their own leisure is not only rude, but it's also obnoxious. Thank you...

[mathgrant]

dravid: I once saw a paradox somewhere on the GL where you are playing the Monty Hall game and and on the other side, unbeknownst(?) to you, your friend is also playing it with the prize behind the same door as on your side. This continues until you and your friend choose different doors including one which is the prize. Monty Hall then opens up the door that neither of you chose. The paradox is "Both me and my friend will switch doors but how can we each have a 2/3 probability at the same time?"

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[This message is being edited by mathgrant (being edited right now).]

[Dread Pirate Westley]

Unbeknownst to A and C, B also secretly approaches the guard. While we know he has no chance of being executed, he gets the name of the other prisoner who is going to walk. He also goes back to his cell believing he has a 67% chance of drawing a get out of jail card.

Now that I think about this some more, each of them has a 1 in 3 chance of being killed. This leaves a 2 in 3 (67%) chance of walking free. Learning the name of one of the people who will live doesn't change anything.

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[This message will have been edited by Dread Pirate Westley (to be edited 12-13-2003).]

[Andy]

Prisoner's paradox(?):
One of A, B, and C will be executed and the other two will walk free. The guard must tell the truth and must give a random answer if one is possible.
A says to the guard, "Of the other two prisoners, tell me the name of one whowill walk." The guard answers, "B will walk."
C asks the guard the same question and gets the same answer.
D (a prisoner whose sentence is life in prison) says to the guard, "I have a bet with some of the other guys in here, excluding A, B, and C. Tell me the name of one of the two of those three who will walk." The guard says, "B will walk."

From D's viewpoint, A and C now each have a 1/2 probability of being executed as neither was excluded from consideration by the guard.

From A's viewpoint, he (A) still has the same 1/3 probability of being executed as he had before. Originally, his belief was that each of the three had a 1/3 chance of being executed. This meant that there was a 2/3 probability that B or C would be executed. The guard has given A no information about the pair as a pair (one of them would certainly go free, and B is now known to be the one). A has no reason to suppose that B and C together have any other than the original 2/3 probability of being executed, but now that 2/3 probability is entirely with C.

C has similar new information, and concludes that the probability of A being executed is 2/3.

How can A and C reach different and seemingly conflicting conclusions? Simply, they have different information. A knows that of B and C, B will walk. C knows that of A and B, B will walk. A doesn't know what the guard told C, and vice versa. If they get together and compare notes, they will both receive new information and will then conclude (as D has already) that A and C each have 1/2 chance of being executed.

[Ghost Post]

Interesting explanation. So the true probability is 1/2, but the subjective probability, according to the best information that the prisoners have, is 2/3.

[Wonko the Sane]

I like DPW's response. Good call.

[Vanyo]

dravid wrote:

[GMFPhoenix]

Mathematics operates on one basic principle. Any way you carry out an operation, you get the same answer (assuming all steps are legal)

Lets assume you choose door #1 and Monty opens door #2. The chances that #1 has a good prize is supposedly 1/3. Therefore, #3 has a 2/3 chance of having a good prize. What happens if we calculate #3 first? We must say that it has a 1/3 chance of having a good prize. We have reached a conundrum. Obviously something must give since we get 2 different answers for the doors.

The error is in the fact that we consider the denominator of the first door to be a 3. In fact, the denominator should be 2, since there are only 2 possible result left.

It is the same as saying that the second born child's sex is influenced by the first child. If the first is a girl, the combo can either be g/b or g/g. Still a 50% chance of either. This riddle's false solution is saying that there is a 25% chance of one sex
(i.e. g/b g/g b/b b/g)

Back to the riddle:
Possible solutions originally
g/b/b
b/g/b
b/b/g

after the first door is bad, there are:

b/g/b
b/b/g

therefor, the chance that the good prize is behind either door is still 50/50.

