Talk:Three Prisoners problem

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[edit] Rebuttal To Solution

One prisoner is scheduled to be executed. His chance of being executed is 100%, not 1/3. One of the prisoners (not knowing who has been selected) wishes to estimate his chance of having been the one selected. With the current information, he can estimate his chance at 1/3. (Note that his real chance is either 0% or 100% depending on whether or not he has been selected). The warden releases one prisoner that was not selected for execution. The prisoner estimating his chance now has new information. He now scraps his original estimate since he has new information, there are now two prisoners instead of three. He now estimates his chance at 1/2 since there are only two prisoners left. Note, his real chance has not changed. If he was the chosen one, his real chance has always been 100%. There would be no advantage to switching since there is a 50% chance of making the wrong decision.--StevenQ 04:08, 25 June 2006 (UTC)

  • Restored this thing because it brings an interestin point -- once Prisioner B is ot of the game, there's a 50-50% chance of either A or C be executed. Changing places doesn' tweaks the statistics in any way. Kobayen 21:42, 17 July 2006 (UTC)

The flaw that is displayed in the wording of the problem is that it is not revealed to us whether the warden is telling the truth or potentially lying. The problem presumes that the warden will never say that A will be released. If the warden is telling the truth, then B will most certainly be released, therefore making it a 50/50 chance that A will be executed (since B is removed from the situation). If the warden is possibly lying, than the 6 scenario diagram is incomplete, because there should be several scenarios where the warden says that A will be released. The question again is, why should we presume that he will never say that A is to be released?

To answer this I will use a similar diagram as the orginal, following the same instructions for how to read it. The diagram should be as follows (granting the warden the ability to lie and to give all possible responses):

1. A b c [a] 2. A b c [b] 3. A b c [c] 4. a B c [a] 5. a B c [b] 6. a B c [c] 7. a b C [a] 8. a b C [b] 9. a b C [c]

So if the warden tells A that B will be released (which he will do in 3 of the cases) than, if the warden is lying, there is still a 33 % chance that A, B, or C will die. If the warden is telling the truth, than only 2 cases remain, one where A dies, and one where C dies. Therefore it becomes a 50/50 chance for either one. The important thing to remember here is that the initial condition (the warden lying or telling the truth) must be given in order to attain the correct probability. Also, we must not assume that the warden will not tell A that A will be released, because we have no information to justify that initial condition. --AFpilot157 11:03, 31 October 2006 (UTC)

(Ummm, yes, we do have information to justify that initial condition--that's one of the specific points in the problem: "A asks the warden to tell him the name of one of the OTHERS in his cohort who will be released." [emphasis mine] 66.32.28.191 16:47, 25 May 2007 (UTC)


Strictly speaking, the truth assumption is indeed necessary. But that is being somewhat nitpicky; one could also question the premise of one execution ("no wait, maybe there are really two"), or the absence of the possibility for a last minute pardon, etc etc.
I added some wording in the intro to reflect the assumption. It can be improved...
Update - I did improve it. Feel free to continue... Baccyak4H 16:29, 1 November 2006 (UTC)
Baccyak4H 16:23, 1 November 2006 (UTC)

Thanks Baccyak4H for making the problem a little easier to understand, but I will still continue with my solution. So basically, A is only asking about his cohorts (not about himself), and the warden is telling the truth whichever answer he gives. So in this problem there are only two choices the warden can make, either B or C. We assume he will answer each one 50 percent of the time. My question concerns the 6 scenario diagram that is listed in the original solution. Why are there 6 scenarios? More directly, why are the two scenarios (a B c [c]) and (a b C [b]) listed twice? Allow me to describe what I mean. We agree that the warden will say 'B will be released' half the time, and say 'C will be released' the other half of the time. When the warden says that B will be released, there are two possible scenarios, either A will die or C will die. When the warden says that C will be released, there are again two possible scenarios, either A will die or C will die. Here are the 3 scenarios from the orginial solution where the warden says that C will be released:

