One out of three chance: 50%?

[DyslexicHobo]

A little statistics problem for you guys to ponder:
Yourself and two prisoners are on death row. You are told that one out of three of you are going to survive, but you are unsure which one (this means that two of you are going to die).
This means your chances of survival are 33%?

BUT! You know for sure that at least one of the other prisoners is going to die. This leaves the last spot in the electric chair for either you or the other prisoner. This is a one in two chance.

Are your chances of survival 33% or 50%?

 

[CSMR]

Originally posted by: DyslexicHobo
A little statistics problem for you guys to ponder:
Yourself and two prisoners are on death row. You are told that one out of three of you are going to survive, but you are unsure which one (this means that two of you are going to die).
This means your chances of survival are 33%?

No, it could be anything from 0 to 100% given what you have said.

 

[Navid]

Originally posted by: DyslexicHobo
A little statistics problem for you guys to ponder:
Yourself and two prisoners are on death row. You are told that one out of three of you are going to survive, but you are unsure which one (this means that two of you are going to die).
This means your chances of survival are 33%?

BUT! You know for sure that at least one of the other prisoners is going to die. This leaves the last spot in the electric chair for either you or the other prisoner. This is a one in two chance.

Are your chances of survival 33% or 50%?

After one of the prisoners is executed, you have a new situation. Now, there are only two prisoners left. At that point each of them has a 50% chance of survival. But, not before. So, after the first prisoner is executed, the chance of survival for the remaining two goes up from 33% to 50%.

Think of a competition between a number of contestants. At the beginning, the chances of winning for each contestant is 1/n, where "n" is the number of active contestants. As time goes on and contestants are eliminated, n decreases, and the chance goes up for the remaining ones. Of course, the chance for the eliminated contestants changes to 0.
At the very last stage, when there are only two contestants left, each one has a 50% chance.

I have highlighted the erroneous statement in your second paragraph. You could be the first to be executed! That statement becomes true only if and only after one of the other prisoners gets executed first. There is 66% chance of that happening. If that happens, you will have 50% chance. So, at the beginning, you have 66% x 50% = 33% chance of survival.

 

[sao123]

Originally posted by: DyslexicHobo
A little statistics problem for you guys to ponder:
Yourself and two prisoners are on death row. You are told that one out of three of you are going to survive, but you are unsure which one (this means that two of you are going to die).
This means your chances of survival are 33%?

BUT! You know for sure that at least one of the other prisoners is going to die. This leaves the last spot in the electric chair for either you or the other prisoner. This is a one in two chance.

Are your chances of survival 33% or 50%?

You are thinking of the lets make a deal problem... but that doesnt seemingly apply in this case.

 

[Insomnibyte]

There are to many factors here in my view.

Is this for political reasons, aka politcal gain?
Do the guards particuraly hate any of the other two prisioners with you

these factors can exponentialy increase or even decrease your chances of survival

you may have a 100% chance of being the first to go, or you can be 100% guarenteed of survival, so as the other posters, especially the first poster said, it can vary quite heavily from 0% to 100%

 

[DyslexicHobo]

The reason for execution does not matter. You are garunteed that at least two of the three prisoners are going to die. This is COMPLETELY random.

I fully understand that, obviously, there is a 1/3 chance of survival (looking at it from one perspective).

But my problem is, that when the situation is posed in a certain light, it appears that there is a 50% chance of survival. In this certain perspective, what is wrong?


To restate the problem perspective: yourself and two prisoners are on death row. The guard says to the three of you "Two of you are going to die". You ask the prison guard "I know at least ONE of those other two prisoners is going to die. Tell me which one." The guard points to one of them and you can now think to yourself "Yes! Now I have a 50% chance of survival!". However, no new information is gained when he tells you that one of them is going to die. When you ask him the question, you know for a fact that at least one of them is going to die; the fact that the guard tells you gives you no more information. Because your chances of survival after he tells you which of the two are going to die obviously increases your chances of survival to 50%, and the fact that the guard told you which prisoner is going to die introduces no new information, why can't you think that you have a 50% chance of survival from the start?

 

[CSMR]

Why is it obvious that the probability of survival is not, say 47%?
You must be missing out hidden assumptions. Maybe it is implied in the meaning of the capitalized word "COMPLETELY"?

 

[Navid]

Originally posted by: DyslexicHobo
The reason for execution does not matter. You are garunteed that at least two of the three prisoners are going to die. This is COMPLETELY random.

