View Full Version : Twist on an old favorite

JSUCamel

09-03-2009, 12:23 AM

Usually this is told within the context of a gameshow, but.. still. It's pretty controversial, and I'd like to see what you guys have to say about it:

Three condemned prisoners share a cell. A guard arrives and tells them that one has been pardoned.

"Which is it?" they ask.

"I can't tell you that," says the guard. "I can't tell a prisoner his own fate."

Prisoner A takes the guard aside. "Look," he says. "Of the three of us, only one has been pardoned. That means that one of my cellmates is still sure to die. Give me his name. That way you're not telling me my own fate, and you're not identifying the pardoned man."

The guard thinks about this and says, "Prisoner B is sure to die."

Prisoner A rejoices that his own chance of survival has improved from 1/3 to 1/2. But how is this possible? The guard has given him no new information. Has he?

Discuss.

Frenzy

09-03-2009, 12:26 AM

Hey Prof. Snow! Talk to Prisoner A about the law of averages!!!

Zanguini

09-03-2009, 08:18 AM

What is really happening is that they are all really going to die and the guard is just messing with them for a couple of laughs.

Sei'taer

09-03-2009, 08:34 AM

I don't get the question, I guess.

1 chance to survive/3 prisoners (A,B,C).

Take 1 prisoner (B) out of the equation, because you know he isn't on the list of people who will survive anymore, given the guards answer. Which leaves

1 cts/2 prisoners (A,C)

I kind of read it like figuring odds on a horse race...maybe that was the wrong way to look at it.

If prisoner A shanks prisoner C and makes it look like prisoner B did it, he might be able to seal the deal and get the pardon. Either that or everyone dies, so no one gets special treatment, which is really the correct way to deal with things in the long run.

GonzoTheGreat

09-03-2009, 08:59 AM

All right, let's extend the story a bit.

Prisoner B takes the guard aside, and the guard tells him the name of a prisoner who is going to die too. Same with prisoner C.

So now, according to the claimed logic, every prisoner only has a chance of one out of two to die, yet still two of the three prisoners are going to be executed.

Continue with the discussion. :D

Brita

09-03-2009, 09:40 AM

But he didn't say executed, he said "is sure to die". Well, of course, we all will surely die. So he didn't tell him anything of value at all.

I'm with Zanguini on this one.

Jokeslayer

09-03-2009, 09:50 AM

I think we need to know more about prisoner A first. If he's rich and white while the others are black and poor, I know where I'd put my money.

Brita

09-03-2009, 11:42 AM

I think we need to know more about prisoner A first. If he's rich and white while the others are black and poor, I know where I'd put my money.

LOL! Touche.

Davian93

09-03-2009, 11:44 AM

I think we need to know more about prisoner A first. If he's rich and white while the others are black and poor, I know where I'd put my money.

Yeah, but if he were rich and white, what's he doing in prison?

Brita

09-03-2009, 11:50 AM

Yeah, but if he were rich and white, what's he doing in prison?

He's Conrad Black.

Davian93

09-03-2009, 11:59 AM

He's Conrad Black.

I just had to look him up on Wikipedia (sad, I know). Sounds like a real winner.

John Snow

09-03-2009, 12:01 PM

Hey Prof. Snow! Talk to Prisoner A about the law of averages!!!

or the Law of Large Numbers, eh?

(or was that Numbnuts?)

JSUCamel

09-03-2009, 03:24 PM

OK, here are the two arguments:

1) One camp claims that the information given has no effect on the outcome -- he's still got a 1/3 chance of being pardoned.

2) Another camp claims that the information given has increased his odds for a pardon from 1/3 to 1/2.

The reasoning is as follows:

You've got two prisoners who will die and one prisoner who will be pardoned. Since Prisoner A is asking, the guard can't reveal Prisoner A's fate.

Which prisoner is going to die is irrelevant -- the fact is that the guard says that another prisoner will die. The fact that Prisoner B will die means that there are only two choices left: Prisoner A will be pardoned or Prisoner C will be pardoned.

Thus, there is a 50/50 chance that Prisoner A will be released.