  1. A b c [c]
  2. a B c [c]
  3. a B c [c]

Why is it more likely that B will die than A? Does the warden not like B as much? What I am saying is that when the Warden truthfully eliminates one of the prisoners from the scenario, there are only two possible choices, either A will die or B will die. Where is the justification for (a B c [c]) to count as double weight? To explain my reasoning, let me start the whole scenario without any question being asked by A. There are 3 prisoners, two of which will live, and one of which will die. Here are the possibilities (uppercase letter is for the prisoner that will die):

  1. A b c
  2. a B c
  3. a b C

Let us assume that A asks the warden which one of the three prisoners is to be released (this is different than what A asks in the original problem). Here are the possibilities (the bracketed letter describes what the Warden truthfully answers with).

  1. a B c [a]
  2. a b C [a]
  3. A b c [b]
  4. a b C [b]
  5. A b c [c]
  6. a B c [c]

Here there is a 1/3 chance for the warden to say a particular prisoner will be released, and a 1/3 chance for each prisoner to die. Now we must make one more step to have the same question as the original problem. We must eliminate the scenarios where the warden answers 'A will be released'. Once we do that, we get:

  1. A b c [b]
  2. a b C [b]
  3. A b c [c]
  4. a B c [c]

From these four scenarios, A is more likely to die than B or C (before the warden answers the question). The reason is because A has no chance of being promised release (but he still could be released), whereas B and C both have a 50/50 chance of being promised a release. This is because the warden is limited in that he cannot answer with the words 'A will be released'. Understand that the situation would change if it was a different prisoner asking the question 'which of my cohorts will be released'. The interesting thing is that once the warden makes his decision on who will be released, the percentage that A will die goes back to 50 percent. Here are the final two scenarios when the warden says 'B will be released'.

  1. A b c [b]
  2. a b C [b]

I believe this to be the correct solution, showing that there is a 50/50 chance that A or C will be executed once the warden says that 'B will be released'. My justification in this is that I feel it is incorrect to give the scenarios (a B c [c]) and (a b C [b]), from the original solution, double weight. AFpilot157 12:36, 1 November 2006 (UTC)

I haven't digested your whole discussion, but your question:
why are the two scenarios (a B c [c]) and (a b C [b]) listed twice?
early on may hold the key to any misunderstanding. Part of it may be that, upon checking, the wording of the problem fails to provide a key piece of information, that A asks the warden to tell him one of the other two which will be released. This again reflects the equivalence to the Monty Hall problem where Monty will never reveal the contestent's door first. I just made a change to reflect this, so now let me try to answer your question and see if that clears up the rest of your discussion.
Each scenario in the article is generated so that each represents an equal unit of probability of outcome, 1/6 to be exact, at the start of the problem (before the decision to execute, and questioning the warden, have been made). Let's break the scenarios down using conditional probability.
There are three possible executees, each equally likely. So ignoring the warden, we have
  1. A b c
  2. a B c
  3. a b C
each equally likely (1/3).
Now, given the executee is B (1/2 chance), the is 100% chance the warden will answer C. So the probability of {a B c [c]} is (1/3) * 1 = 1/3.
Likewise, the probability of {a b C [b]} is (1/3) * 1 = 1/3 (1/3 that C is executed, 1 that B named given C executed).
But if A is on the block, the warden has 50/50 chance of answering either B or C. So {A b c [b]} and {A b c [c]} both have probability (1/3) * (1/2) = 1/6 (1/3 that A is executed, 1/2 for either naming).
So the four unique scenarios, with respective probabilities are