I fully understand that, obviously, there is a 1/3 chance of survival (looking at it from one perspective).

But my problem is, that when the situation is posed in a certain light, it appears that there is a 50% chance of survival. In this certain perspective, what is wrong?


To restate the problem perspective: yourself and two prisoners are on death row. The guard says to the three of you "Two of you are going to die". You ask the prison guard "I know at least ONE of those other two prisoners is going to die. Tell me which one." The guard points to one of them and you can now think to yourself "Yes! Now I have a 50% chance of survival!". However, no new information is gained when he tells you that one of them is going to die. When you ask him the question, you know for a fact that at least one of them is going to die; the fact that the guard tells you gives you no more information. Because your chances of survival after he tells you which of the two are going to die obviously increases your chances of survival to 50%, and the fact that the guard told you which prisoner is going to die introduces no new information, why can't you think that you have a 50% chance of survival from the start?

You are in a contest with 9 others. There is only one prize for the winner. The chance of you winning is 10%.
But, if we think about it the way you do, we say, from the other nine people, we are sure that 8 of them will lose. Then, between the remaining one and you, the chance of winning is 50%. So, your chance of winning is 50%.

I hope you see how invalid this logic is. If you do, you should know that it is identical to your logic.

 

[Special K]

Originally posted by: DyslexicHobo
The reason for execution does not matter. You are garunteed that at least two of the three prisoners are going to die. This is COMPLETELY random.

I fully understand that, obviously, there is a 1/3 chance of survival (looking at it from one perspective).

But my problem is, that when the situation is posed in a certain light, it appears that there is a 50% chance of survival. In this certain perspective, what is wrong?


To restate the problem perspective: yourself and two prisoners are on death row. The guard says to the three of you "Two of you are going to die". You ask the prison guard "I know at least ONE of those other two prisoners is going to die. Tell me which one." The guard points to one of them and you can now think to yourself "Yes! Now I have a 50% chance of survival!". However, no new information is gained when he tells you that one of them is going to die. When you ask him the question, you know for a fact that at least one of them is going to die; the fact that the guard tells you gives you no more information. Because your chances of survival after he tells you which of the two are going to die obviously increases your chances of survival to 50%, and the fact that the guard told you which prisoner is going to die introduces no new information, why can't you think that you have a 50% chance of survival from the start?

Your problem is actually very simple to understand if you approach it rigorously.

Let A = event of you being picked for execution
Let B = event of one of the other prisoners (call it prisoner B) being picked for execution

P(A|B) = P(AB)/P(B)

P(B) = 2/3

P(AB) = 1/nCr(3,2)

so P(A|B) = 1/2

If the guard doesn't tell you that one of the other prisoners will die for sure, then your probability of being chosen can be calculated as follows:

P(you are picked to die) = nCr(2,1)/nCr(3,2) = 2/3

So you can see that the guard's information actually does change your probability of being picked for execution.

 

[Matthias99]

The reason for execution does not matter. You are garunteed that at least two of the three prisoners are going to die. This is COMPLETELY random.

Actually, you said that exactly two of the three would die originally. Saying that at least two of the three will die changes things. You seem to have gone back to exactly 2/3 below, so I will assume this is a typo.

I fully understand that, obviously, there is a 1/3 chance of survival (looking at it from one perspective).

Yes.

But my problem is, that when the situation is posed in a certain light, it appears that there is a 50% chance of survival. In this certain perspective, what is wrong?

The problem is basically that you're looking at a conditional probability, trying to work backwards, and not interpreting it properly. This is actually a harder problem to understand than you might first think.

At the start, there's a 33% chance for each of you to not die (or a 66% that you will die, depending on your perspective). If a guard walks up and shoots one of the other prisoners (or tells you that he is going to be executed), it looks on the surface like you now have a 50% chance of survival! In fact, if they were going to kill one prisoner and then randomly select one of the survivors to be killed, you actually would have a 50% chance of making it. But the decision has already been made, and the new information doesn't change that.

The problem is that the guard could have told you that either of the other two prisoners is going to die. If the three prisoners are A, B, and C, consider the following:

(1/3 chance) A lives, B dies, C dies
(1/3 chance) A dies, B lives, C dies
(1/3 chance) A dies, B dies, C lives

A asks the guard "tell me if B or C is going to die."

Now, let's assume the guard knows the outcome already, and if both B and C are going to die he chooses randomly.