GonzoTheGreat

09-03-2009, 03:41 PM

Camel, I already answered that with a scenario where every one of the prisoners asks the question, and is told that one of the other prisoners will be executed. So, if you are correct, that means that all three of the prisoners have 50% chance of being released. Thus, either your reasoning is incorrect, or 1.5 prisoners will be released, and the other 1.5 prisoners will be executed. Naturally, the fairest way of doing that would be chop each of the prisoners in two, and then pardon their right (or left) sides. But that contradicts the "one will be pardoned thing".

JSUCamel

09-03-2009, 03:52 PM

Camel, I already answered that with a scenario where every one of the prisoners asks the question, and is told that one of the other prisoners will be executed. So, if you are correct, that means that all three of the prisoners have 50% chance of being released. Thus, either your reasoning is incorrect, or 1.5 prisoners will be released, and the other 1.5 prisoners will be executed. Naturally, the fairest way of doing that would be chop each of the prisoners in two, and then pardon their right (or left) sides. But that contradicts the "one will be pardoned thing".

Well, the thing is that Prisoner B has a 100% chance of dying (1/1) whereas the others have a 50% chance, so your logic fails. The only way your scenario would be true is if, when asked by each prisoner, the guard picks a different prisoner (i.e. when A asks, he says B will die, when B asks, he says C will die, and when C asks, he says A will die). Only in that case does the 50% logic fail.. but it wouldn't happen (assuming the guard is telling the truth) because ONE of those prisoners will be pardoned.

bowlwoman

09-03-2009, 03:57 PM

Which prisoner is going to die is irrelevant -- the fact is that the guard says that another prisoner will die. The fact that Prisoner B will die means that there are only two choices left: Prisoner A will be pardoned or Prisoner C will be pardoned.

No, the problem with this is the wording the guard used about the fate of Prisoner B. He said, "Prisoner B is sure to die." Like Brita said, we're all going to die. A is going to die, B is going to die, C is going to die. Two of them will be executed, and one will die from other means. But, they all will die at some point. Whether or not they are pardoned is completely independent of the fact that they will die. So, no new info here.

It's kinda like the brainteaser: how many months have 28 days? They all do. Not just February.

Gilshalos Sedai

09-03-2009, 04:04 PM

I agree with Bowlwoman. The guard gave no new information.

GonzoTheGreat

09-03-2009, 04:07 PM

Well, the thing is that Prisoner B has a 100% chance of dying (1/1) whereas the others have a 50% chance, so your logic fails. The only way your scenario would be true is if, when asked by each prisoner, the guard picks a different prisoner (i.e. when A asks, he says B will die, when B asks, he says C will die, and when C asks, he says A will die). Only in that case does the 50% logic fail.. but it wouldn't happen (assuming the guard is telling the truth) because ONE of those prisoners will be pardoned.Nope, the problem is that you are looking at it from the outside, but you are asking what the prisoners can know.

Suppose, just for the sake of argument, that prisoner C has been pardoned.

Prisoner A asks the question, and is told that B will be executed.

Prisoner B asks, and is told A will die.

Prisoner C asks, and is told that A will die.

Now, would it really be correct for every one of those three to conclude that they have 50% survival chance?

We know that, after they've gotten the (truthful) answer to their question, two out of three prisoners will die. So given those actual odds, how do you defend the one out of two odds you are claiming?

Of course, in my scenario, if the three prisoners compare their answers, they can figure out which of them will survive.

Sei'taer

09-03-2009, 04:23 PM

OK, here are the two arguments:

1) One camp claims that the information given has no effect on the outcome -- he's still got a 1/3 chance of being pardoned.

2) Another camp claims that the information given has increased his odds for a pardon from 1/3 to 1/2.

The reasoning is as follows:

You've got two prisoners who will die and one prisoner who will be pardoned. Since Prisoner A is asking, the guard can't reveal Prisoner A's fate.

Which prisoner is going to die is irrelevant -- the fact is that the guard says that another prisoner will die. The fact that Prisoner B will die means that there are only two choices left: Prisoner A will be pardoned or Prisoner C will be pardoned.

Thus, there is a 50/50 chance that Prisoner A will be released.

So I got it right! Wooohooo!

JSUCamel

09-03-2009, 04:33 PM

Nope, the problem is that you are looking at it from the outside, but you are asking what the prisoners can know.

Suppose, just for the sake of argument, that prisoner C has been pardoned.

Prisoner A asks the question, and is told that B will be executed.

Prisoner B asks, and is told A will die.