  1. A b c [b] (1/6)
  2. A b c [c] (1/6)
  3. a B c [c] (2/6)
  4. a b C [b] (2/6)
and the schematic "breaks" the 2/6 into two identical events of 1/6 probability for convenience. That's why those two are repeated.
If you still have trouble with that splitting, let's make a slightly more elaborate scheme which however gives identical outcomes.
  1. Select the executee (3 equally likely events)
  2. Given the executee, randomly select one of the names of those to be released (2 equally likely events; these two will not be the same two for different executees)
  3. Given the executee and the name, if the name is A, switch the name to another name of one to be released; if not A, keep name. This gives an unambiguous action which happens 100% of the time, given the executee and name.
Adding another layer to our notation, let *X* be the name given by the warden after applying the third step. So the scenarios now are
  1. A b c [b] *b*
  2. A b c [c] *c*
  3. a B c [a] *c*
  4. a B c [c] *c*
  5. a b C [a] *b*
  6. a b C [b] *b*
Each of these lines clearly has probability 1/6. There are 3 equally outcomes which the warden says "B". Of these, A is released in two of them.

But I hope my change to the original article will make things clear too. Baccyak4H 19:32, 1 November 2006 (UTC)

Thank you for responding so quickly Baccyak4H. After reading your response, there are two things I wish to bring up when it comes to this problem. First is the probability of independent events (or dependent events). Second is the actual question we are striving to answer. As stated in the original article, the question is 'what are A's and C's respective probabilities of dying now?'. In your response, it seems to me that your first step is to select the executee, and then follow through in finding the probability of being picked by the warden to be released. How I look at it is through a different direction. I begin by selecting the person the warden picks, and then finding the probability of the others being executed. In your response you multiply the probabilities of two events together, one event being the probability of being executed, and the other event being the probability of being picked by the warden to be released. The key here that I wanted to bring up is that those two events are NOT independent of each other, and therefore cannot be multiplied together. They are dependent events, and have an affect upon each other's probability. If someone is picked to be released, they cannot be executed, and if someone is executed, they cannot be picked to be released. An example that can be used is dice and coins. The probability of me flipping a coin and getting a tails is 1/2. The probability of me rolling a die and getting an even number is also 1/2. The probability of getting tails on a coin AND getting an even number on a die is 1/4, because they are independent of each other and can be multiplied together. When you multiplied the probability of getting picked by the warden, and the probability of getting executed, the result can be worded as:

The probability that prisoner ____ gets released AND prisoner ______ gets executed is _____.

In the case of your calculations, it was 1/6 for some, and 2/6 for others, depending on which prisoner was chosen as released, and which one was chosen as executed. In conclusion, it does depend on which direction we take when approaching the solution. Do we start off with choosing which prisoner gets picked (which will then affect chances of being executed for everyone), or do we begin by choosing who gets executed (which directly affects who can get picked). Since we are told the warden selected B to be released, and then we are asked the chances of A and C being executed, we must start off with the probability of being picked. As stated in my previous solutions, A cannot be picked. This therefore makes it so that there is a 50/50 chance for B or C to be picked respectively. If B is picked, the only options are that A is executed, or C is executed. If C is picked, the only options are that A is executed, or B is executed. This again leaves us with the solution that A's and C's chances for being executed are exactly the same once the warden says that 'B will be released'. As a note, if any of the wording in the problem is changed, or if the question that it asks us is changed, then everything changes, because of the probability of dependent events. To further illustrate this, you mentioned the following in your response:

'There are three possible executees, each equally likely'

The fact is that the likeliness of anyone being executed is directly affected by the fact that the warden picks one to be released. Even if we do not know who he picks, as soon as we know that he cannot pick A (deduced by the fact that A is asking ONLY about his cohorts), then the probability of being executed is no longer equal, but becomes greater for A (since A has no chance of being picked by the warden). Yes it's true that in the very beginning they all have a 1/3 chance of being executed. But that 1/3 must change as soon as we are told that either B or C will definitely be released (and A is not given the opportunity to be picked by the warden). This is why the probability of dependent events is affecting the solution to this problem, and it is necessary to pick the correct direction to start off in. AFpilot157 16:11, 1 November 2006 (UTC)