There's a 1/3 chance that A lives and B and C die. In this case there's a 50% chance the guard says B will die, and a 50% chance he answers that C will die.
There's a 1/3 chance that B lives and A and C die. In this case there's a 100% chance the guard says C will die.
There's a 1/3 chance that C lives and A and B die. In this case there's a 100% chance the guard says B will die.

So, overall, there's a (1/3 * .5) + (1/3 * 1) = (1/2) chance he'll say B is going to die, and a (1/3 * .5) + (1/3 * 1) = (1/2) chance he'll say C is going to die. No surprise there.

If he says B will die, the 1/3 chance that B lives is eliminated. This means:

There's a 50% chance that A lives and B and C die. In this case there's a 50% chance the guard would say "B" and a 50% chance the guard would say "C".
There's a 50% chance that C lives and A and B die. In this case there's a 100% chance that the guard would say "B".

What you're trying to do is then to work backwards from him answering "B" and figure out the probability that A lives. If you look at the outcomes, overall there's a 3/4 chance that he'll answer "B" and a 1/4 chance that he'll answer "C":

C lives (50%), guard always says "B" (100%) = answer "B", 50% probability.
A lives (50%), guard chooses to say "B" (50%) = answer "B", 25% probability.
A lives (50%), guard chooses to say "C" (50%) = answer "C", 25% probability.

The guard answered "B", so you must be in one of the first two states. However, it's twice as likely that you're in the state where C lives as the one where A lives. Essentially: out of the 3/4 of the time that the guard answers "B", 2/3 of those times it's because A is going to die. It's half as likely that the guard is answering "B" because A lives as because C lives. Therefore A still has only a 33% chance of survival.

 

[esun]

You can just list the possibilities. You are A, the others are B and C. Two must be executed.

A B
A C
B C

If all of these have equal probability, then in two out of three cases you will die.

If you are told that, for example, B will definitely die, then the problem fundamentally changes. You are now asking the probability you will live given that B is going to die. Well we can eliminate the A C choice from the set of choices, so we're left with A B and B C. So now you have a 50% probability to die.

 

[Navid]

Originally posted by: esun
You can just list the possibilities. You are A, the others are B and C. Two must be executed.

A B
A C
B C

If all of these have equal probability, then in two out of three cases you will die.

If you are told that, for example, B will definitely die, then the problem fundamentally changes. You are now asking the probability you will live given that B is going to die. Well we can eliminate the A C choice from the set of choices, so we're left with A B and B C. So now you have a 50% probability to die.

Nice!

 

[Matthias99]

Originally posted by: Navid

Originally posted by: esun
You can just list the possibilities. You are A, the others are B and C. Two must be executed.

A B
A C
B C

If all of these have equal probability, then in two out of three cases you will die.

If you are told that, for example, B will definitely die, then the problem fundamentally changes. You are now asking the probability you will live given that B is going to die. Well we can eliminate the A C choice from the set of choices, so we're left with A B and B C. So now you have a 50% probability to die.

Nice!

...except it's not actually correct. This is only true if they're going to execute one prisoner, then randomly choose one of the survivors to be executed next. If the decisions are set in stone at the beginning then one prisoner (who's not you) being killed does not change your odds of survival.

You can't take the conditional probability in reverse. See my post above. The logic you have to follow is rather tricky and it's easy to not see the mistake.

 

[eLiu]

For those of you arguing 50%... try thinking about this problem; it may help you understand why 50% is wrong.

Suppose there are 1000 prisoners (including you), and 999 of them are going to die. So you have a 1/1000 chance of living.

The next morning, you realize that 998 prisoners have been killed. What is your chance of living? Hint: it isn't 50%. The idea is that by not killing you, the guards have implicitly provided you with an enormous amount of information.

On the other hand, if the next morning you realize only 1 person has been executed (so 999 people left, of which 998 will die), then your level of information has hardly changed at all.

Edit: if any of you have heard of the "Monty Hall" (Let's Make a Deal) problem, this is exactly the same thing. That is, if you were offered to swap places on the kill-list with the other remaining prisoner, you should totally do it.

That is, dying is analagous to losing. 2 out of 3 "doors" lose here and you are provided information about one of them.

 

[Navid]

It sounds like almost everyone agrees that the chance of survival is 33%.
Except, esun provided the simplest and clearest explanation.

 

[Check]

it's 33%, but the way you worded it makes it seem tricker than it is.
If there were four names in a hat and only one was picked, you know that at least two people (other than you) won't be picked. Knowing that two other people won't be picked, doesn't make your odds off getting picked 1/2. They are still 1/4, you just know that you are going to have two other people for company when you don't get picked.