Prisoner C asks, and is told that A will die.

Now, would it really be correct for every one of those three to conclude that they have 50% survival chance?

We know that, after they've gotten the (truthful) answer to their question, two out of three prisoners will die. So given those actual odds, how do you defend the one out of two odds you are claiming?

Of course, in my scenario, if the three prisoners compare their answers, they can figure out which of them will survive.

Hmmm...

GonzoTheGreat

09-03-2009, 04:36 PM

Hmmm...Yes, indeed. :D

Belazamon

09-03-2009, 04:54 PM

I think one of the problems is, we're dealing with two different sets of odds here.

Let's replace the prisoners with pennies. If you flip all three pennies, each penny has a 50-50 chance of landing on heads. But the odds of two pennies landing heads-up and one landing tails-up is not 50-50, because you're not talking about the same outcome any longer. That doesn't change the outcome for any individual flip though.

Of course, there's also the "sure to die" thing. But hey, I like statistics. :D

This is one of the problems we covered in probability during my math degree.

I assume that by the guard saying "B is sure to die" that B is one of the ones to be executed. You can show mathematically using probability theory that A gains no new information regarding his own fate. His chances are still 1/3 that he will be pardoned. Ironically, he now knows that C has a 2/3 chance to be pardoned.

I'm not going to post the actual math, but looking at my notes, you can think of it like this:

Small letters mean the prisoner is pardoned, capital letters mean the prisoner is executed. The letter in brackets is the guard's response as to who is to be executed. There are six possibilities, each with 1/6 chance of happening:

a B C (B)

a B C (C)

A b C (A)

A b C (C)

A B c (A)

A B c (B)

But A is doing the asking, and the guard cannot tell A if he is being executed. This modifies the guard's possible responses and the list is now (all with 1/6 chance still)

a B C (B)

a B C (C)

A b C (C)

A b C (C)

A B c (B)

A B c (B)

Now you notice that there are three cases in which the guard answers "B is sure to die", and only in one of those cases is A pardoned. So the chance of A being pardoned given B is executed is still only 1/3. But in the other two cases C is pardoned, so C's chances are now 2/3 not to be executed.

The problem is the way the question is asked. By giving the guard the option of who to say is being executed A doesn't gain new information. If A had asked "Will B be executed" and the guard said "Yes", then A knows that his chances are now 1/2 (That's equivalent to the first list, only two of the cases does the guard answer B, and in one of them, A is pardoned).

JSUCamel

09-03-2009, 05:05 PM

The problem is the way the question is asked. By giving the guard the option of who to say is being executed A doesn't gain new information. If A had asked "Will B be executed" and the guard said "Yes", then A knows that his chances are now 1/2 (That's equivalent to the first list, only two of the cases does the guard answer B, and in one of them, A is pardoned).

Yes, er.. that's what I meant.

I get Gonzos point, but that is only if you take it further than the problem states. From the prisoner A's view, he at least believes that his odds have gone from 1/3 to 1/2. Which is correct from his point of view. Objectively, you might have to consider that Prisoner A is actually the one who's gonna die so the guard is lying to A. A just doesn't realizes this fact :)

There is another classic example of this kind of probabilty, or decision theory if you will. It goes like this:

You're on a game show and you are to select a door out of three. Behind two of the doors are a worthless price and behind the third one, there's a huge price, let's say a car.

You pick, let say, door no 1. The game show host doesn't open door no 1 but opens up one of the wrong doors, lets say door no 2. He then asks you if you want to stay with your selected door or if you want to change door.

Should you change door?

Probabilities are fun, totally wack at times, but fun :)

Edit: Actually, the prison example isn't complete. When prisoner A gets to know that Prisoner B will die, prisoner A should get the chance to switch places with prisoner C.

Terez

09-03-2009, 05:16 PM

There is another classic example of this kind of probabilty, or decision theory if you will. It goes like this:

You're on a game show and you are to select a door out of three. Behind two of the doors are a worthless price and behind the third one, there's a huge price, let's say a car.

You pick, let say, door no 1. The game show host doesn't open door no 1 but opens up one of the wrong doors, lets say door no 2. He then asks you if you want to stay with your selected door or if you want to change door.

Should you change door?

Yes, that's the 'old favorite' he was talking about. It's explained rather well here (http://www.marilynvossavant.com/articles/gameshow.html).