When you stated
In your response you multiply the probabilities of two events together, one event being the probability of being executed, and the other event being the probability of being picked by the warden to be released. The key here that I wanted to bring up is that those two events are NOT independent of each other, and therefore cannot be multiplied together. They are dependent events, and have an affect upon each other's probability.
you apparently missed what I meant by the word "given". This word was not intended to connote dependence or lack thereof. It was meant to connote conditionality. (See this page for descriptions of this term and others coming up.) This is an entirely different concept. You are right that multiplying the overall probability of an execution with the overall probability of a name does not give the right answer (for precisely the reason you give). But I was not referring to the overall probability of the name (that would be the marginal probability), but the conditional probability, the probability of the name given we know the particular executee. The idea behind conditionality is that the universe of events that could occur has been reduced in some way, indeed by knowing (or just assuming) that a certain event(s) have or have not occurred (in this case, the particular executee, which while we don't know, we can assume as if we knew then our results apply if we turn out to be right). Then considering this part of the universe of events the whole universe for probability calculus all over again.
That concept is crucial. On the page given above (I am changing symbols to avoid conflicts with this page), the conditional probability of X given Y (written P(X \mid Y)) is 'defined' to be the probability of both X and Y happening (written P(X \cap Y)), divided by the (marginal) probability of Y happening (written P(Y)). Understand that P(X \mid Y) only has relevance about X if Y actually happens. It says nothing about what X is if Y doesn't happen, and (to your point) nothing about what X is when there is no reference to Y at all; that last one would be the marginal probability of X, what you thought I was multiplying, but I wasn't. Let's write out the formula:
P(X \mid Y)= \frac{P(X \cap Y)}{P(Y)}\,\!.
Now let Y be the event that a particular prisoner is on death row, and X the event of a particular prisoner being named by the warden. But what happens when we multiply both sides by P(Y)?
P(X \cap Y) = P(X \mid Y) P(Y)\,\!
What is this saying? The probability of both the particular executee and the particular naming is this product. This is the multiplication I was doing. So long as Y's probability is not zero (it's always 1/3 here), this computation is on about as solid ground as it comes.
You actually acknowledged this concept without (apparently) knowing it when you said "The fact is that the likeliness of anyone being executed is directly affected by the fact that the warden picks one to be released." This is indeed true; you are saying P(Y) \ne P(Y \mid X) which is not (in general) true. More about this below.
And you actually have it backwards about knowing A cannot be named. Once we know that, it follows that since the warden can always come up with another name, whatever that name is provides no information about A 's fate. It only provides a certainty about the one named, and sobering news for the one not named. Consider if C was listening through a duct unbeknownst to A and the warden. C knows that the warden could say C, and wants to hear it. Knowing C could have been said, but wasn't, provides information about C 's fate.
Your idea about starting with enumerating what the warden says is certainly possible to try; you might need Bayes' theorem to do that, I haven't tried. The idea of conditioning on the executee makes the probability calculus turns out to be quite straightforward however; anything valid you do from the other direction will by necessity agree with the current way. Baccyak4H 04:00, 2 November 2006 (UTC)

I see what you mean with your response. All I wish to do now is introduce two ideas, and see what you think about them. First, since the problem is asking 'what are A's and C's respective probabilities of dying now?', that brings up the interesting use of the term 'now'. By using the term 'now', it is implying that the warden already made his choice about one of the people who will live by saying 'B will be released'. This means that we are being asked about the probability of death for A after we know that B will live. Since we know beyond a doubt what B's fate is, perhaps he can be removed from the situation, since we are trying to find the probability of A's and C's death respectively (without looking at B). Secondly, I want to mention that it seems the perspective of ourselves, the perspective of A, and the method we use in attacking this problem all have different effects on the situation. To better illustrate my two ideas, please go ahead and read this article. I am not well-versed in the mathematics to put it in my own words, and so I wish to give credit to the authors of this article:

http://www.stat.auckland.ac.nz/~iase/publications/17/3I3_LOSC.pdf

Hopefully this will show that there are multiple solutions to this problem, depending on which method you use, and how you look at the information being presented in the problem. AFpilot157 08:32, 2 November 2006