[GMFPhoenix]

Mathematics operates on one basic principle. Any way you carry out an operation, you get the same answer (assuming all steps are legal)

Lets assume you choose door #1 and Monty opens door #2. The chances that #1 has a good prize is supposedly 1/3. Therefore, #3 has a 2/3 chance of having a good prize. What happens if we calculate #3 first? We must say that it has a 1/3 chance of having a good prize. We have reached a conundrum. Obviously something must give since we get 2 different answers for the doors.

The error is in the fact that we consider the denominator of the first door to be a 3. In fact, the denominator should be 2, since there are only 2 possible result left.

It is the same as saying that the second born child's sex is influenced by the first child. If the first is a girl, the combo can either be g/b or g/g. Still a 50% chance of either. This riddle's false solution is saying that there is a 25% chance of one sex
(i.e. g/b g/g b/b b/g)

Back to the riddle:
Possible solutions originally
g/b/b
b/g/b
b/b/g

after the first door is bad, there are:

b/g/b
b/b/g

therefor, the chance that the good prize is behind either door is still 50/50.

[Dragon Phoenix]

* shoots GMFPhoenix *

[Chuck]

One way of understanding the situation is to make it more extreme. Let's say there are 100 doors instead of 3. You choose a door. Monty then opens 98 other doors, leaving yours and one other closed. Should you switch?

[mwf]

I like people like GMFPhoenix.

GMFPhoenix I think you should put your money were your mouth is. We should get together and play this game for real money. You play Monty Hall, we get a deck of cards. Using 1 red and 2 black cards. You place the 3 cards down and I will draw one without looking at it. You then look at the other two and turn a black one over. I then have the chose of flipping mine or flipping the face down one in front of you. I will then flip the face down card in front of you. If it is red you pay me $900, if not and it is black I will pay you $1100. We would have to play a minum on a 100 hands this way.

So are you up to it? If you are so sure that you are right about this, you are gauranted $10K by the time we are done. If not you would owe me $27K.

I would suggest that you sit down play this game with your self or a friend with small amounts of money first.

[QuikSand]

I think the brilliance of the Monty Hall puzzle is that it so often leaves outwardly educated, intelligent, rational people with at least a working knowledge of probability clinging to an incorrect answer by sometimes desperate means. (See GNMFPhoenix above, though I confess to not follow the "proof")

To me, the most engaging thing is seeing what argument will successfully convert a 50/50 believer. I've seen the "whole deck of cards" argument (used many times in this thread) do the job sometimes, but typically the 50/50 folks are too entrenched, and suspect that there is somehow a flaw in the parallel. Same goes for the "lottery ticket" argument, which is essentially the same thing on a grander scale.

I've found that I tend to go a different route, which has worked a few times (not many, though). Here's an alternate explanation of the "switch for 2/3" solution (I confess I didn't read all 3 pages of this thread, it may have been posted elsewhere).

You select one of the three doors. There are two conditions that describe which door you have chosen:

A - you chose the winner (1/3 chance)
B - you chose the loser (2/3 chance)

We also shouldn't have any disagreement about the probabilities that I've assigned above.

Whichever door you selected, Monty now has two doors to choose from. There are two possible conditions for that set of "Monty's two doors." We should all agree that the two possible conditions are:

A - both doors are losers (1/3 chance)
B - one door is the winner (2/3 chance)

We also should agree that the cases A&B described twice above match one another: IFF you chose the winner, then both of Monty's doors are losers. IFF you chose a loser, then one of Monty's doors are winners.

Now, if we are in case A (you chose a winner, Monty's doors are both losers) -- what will the door be that he "leaves" for you? Since both of his doors are losers, the one he does not reveal will, in every single case, be a loser. Period. There shouldn't be any argument in that step either.

If we are in case B from above (you chose a loser, so Monty's doors include one winner and one loser), what will the door be that he "leaves" for you? Since his rules are to not show you the winning door, he will in every such case reveal a loser, thereby "leaving" you a winner. Again, this has one extra step, but nobody should argue the logic there.