I get Gonzos point, but that is only if you take it further than the problem states. From the prisoner A's view, he at least believes that his odds have gone from 1/3 to 1/2. Which is correct from his point of view. Objectively, you might have to consider that Prisoner A is actually the one who's gonna die so the guard is lying to A. A just doesn't realizes this fact :)

A might believe that his chances are now 1/2, but he'd be wrong. A's main problem is he doesn't know what question to ask :D

Yes, that's the 'old favorite' he was talking about. It's explained rather well here (http://www.marilynvossavant.com/articles/gameshow.html).

The prisoner example is actually older than the monty hall example. Although I find the monty hall example easier to grasp, but maybe because I like game shows :)

The really "old favorite" is actually something Bertrand Russel came up with in the late 1800s btw. Fun trivia :P

Maybe I'm not quite getting it, but in actuality, two prisoner have a 100 percent chance of dying, the last one zero. After all, the decision of who lives and who dies has already been made. The only thing that changes when A learns that B is sure to die is that A gets some more information, and that doesn't affect his chances of survial one way or the other. Chance doesn't really come into it, because the winner has already been selected.

Maybe I'm not quite getting it, but in actuality, two prisoner have a 100 percent chance of dying, the last one zero. After all, the decision of who lives and who dies has already been made. The only thing that changes when A learns that B is sure to die is that A gets some more information, and that doesn't affect his chances of survial one way or the other. Chance doesn't really come into it, because the winner has already been selected.

Like I said, although with an edit, in my post. Prisoner A should actually be given the chance to swap places with Prisoner C, it was missed out in the example. Then A gets the choice, and can then improve his chances.

tworiverswoman

09-03-2009, 05:40 PM

But we're not looking for his chances of survival - only his KNOWLEDGE of his chance -- and that HAS changed from 1 in 3 to 1 in 2.

But we're not looking for his chances of survival - only his KNOWLEDGE of his chance -- and that HAS changed from 1 in 3 to 1 in 2.

That's only in the prisoner's mind. His knowledge has changed, but his knowledge is incomplete. If he's been pardoned, he may think he has a 50 percent of dying, but he actually doesn't; if he's condemned he may still think he's got a 50 percent change of living, but he doesn't.

To him it may seem that there's chance involved, but if the decision has been made and will stand no matter what, there really isn't. His perception of what's going on doesn't change his or anyone else's odds, though he might feel that it does.

This is assuming that he can't change place with one of the others, of course.

But we're not looking for his chances of survival - only his KNOWLEDGE of his chance -- and that HAS changed from 1 in 3 to 1 in 2.

No, no, as I pointed out, the math shows that A's knowledge of his chance of survival has NOT changed. It's still 1/3. However, his knowledge of C's chance of survival HAS changed. He now knows that C has a 2/3 chance of survival. That is why if he is given the option to switch with C, he should take it.

Birgitte

09-03-2009, 10:40 PM

B is sure to die

Am not. I'm a cylon.

JSUCamel

09-03-2009, 10:48 PM

Hmm.. This thread didn't go anywhere near the way I expected it to. Awesome.

GonzoTheGreat

09-04-2009, 04:54 AM

You're on a game show and you are to select a door out of three. Behind two of the doors are a worthless price and behind the third one, there's a huge price, let's say a car.

You pick, let say, door no 1. The game show host doesn't open door no 1 but opens up one of the wrong doors, lets say door no 2. He then asks you if you want to stay with your selected door or if you want to change door.

Should you change door?And this is why the prisoner's odds don't change: he can't change doors.

Awesome.I know that I am that.

The question is: does this thread change the odds of other people figuring it out?

DahLliA

09-04-2009, 08:20 AM

Yes, that's the 'old favorite' he was talking about. It's explained rather well here (http://www.marilynvossavant.com/articles/gameshow.html).

that was actually a good read. if I ever find myself on a gameshow I'll be sure to switch :D

Brita

09-04-2009, 08:31 AM

OK- I was thinking about this on the drive in.

Simply put, the prisoner knows that at least one of the other prisoners will be executed. So to be told prisoner B's fate really doesn't change his own chances at all. He gains no new knowledge on his own fate because he already knew that at least one of the other prisoners would be executed.