Second point first. The problem is the problem. That article is using the context of the problem as a pedagogical tool to demonstrate some more sophisticated probabilistic techniques. But in setting up things for purposes of demonstration, often it generates a scenario which is no longer the problem, for example an page 3:
"On the other hand, if A thinks the warden would never tell the name of C, in case A were the survivor, ..."
The difficulty is, we can't use this type of exercise to answer your questions about survival probabilities for our problem here because they are no longer the same problem. It is permissible to use different methods to arrive at the answer, but if done correctly, the answers will always agree.
For example, what is not happening here is that the warden has not yet decided who to execute, but when asked by A who will be released, decides on the spot randomly between B and C to be released and then reports that (B in the statement of the problem) to A, only afterwords deciding who to execute among the remaining two. Remember this problem is analogous to the Monty Hall problem, and in that problem, Monty knows what is behind each of the three doors; they are already fixed and he can only react to them. If you or someone chooses to condition on who the warden names, this point will be very important to acknowledge.
Now the first point. You have actually (again) described conditional probability without apparently realizing it. "the probability of death for A after ["given" -B4H] we know that B will live. Since we know beyond a doubt what B's fate is, perhaps he can be removed from the situation" You are describing probability calculations which are conditional on the warden naming B. Those are done exactly as I described above with those pretty ( ;-) ) formulas. Baccyak4H 15:26, 2 November 2006 (UTC)
Oops, meant to add: A minor note about that article. Be careful reading it and comparing to the WP article; the statement of the problem there differs in that one will be released, as opposed to executed; there the warden reveals that B will be executed. Baccyak4H 15:28, 2 November 2006 (UTC)

I noticed that the scenario in the article was worded differently than the one in Wikipedia, but it should not matter. The reason is because Wikipedia says 'two of the three will be released' but the article says 'two of the three will be shot'. The object of the question was changed, but the scenario was changed in a way to compensate, so that the answers for both should be the same, and can be treated the same. Even though this problem is supposed to be analogous to the Monty Hall problem, we must treat each paradox individually, and not say that since one is supposedly analogous to the other, that there is no way they could have different outcomes. Is the article incorrect in the views it portrays? Is it possible that the problem as worded does not give us enough information to know for sure which direction to take? Also, would it matter if we made the following scenario:

'There are three prisoners, A, B, and C. B is released from prison. One of the remaining prisoners (A or C) will be executed. What are the chances of A being executed?'

Would the outcome be any different for this scenario, and if so, then why? I mean, for the original three prisoners scenario, once the warden says that B will be released, how does the warden go about picking who gets executed afterwards? Let us say that the warden (after saying B is released), flips a coin to see which one (A or C) gets executed. Is it true that there is more of a chance for the coin to land on tails than heads? AFpilot157 11:23, 2 November 2006

Right. Only the labels ("released"/"shot") change. And saying this problem is analogous to the Monty Hall problem is true, but indeed not as strong as it could be. It is isomorphic to the Monty Hall problem. A considerably stronger statement, since it implies their solutions will be isomorphic as well. Baccyak4H 17:40, 2 November 2006 (UTC)
(crossed posts) About a third or so of the article is over my head, and much is too dense for a casual read, so I cannot vouch for much of it, but that is beside the point. It essentially says "By changing some of the conditions of this problem, we can illuminate some techniques of probability theory." And I will add, some of the content being illuminated comes as close to philosophy as it does to probability.
Your scenario as written does not portray the information about B in the same way. It was not chosen to be revealed in a way so as to explicitly avoid being A instead. In other words, based on only your statement of the problem, we could, if we chose to, logically augment the problem by adding "B asked, and found out, that s/he was to be released", which we could not do in the original problem without changing its very premises. Here they haven't changed, they have only been added to. So it does matter. A lot.
And I repeat what I alluded to earlier. The warden already knows who is to be executed when asked by A about a release. If the warden doesn't know, then "chooses" to release B and then figure out who to execute, that is a different problem. Baccyak4H 18:02, 2 November 2006 (UTC)