So, when we start out with the obvious probability tree that we all agree on:

A - you chose the winner (1/3 chance)
B - you chose the loser (2/3 chance)

We can now safely expand these two case descriptions to read like this:

Case A: (1/3 probability)
You originally chose the winner
therefore
Monty had 2 losers to leave for you
therefore
The door he's leaving for you is a loser
therefore
Switching will cause you to lose

Case B: (2/3 probability)
You originally chose a loser
therefore
Monty had 1 winner and 1 loser to choose from
therefore
The door he's leaving for you is a winner
therefore
Switching will cause you to win

Which we would then shorten to:
Case A - 1/3 chance - switching wins
Case B - 2/3 chance - switching loses
Therefore, switching wins 2/3 of the time.

- - -

Does the argument work? Usually not. The true believers come up with some generic attack on the logic and come back to something along the lines of "there's only two options, it has to be 50/50."

Oh, well... I suppose it never really ends.

mwf, if you need someone to help front the money for your wager, count me in.

[BourbonBoy]

I really hate to beat this one to death, but I was arguing with some folks at work who still cling to their 50/50 arguements despite all contrary evidence, Steve Selkins mathematical proof, and God forbid a bit of common sense. The funny thing is that most of these people claim to understand the logic of the 1/3 arguement but still won't let go of their posistion (one of my co-workers even tried to use Schrodinger's cat to obliterate the 2/3 supporters...what will they think of next?). It's an interesting testiment to human nature regarding giving up ideas and beliefs. The problem is that this problem relies heavily on its English semantics (we all seen this in the arguements relating to Monty Hall opening doors at random, etc.) so there tends to be quite a bit misunderstanding. I take a cue from Godel who implicitly stated that the notion of truth is a much more certain notion than proof. So using logical English sentences to explain this triffle of a problem leaves to many loopholes for semantical arguements to attack. Using the matrix below we can effectivly remove the English influence on the logic.

doors
A B C In this example let's say
1 0 0 1 we pick door A.
cases 2 0 1 0
3 1 0 0

1 P 0 1 For all cases we pick door
2 P 1 0 A, so our outcome set of
3 P 0 0 door A have a prize is
{0 , 0 , 1}

1 P X 1 The X is the door Monty
2 P 1 X would show us for each case
3 P X 0 in the event that we picked
door A.(this works exactly the same if we picked door B or C..duh?) So our new outcome set is {1, 1, 0}.

So our possible outcomes of staying (0,0,1}
Those of switching... {1,1,0}
Hmmmmm...that's a bit long winded but gets the point across without using a formal proof, it simply states what happens. No matter how you try to proove it, the truth speaks for itself.

[BourbonBoy]

sorry the matrices didn't line up right...
if you don't get it i'll repost them.

[Rhino]

god if this works i'd be really surprised. i haven't looked at this site for months b/c i've been so busy, but i think it's outfrickinstanding that the monty hall puzzle is still debated. oh, and for those of you that remember me, i know it's "out of character" to write in lower case, but i guess it's my true self shining through

[Duncan]

A very long thread. Here's my preferred explanation of Monty Hall in case it helps anyone. The player starts with a 1/3, 1/3, 1/3 belief over doors A, B, C. A is initially chosen. Then C is shown to be empty. So how does this information change the beliefs of the player? Well the belief for A is unchanged. Another door was always going to be revealed as empty. The player knew that before the game started and could have conditioned on that evidence from the start. It would still have been a simple choice, 1 from 3. But the belief for C is now 0. That leaves a belief of 2/3 for B.

Alternatively, the only way you win if you stick with the first choice, is if your first choice was correct, 1/3. The only way you can lose if you switch, is if your first choice was correct, 1/3, implying the probability of winning after a switch is 2/3.

This is the same as the prisoners question as long as you take the point of view of prisoner A. Each player's beliefs must sum to 1. But other prisoners and the guard will have different beliefs based on their knowledge. Before selecting the prisoner to be executed it is 1/3 ,1/3, 1/3 for all concerned (assuming the selection is random).