No, no, as I pointed out, the math shows that A's knowledge of his chance of survival has NOT changed. It's still 1/3. However, his knowledge of C's chance of survival HAS changed. He now knows that C has a 2/3 chance of survival. That is why if he is given the option to switch with C, he should take it.

Logically, this doesn't make sense to me, because prisoner C also knows that at least one of the other prisoners will be executed, so for him to know that B will be executed doesn't change the knowledge of his own fate at all.

If he gets the option to switch though....

Zanguini

09-04-2009, 08:49 AM

you see actually the guard lied because the one who was pardoned is actually B. You see he works in a woman's prison. And he hopes by dropping B down to the lowest level she will be grateful enough to where he will get some bonus from her after the other two die. Now it is interesting because like all prison guards this prison guard is a sadistic asshole and he wants to give both a and c just that little bit of hope so when they are collected for execution it will be just that much sweeter for him.

Logically, this doesn't make sense to me, because prisoner C also knows that at least one of the other prisoners will be executed, so for him to know that B will be executed doesn't change the knowledge of his own fate at all.

If he gets the option to switch though....

At this point prisoner C knows nothing because nobody told him anything. Say prisoner C also asks the same question and gets the same answer "B is sure to die". So from the math C now has a 1/3 chance of being pardoned and A has a 2/3 chance of being pardoned... from his point of view. But now there are twice as many probabilities in the system (unbeknownst to both A and C) and probabilities have to add to one, so everything being of the same weight, all the probabilities get divided by 2.

So A's chance of being pardoned is now:

1/3 x 1/2 (from A's knowledge)

+ 2/3 x 1/2 (from C's knowledge)

= 1/2

C gets the same equation, so their chances are now 1/2 that they each get pardoned, which is what you'd expect. There are only 2 cases out of six where the guard can tell both A and C that "B is sure to die", and in one of them A is pardoned, and in the other C is pardoned.

The math all depends on what knowledge is introduced into the system.

Brita

09-04-2009, 08:52 AM

you see actually the guard lied because the one who was pardoned is actually B. You see he works in a woman's prison. And he hopes by dropping B down to the lowest level she will be grateful enough to where he will get some bonus from her after the other two die. Now it is interesting because like all prison guards this prison guard is a sadistic asshole and he wants to give both a and c just that little bit of hope so when they are collected for execution it will be just that much sweeter for him.

Eww.

Sarevok

09-04-2009, 09:11 AM

I love these kinds of problem, I just lack the practice to solve them. Thanks Orc! :)

I love these kinds of problem, I just lack the practice to solve them. Thanks Orc! :)

I knew there was a reason I got my Master's degree in Mathematics :D

GonzoTheGreat

09-04-2009, 11:38 AM

I'll quote myself:

Suppose, just for the sake of argument, that prisoner C has been pardoned.

Prisoner A asks the question, and is told that B will be executed.

Prisoner B asks, and is told A will die.

Prisoner C asks, and is told that A will die.

Now, would it really be correct for every one of those three to conclude that they have 50% survival chance?

We know that, after they've gotten the (truthful) answer to their question, two out of three prisoners will die. So given those actual odds, how do you defend the one out of two odds you are claiming?Now, what would happen if they were allowed to switch after getting the information?

Prisoner A would switch with C and then survive.

Prisoner B would switch with C and then survive.

Prisoner C would switch with B and die.

So what has changed is not their knowledge of their own fate, but their knowledge of the fate of the others.

Just as in the Monty Hall Problem, really.

I'll quote myself:

Now, what would happen if they were allowed to switch after getting the information?

Prisoner A would switch with C and then survive.

Prisoner B would switch with C and then survive.

Prisoner C would switch with B and die.

So what has changed is not their knowledge of their own fate, but their knowledge of the fate of the others.

Just as in the Monty Hall Problem, really.

Yes, exactly. The logics of the examples are EXACTLY the same. Why people got confused is because there SHOULD be a choice in the prisoner example (like in the monthy hall), but Camel didn't have it in his example, I said so in an earlier post. In the original prisoner problem, Prisoner A gets to make a choice of switiching to Prisoner C. Only when you introduce the choice of switching, prisoner A can change his probability of survival.

vBulletin® v3.8.4, Copyright ©2000-2017, Jelsoft Enterprises Ltd.