Perhaps the Monty Hall problem should be analyzed in the same way the article analyzed this problem. Maybe there are slight differences, or maybe Monty Hall has different solutions. AFpilot157 11:55, 2 November 2006

So long as the problem analyzed is still the Monty Hall problem, it doesn't matter how it's analyzed: the solution is the same. To get different solutions, one would need to alter the Monty Hall problem to something better called "a variant of the Monty Hall problem." Baccyak4H 18:02, 2 November 2006 (UTC)

How can we determine that the warden already knows who is to be executed when asked by A about a release? You mentioned that if the warden doesn't know, then 'chooses' to release B, that is a different problem. Where in the original article or problem does it specify that the warden already knows who is to be executed? This is what I meant in that there is more than one way to approach it. The problem as stated says:

'There are three prisoners, A, B, and C. Two of them will be released and one will be executed. A asks the warden to tell him the name of one of the others in his cohort who will be released. As the question is not directly about A's fate, the warden obliges and says, "B will be released." Assuming the warden's truthfulness, what are A's and C's respective probabilities of dying now?'

There is nothing in this to justify that the warden already knows who will be executed. There is something at least to justify that the warden is picking to release a prisoner first, since the warden is truthful and says the words 'B will be released'. It is after this that we are asked to find the probability of being executed (maybe it is even after B took his bags and went home). AFpilot157 15:10, 2 November 2006

If you wish to take the description to that level of literalness, feel free to do so; perhaps after the original description spell out the assumptions explicitly as a lead into a calculation of A and C's survival probability. I would recommend not putting this much detail in at the top: nearly all readers will find such additions unnecessary and making for a poor and murky read. See, for example, this WP article which demonstrates the pitfalls of letting literalness supercede common sense.
If you wish to suggest the problem is anything but what I have repeatedly described it is with respect to the warden's thought process, perhaps you can start an article about variants of the problem and put that material there. Others may think it too derivative for WP, although I for one will not. What I will do is resist any efforts to describe this problem in any way which is inaccurate.
That said, I am a golfer, not a fisherman, yet I have bigger fish to fry (rough translation: don't want to stay this bogged down in details). Carry on. Baccyak4H 03:54, 3 November 2006 (UTC)

[edit] The solution is sound

I wrote a PHP script to investigate this problem. Just like the Monty Hall problem, the solution is totally counterintuitive but at the same time apparently correct. Here is my script:

 for ($i = 0; $i < 1000; $i++) {
        srand((double)microtime()*1000000);
        $n = rand(1,3); //Number of prisoner who is scheduled to die (1, 2, or 3)
 
        if ($n != 2) { //Warden reveals that prisoner 2 will be released
                if ($n == 1) { //if prisoner 1 is scheduled to die
                        $dead++;
                }
                elseif ($n == 3) { //if prisoner 3 is scheduled to die
                        $alive++;
                }
        }
        elseif ($n != 3) { //Since prisoner 2 is scheduled to die, warden instead reveals that prisoner 3 will be released
                if ($n == 1) { //if prisoner 1 is scheduled to die
                        $dead++;
                }
                elseif ($n == 2) { //if prisoner 2 is scheduled to die
                        $alive++;
                }
        } 
 }
 echo "<br><br>Alive:" . $alive . "<br><br>Dead:" . $dead;

The output was always "Alive: 667 / Dead: 333" (with a slight amount of variation, obviously). It is the "elseif ($n != 3)" (accounting for the possibility that the warden reveals the third prisoner to live, instead of the second) that causes this to occur. Without that part, the results come to "Alive: 333 / Dead: 333", which is the more intuitive result. Every time that "elseif ($n != 3)" part is run, prisoner 1 lives, since if $n was equal to 1 he would have been counted dead in the first part. This should be considered the key to understanding how this works. — GT 02:57, 22 July 2006 (UTC)

if the warden also tells C that B is free, will both A and C less likely be killed if they switch?