 

[PhatoseAlpha]

Edit: For the original poster: The reason you're getting confused here is that you're making an invalid assumption. You have a 50% chance of living if A dies, and a 50% chance of dying if B dies, yes.

However, the assumption you're making without stating it as such is that you're assuming there is a 50% chance A dies, and a 50% chance B dies, therefore you have a (.5 * .5) + (.5 * .5) chance of living, or 50%. But A and B do not have a 50% chance of dying. There's 3 events concerning them - A might die, B might die, or both might die. That 2/3 chance that event dead A or event dead B happens, not 50% So, you have a (.66 * .5) + (.66 * .5) chance of dying, or 66%.



It's 66% chance of death. The thing to remember, is that while 50% of the time you survive *if* they fry someone else first, that only happens 66% of the time - the other 33% of the time *you* fry first, and never get to the nice 50% stage.



Round 1: 3 guys, 1 fries. 66% chance of survival: Running tab: 33% dead, 66% not dead yet
Round 2: 33% of time already dead. 50% of remaining 66% I fry. 50% of 66% is another 33% in addition to the 33% baked in round 1. 66% total chance of death.


It's really not analogous to the Monty Haul problem, since Monty is a non-random actor - he knows what's behind the doors, so he'll never ever ever open the door with the nice prize behind it. The jailers on the other hand aren't playing hiding and seek. They'll be happy to show you the bad news.

 

[metalone]

There are two stated problems, not one. Before finding out that one of the prisoners will die, your chances are 1/3 that you will live or 2/3 that you will die. The problem is then restated with additional evidence that one of the other prisoners will die, this changes the calculation to 1/2 or 50% that you will live or die. The information that one prisoner will die is given , so it must be used before calculating the % or survival.

 

[Check]

Originally posted by: metalone
There are two stated problems, not one. Before finding out that one of the prisoners will die, your chances are 1/3 that you will live or 2/3 that you will die. The problem is then restated with additional evidence that one of the other prisoners will die, this changes the calculation to 1/2 or 50% that you will live or die. The information that one prisoner will die is given , so it must be used before calculating the % or survival.

Yeah, after you watch one of the other prisoners fry then it's a 50% chance when it's just the two of you. But you originally have a 33% chance.

 

[imported_Tick]

There is an important distinction:

1) Two prisoners are selected to die, on after the other.

Or

2) One prisoner dies, THEN the other is selected.

The OP needs to say which one he meant.

 

[MikeyLSU]

if only 2 will die of the 3, then your survival % is 33% which bumps up to 50% when you find out that one of the other 2 will be executed.

It is a rather simple problem...many people are trying to make assumptions. To the poster with the 1/1000 problem, yes if you wake up and there are only 2 of you left and 1 will live, you have a 50% chance of dying. Again, if you are trying to argue that they have already picked who lives and dies, than your % from the beginning was either 100% or 0%, not 1/1000.

If you don't know the specifics, you must assume random probability.

 

[Matthias99]

Originally posted by: MikeyLSU
if only 2 will die of the 3, then your survival % is 33% which bumps up to 50% when you find out that one of the other 2 will be executed.

It is a rather simple problem...many people are trying to make assumptions. To the poster with the 1/1000 problem, yes if you wake up and there are only 2 of you left and 1 will live, you have a 50% chance of dying. Again, if you are trying to argue that they have already picked who lives and dies, than your % from the beginning was either 100% or 0%, not 1/1000.

If you don't know the specifics, you must assume random probability.

*sigh*... which means that your chance of dying doesn't change if they already made the decision. You only have a 50% chance of survival at that point if they are killing one person, then randomly selecting a survivor to die and repeating that process over and over. This is clearly not the situation described in the OP.

It's similar to the Monty Hall problem, except that here you're not offered the chance to switch. The odds that you are going to live (or originally picked the winning door, as stated in the Monty Hall problem) didn't actually change. My long post above goes through the logic that shows why you still only have a 33% chance of living based on the new information. The 1000-prisoner case works the same way -- you still only have a 1/1000 chance of survival, because it's much much more likely that you were doomed to begin with than not.

http://en.wikipedia.org/wiki/Monty_hall_problem
http://en.wikipedia.org/wiki/Three_Prisoners_Problem

 

[MikeyLSU]

Originally posted by: Matthias99

Originally posted by: MikeyLSU
if only 2 will die of the 3, then your survival % is 33% which bumps up to 50% when you find out that one of the other 2 will be executed.