After drawing it is 1/3, 1/3, 1/3 for the prisoners, and either 1, 0, 0 or 0, 1, 0 or 0, 0, 1 for the guard, depending on the outcome of the draw.

After A has asked his question and been told that C will not die we have:

Pris A: 1/3, 2/3, 0
Pris B: 1/3, 1/3, 1/3
Pris C: 1/3, 1/3, 1/3
Guard: He knows with certainty.

If B overhears the question and answer. He has:

Pris B: 1/3, 2/3, 0

as he is in possession of exactly the same evidence as Pris A.

If C is told he is to go free he has:

Pris C: 1/2, 1/2, 0

But if he is also told of A's question and the answer he too has:

Pris C: 1/3, 2/3, 0

Some (but by no means all) posters seem to have had a problem with the apparent inconsistency of these 'probabilities'. But they are not probabilities in the repeated experiments / frequentist sense, rather they are degrees of belief / certainty. The true probabilities are known only to the guard. But any prisoner in any of the situations above could reasonably say, "given that one of us was selected at random, and given the information I have, in X% of such eventualities I will die". The fact that they would disagree on X is not surprising; they have based their calculation of X on different information.

Take Pris C above. His 50:50 belief is based on the simple fact that it is not he who is to die. The subsequent 1/3, 2/3 belief is based on additional knowledge of the constraint implied by A's question. i.e. if A is not to die, then C had to be chosen. And, if A is to die it was a 50:50 choice between B and C.

Consider B asking the same question as A and again hearing that C will go free. He is happier. If the guard had said A, then B's days would have been numbered. He does the maths a figures out that it is now 50:50.

[Porro]

Hang on a mo, Wonko said...

"Probability doesn't mean that something has an x/y chance of happening".

But surely probability can validly be understood to mean that something has an x/y chance of happening.

The following was taken from an American University's web site:

"The probability of an event is a number describing the chance that the event will happen. An event that is certain to happen has a probability of 1. An event that cannot possibly happen has a probability of zero. If there is a chance that an event will happen, then its probability is between zero and 1."

[Ghost Post]

Forget both problems for a second (or as long as this takes you).


You have three things. One of them is THE thing. Yo have two guesses to determine which thing is THE thing. The prob. of guessing right is 2/3 or 66.6% or whatever you want to call it. Oh, and which is THE thing is only revealed after you have guessed.

On your first guess, you have a 1/3 chance.
Your second guess also has a 1/3 chance of being right, since you don't know yet if the first was. You do know that one of your guesses is wrong.

If that made sense, keep reading. If not, think about it until it does, then continue.

Now back to the real problem.

In the MH, you have only one guess. A 1/3 chance. Being shown that one other one was wrong does not change that probability, because you already knew that at least one other choice was wrong.

Imagine that you are aced with this: You have guessed A. MH has not revealed any choice as wrong yet. He says "Today, I'm gonna change it up a bit. I'm not revealing anything, but I'm giving you an option: If you stay with A, I'll reveal it like normal. However, you can renounce your choice and I'll reveal both, that's right, both of the other choices and if either is correct you can keep it." I should add that he knows which choice is right, but says this regardless.

Obviously, the latter is twice as likely to be correct than the former. You take it.

Now he reveals B and C. B is ... wrong.
C is ... correct! So C had been chosen before the show.

In the puzzle he is doing the exact same thing. Your guess of A has 1/3 chance, before and after he reveals B as wrong, because you knew one was going to be when you guessed.

If you take C however, it is as if you have been given two guesses. One was going to be wrong for sure, because only one could be right. Now he has shown you one which was wrong and you can take C with the same probability: 2/3


Whew!

This applies to the priosoner problem in the same manner, as they are essentially the same question.

[Chuck]

Ok, one more time!

I decide in advance that I'm going to choose door 'A' first and always switch.

By doing this, the only way I can lose is if the prize is behind door 'A'.

There is a 1/3 chance that the prize is behind door 'A'.

Therefore, I have a 1/3 chance of losing and a 2/3 chance of winning.

Right?