[edit] Fixing description of problem

The description of the problem needs to emphasis that the guard's choice could be either "B" or "C" (might be best to not mention any letter at all). If the guard's choice is always "B", then clearly the outcome is 50/50 chance for both "A" and "C". You can't get the paradoxical outcome unless what the guard tells "A" is a 'free' choice. Ajapted 12:20, 31 October 2006 (UTC)

That cannot happen if the warden is truthful: if B's number is up, the warden must, and will, say C. Baccyak4H 16:25, 1 November 2006 (UTC)
That's the intended interpretation of the problem. But I think Ajapted is right that it needs to be more explicit. The way it's written now one might conclude that B must never have been scheduled to be executed to begin with, since the warden says "B will be released" and is truthful. However, the intended interpretation is that if B were indeed scheduled to be executed, the warden would say "C will be released" instead.
What the warden is really doing is randomly choosing the name of someone other than A who will be released. But I'm not sure how to work that into the article without breaking the intro and solution, which assume that the warden has already said B is safe. kazayta (talk) 03:42, 11 April 2008 (UTC)

[edit] A or C?

Haha, this made me laugh when I thought about it some more. Besides the fact the odds are 1/1 or 0/1 no matter what like someone mentioned before, the idea in this article is ridiculous to begin with. Think about it.

If A has a 1/3 chance of being executed, then the odds of B or C being executed is 2/3. If he finds out B won't be executed, that means there's a 2/3 chance C will be executed. So far so good.

But now, lets say C is also wondering who will be set free, and he finds out it's B. Since there was a 2/3 chance that A or B would be executed, the chances that A will be executed have just gone up to 2/3.

See what I mean?

(If I'm wrong, be sure to tell me) —The preceding unsigned comment was added by 80.126.65.34 (talk) 16:53, 12 March 2007 (UTC).

Probabilities only make sense in the context of a certain amount of knowledge. To someone (such as the warden) who knows who has been selected (either before or after the execution), the probabilities that any one of the prisoners was/will be executed is 1 or 0. The "paradox" here relies on the specific knowledge that the prisoner has. We (like the prisoner) know more than just "B won't be executed" - if that were all we knew, then as far as we could tell A and C would still be equally likely to be killed. However, A also knows how the information that B won't be executed was obtained, and given this knowledge, the chances of being killed are still 1/3 (he is less likely to be told that B was safe if he himself is the selected one that if C is facing the gallows). If C had listened in on this conversation, then as far as he could tell he would have a 2/3 chance of being executed. If he hadn't heard the conversation with A, but was told that B would be freed in the same way that A was, he would see his odds as 1/3. Say that someone (B?) had heard both these discussions. Then as far as they know, A and C are equally likely to die. In this case, the warden and each of the three prisoners would give us different odds, because they each have different knowledge. This may seem ridiculous, but it is really what probability is all about to start with. If you flip a coin, see that it is heads, but don't let your friend know, then in your knowledge the probability that you got heads is 1, but to them it is 1/2. JPD (talk) 15:26, 15 May 2007 (UTC)


[edit] Another thought

No matter what all three prisoners could ask a question of the guard and get an answer that would lead them to think they have a 2/3 shot at living.

2/3 = 2/3 = 2/3

So they're all equal still... Meaning their chances at life are still an even split (in this case 1/3). This new info changes nothing from the origional odds. Each prisoner could figure this out also too. So where's the paradox? This seems like more of mathmatical vagueness trick than a true paradox. An apples and oranges thing. That 2/3 vs. 1/3 this is meaningless if all 3 have it.

If all of them can have those odds then they're all still equal. If they're all still equal, it's 1/3 again. Correct? Yes? No?