It is a rather simple problem...many people are trying to make assumptions. To the poster with the 1/1000 problem, yes if you wake up and there are only 2 of you left and 1 will live, you have a 50% chance of dying. Again, if you are trying to argue that they have already picked who lives and dies, than your % from the beginning was either 100% or 0%, not 1/1000.

If you don't know the specifics, you must assume random probability.

*sigh*... which means that your chance of dying doesn't change if they already made the decision. You only have a 50% chance of survival at that point if they are killing one person, then randomly selecting a survivor to die and repeating that process over and over. This is clearly not the situation described in the OP.

It's similar to the Monty Hall problem, except that here you're not offered the chance to switch. The odds that you are going to live (or originally picked the winning door, as stated in the Monty Hall problem) didn't actually change. My long post above goes through the logic that shows why you still only have a 33% chance of living based on the new information. The 1000-prisoner case works the same way -- you still only have a 1/1000 chance of survival, because it's much much more likely that you were doomed to begin with than not.

reading over your last post again I see where we differ. I read the OP as nothing has been decided yet. I read it as though 2 will die, they choose by random, once they pick the first person to die on day 1, you now have 2 people left which they will AGAIN choose at random.

Now, overall, yes your chance of living will still be 33%, by using the 66% on the first one, and 50% on the second draw(.66*.5=.33)...but after the first draw if complete, you have a 50% chance of living at that point forward(again, I am assuming a second random draw is taken).

 

[Matthias99]

Originally posted by: MikeyLSU

Originally posted by: Matthias99

Originally posted by: MikeyLSU
if only 2 will die of the 3, then your survival % is 33% which bumps up to 50% when you find out that one of the other 2 will be executed.

It is a rather simple problem...many people are trying to make assumptions. To the poster with the 1/1000 problem, yes if you wake up and there are only 2 of you left and 1 will live, you have a 50% chance of dying. Again, if you are trying to argue that they have already picked who lives and dies, than your % from the beginning was either 100% or 0%, not 1/1000.

If you don't know the specifics, you must assume random probability.

*sigh*... which means that your chance of dying doesn't change if they already made the decision. You only have a 50% chance of survival at that point if they are killing one person, then randomly selecting a survivor to die and repeating that process over and over. This is clearly not the situation described in the OP.

It's similar to the Monty Hall problem, except that here you're not offered the chance to switch. The odds that you are going to live (or originally picked the winning door, as stated in the Monty Hall problem) didn't actually change. My long post above goes through the logic that shows why you still only have a 33% chance of living based on the new information. The 1000-prisoner case works the same way -- you still only have a 1/1000 chance of survival, because it's much much more likely that you were doomed to begin with than not.

reading over your last post again I see where we differ. I read the OP as nothing has been decided yet. I read it as though 2 will die, they choose by random, once they pick the first person to die on day 1, you now have 2 people left which they will AGAIN choose at random.

Now, overall, yes your chance of living will still be 33%, by using the 66% on the first one, and 50% on the second draw(.66*.5=.33)...but after the first draw if complete, you have a 50% chance of living at that point forward(again, I am assuming a second random draw is taken).

The OP did restate it a little more clearly later:

To restate the problem perspective: yourself and two prisoners are on death row. The guard says to the three of you "Two of you are going to die". You ask the prison guard "I know at least ONE of those other two prisoners is going to die. Tell me which one." The guard points to one of them and you can now think to yourself "Yes! Now I have a 50% chance of survival!". However, no new information is gained when he tells you that one of them is going to die. When you ask him the question, you know for a fact that at least one of them is going to die; the fact that the guard tells you gives you no more information. Because your chances of survival after he tells you which of the two are going to die obviously increases your chances of survival to 50%, and the fact that the guard told you which prisoner is going to die introduces no new information, why can't you think that you have a 50% chance of survival from the start?

This is clearly this Three Prisoners Problem: http://en.wikipedia.org/wiki/Three_Prisoners_Problem

It is a variant of the Monty Hall Problem/Paradox: http://en.wikipedia.org/wiki/Monty_hall_problem

If they're making a new random selection after killing each prisoner your chance of survival does rise -- but this is not usually a source of confusion.

 

[DrPizza]

Actually, all three are going to die. Two in the near future, and the other at some point further in the future. Humans aren't immortal, even if their execution is put on hold.
Thus, I say there's a 100% chance he's going to die.