Randy --66.189.25.112 (talk) 01:13, 19 May 2008 (UTC)

No, not correct. The way Gardner phrased the problem is there are 3 condemned men (A, B, and C), one of whom is pardoned. A gets the warden to tell him one of B or C who will be executed (randomizing which one he says if A is the one being pardoned). A's chances of being pardoned are still 1/3, but whichever of B or C is not being executed now has a 2/3 chance of being pardoned - for example if the warden said C is being executed, then B has a 2/3 chance of being pardoned (but only A knows this).
B can also ask the warden the same thing. If the warden ends up telling B that C will be executed then it would seem from B's perspective A should have a 2/3 chance of being pardoned. A and B now seemingly both have a 2/3 of being pardoned, which seems like a paradox since only one of them will be and the total probability must be 1 between them.
The solution to this paradox requires answering the question of how the warden decided what name to give B (after giving C's name to A). If the warden gives C's name to B because he's already given it to A, then this changes nothing and A has a 1/3 chance of being pardoned while B has a 2/3 chance (it's not the case that B has a 1/3 chance and A a 2/3 chance because the warden didn't pick between A and C avoiding B, he picked between B and C avoiding A - B may not know this but it doesn't matter from the 3rd party observer's perspective). On the other hand if the warden goes through the same process with B that he did with A, and happens to have also ended up giving C's name to B then A and B are equally likely to be pardoned. What happens is:
Inmate being pardoned A B C
Probability of being pardoned 1/3 1/3 1/3
What warden tells A B C C B
Probability 1/6 1/6 1/3 1/3
What warden tells B don't care C A C don't care
Probability 1/6 1/6 1/6 1/6 1/3
The don't cares in the "what warden tells B" row are because we're only interested in cases where the warden has given C's name to A. Looking at the cases where the warden gives C's name to A and then also gives C's name to B there are only two cases left, both with probability 1/6. In one, A is pardoned and in the other B is pardoned, so only considering these two cases A and B both have a 1/2 chance. -- Rick Block (talk) 01:45, 19 May 2008 (UTC)

I grapsed th monty hall one... but this one is messing with me. What a way to go. Sitting on death row thinking about a really confusing math problem. Let's hope "A" got out. --66.65.244.151 (talk) 00:25, 20 May 2008 (UTC)

[edit] The solution is not sound

Right or Wrong?

1. the only way this problem can be completely analogous with the monty hall problem is if the prisoner doing the questioning gets the chance to change his fate after questioning the warden. even then he is not reducing his chance of being executed, he is only able to prevent it from being increased.

Which version are you talking about, where only one is pardoned or only one is executed? In the version Gardner presented only one is pardoned. Minus the ability to switch this is completely analogous to the Monty Hall problem where "pardoned" is equivalent to "wins the car". The prisoner doing the asking has a 1/3 chance of being pardoned, while the remaining cohort has a 2/3 chance of being pardoned.

2. by asking the warden which of his 2 cohorts will be released (removing himself from the equation), he is effectively reducing the odds to 50/50 between his 2 cohorts. once the warden responds this changes the equation to 1. now the prisoner must reintroduce himself to the equation thereby splitting the odds 50/50.

Not exactly. Again, in the one is pardoned version, the chance of the pardon is 1/3 with the asker and 2/3 between the other two. If by "removing himself from the equation" you mean "we're only looking at the 2/3 case involving the 2 cohorts (where one of them is the one being pardoned)", then yes there's a 50/50 chance between them - but this is only in the 2/3 case where one of them is the one being pardoned. When the warden's response changes this to 1/0, what's really going on is we're changing the ( 1/3 asker)  : ( 1/3 : 1/3 cohorts) ratios to ( 1/3 asker) : ( 2/3 : 0 cohorts), which is not 50/50 between the asker and the remaining cohort but 1/3 : 2/3.

If all the prisoner is trying to do is make himself feel better about the situation he's better off closing his eyes and sticking his fingers in his ears until showtime to keep his (perceived) chance of execution at 1/3. Brad38 (talk) 11:14, 30 May 2008 (UTC)

If there's only one pardon, and the prisoner can switch fates, doing so doubles his chance of surviving. -- Rick Block (talk) 14:45, 30 May 2008 (UTC)