One out of three chance: 50%?

[Nathelion]

My take on this... (I'm assuming that it's already decided from the beginning who's going to die)

There are three prisoners, A, B, and C.
Now there are 3!=6 orders in which they can call out the prisoners, for convenience let's assume that the prisoner that is spared is called out last. Now these are the possibilities:
A,B,C
A,C,B
B,A,C
C,A,B
B,C,A
C,B,A

All these possibilities are equally probable.
Now the guard comes in and says "two of you will die". You ask which one of you fellow prisoners will die, to which he points out (and this is arbitrary) either B or C. Now notice that there was never any chance that he would point at you!
Case 1 (B dies, 50% probable):
Now all possibilities where B survives are out. Remaining 4 possibilities:
A,B,C
B,A,C
B,C,A
C,B,A

Case 2 (C dies, 50% probable):
Now all possibilities where C survives are out. Remaining 4 possibilities:
A,C,B
C,A,B
B,C,A
C,B,A

Now this is important. We all know that all cases are equally likely, right? So it would seem like there's a 50% chance that you'll die in both cases, and both cases are 50% likely, which would give .5^2+.5^5=50%. WRONG! Look again at the two cases. Notice something? Every order in which YOU survive is possible in both cases! Now the total probability of all orders are the same. The probability of the two cases being 50/50, the orders in which you survive are only half as likely to happen under any given case. So a better representation would be:

Case 1 (B dies, 50% probable):
A,B,C,33%
B,A,C,33%
B,C,A,16.667%
C,B,A,16.667%

Case 2 (C dies, 50% probable):
A,C,B,33%
C,A,B,33%
B,C,A,16.667%
C,B,A,16.667%

So as you can see, the probability that you die is 2/3 after all.

 

[Nathelion]

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.

This is actually not true. If the guard randomly walks up and kills one of you and then walks off, you have a 50% chance of survival - whether it's decided in advance or not. While who dies and who lives must be decided in advance for the situation we are discussing to apply, the key point is that when you ask the guard which of the OTHER prisoners is going to get killed, his answer implicitly excludes you. So he gives you information by telling you which other prisoner is going to die - but implicitly he also tells you that of the two remaining candidate survivors, you are the one more likely to die. If you DIDN'T ask the guard which of the other two was going to die, and he simply announced it without being prompted (or shot the prisoner, whichever way you prefer), then you would actually have a 50% chance of surviving - regardless of whether they rolled dice in the barracks yesterday to decide who's going to die or whether they're rolling the dice as we speak.

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.

Actually your chance of survival is 50% at that point. Now if they announced with a megafone the night before that you, specifically, would not be killed during the night but that you may very well be killed in the morning, then it is true that you would still have a chance of survival of 1/1000 if 998 people are killed during the night. But if they don't tell you that you have "special immunity" through the night, then it is just as likely that they would have killed you as anyone else, in which case you would not have woken up in the morning. If they don't give you "special immunity", then the fact that you wake up at all tells you that you have a larger chance of not dying.

 

[darkhorror]

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 have a 33% chance. it says you know for sure that ATLEAST one of the other prisoners is going to die. That is just a statement trying to confuse you, as you know this by knowing that one of the three of you are going to survive. which means atleast one of the other prisoners is going to die. And if your survive both will die, if not then only one will die.

There would be a way to word it that they would give you a 50% chance this would be if they say before you are exicuted that first they choose one of the other prisoners to exicute then between you and the remaining guy they choose one. Then in that first choice you arn't a part of and the second part it's only 50% chance.

You can interperent the last part of this to believe what you want about the situation. it seems faulty, it doesn't say how things are being chosen, then it says "This leaves the last spot for you or the other prisoner", it seems that they take the part before it which doesn't say anything special and make it to mean something which it doesn't.

 

[Matthias99]

Originally posted by: Nathelion

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.

This is actually not true. If the guard randomly walks up and kills one of you and then walks off, you have a 50% chance of survival - whether it's decided in advance or not. While who dies and who lives must be decided in advance for the situation we are discussing to apply, the key point is that when you ask the guard which of the OTHER prisoners is going to get killed, his answer implicitly excludes you. So he gives you information by telling you which other prisoner is going to die - but implicitly he also tells you that of the two remaining candidate survivors, you are the one more likely to die. If you DIDN'T ask the guard which of the other two was going to die, and he simply announced it without being prompted (or shot the prisoner, whichever way you prefer), then you would actually have a 50% chance of surviving - regardless of whether they rolled dice in the barracks yesterday to decide who's going to die or whether they're rolling the dice as we speak.

The situations are the same (as long as you know that exactly two of the three of you will be executed.) Him killing one of the other prisoners also "implicitly excludes you". It's the 'Monty Hall Problem' -- they are opening a 'losing door' that is not the one you picked. That doesn't increase the chance that you picked the winning door originally.

Given that you actually gave a correct analysis of the problem in your previous post, I'm not sure why you are seemingly not understanding this.

 

[BrownTown]

You know, normally I consider that people on AT are generally educated and bright and such, but this thread (along with .999 = 1 and treadmill plane) just really make me wonder. This just seems so absurdly simply I cannot understand how anyone can miss it I mean honestly. It says two of the three people are going to die, so that mean that one lives and there is a 1/3 chance its gonna be you. I think we can all agree on that. The next sentence is just the classic old misdirection that you see so often, it provides you with NO information whatsoever. It says at least one of the two other people will die, but guess what WE ALREADY KNEW THAT, if two people die you cant die twice, one of the other two prisoners HAS to die irregardless of the second sentence. The answer is 1/3, there are no other alternative interpretations, if you say 1/3 you are right, if you say 1/2 you are WRONG. But don't worry, thats OK, we have all screwed up easy a$$ problems before, and yeah, the Monty Hall problem was the first thing to pop into my mind when is started reading (this is what the OP wants btw, for you to remember the old problem and answer like this is the same when it isn't).

 

[Nathelion]

quote:
Originally posted by: Nathelion

quote:
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.



This is actually not true. If the guard randomly walks up and kills one of you and then walks off, you have a 50% chance of survival - whether it's decided in advance or not. While who dies and who lives must be decided in advance for the situation we are discussing to apply, the key point is that when you ask the guard which of the OTHER prisoners is going to get killed, his answer implicitly excludes you. So he gives you information by telling you which other prisoner is going to die - but implicitly he also tells you that of the two remaining candidate survivors, you are the one more likely to die. If you DIDN'T ask the guard which of the other two was going to die, and he simply announced it without being prompted (or shot the prisoner, whichever way you prefer), then you would actually have a 50% chance of surviving - regardless of whether they rolled dice in the barracks yesterday to decide who's going to die or whether they're rolling the dice as we speak.



The situations are the same (as long as you know that exactly two of the three of you will be executed.) Him killing one of the other prisoners also "implicitly excludes you". It's the 'Monty Hall Problem' -- they are opening a 'losing door' that is not the one you picked. That doesn't increase the chance that you picked the winning door originally.

Given that you actually gave a correct analysis of the problem in your previous post, I'm not sure why you are seemingly not understanding this.

Well what I'm saying is that it's not the monte hall problem unless you know that you are exempt in the first shooting.

In the monte hall problem you have two losing doors and one winning door. You pick one door, and then you know that Monte Hall will point out to you one of the other two doors.

Let's make the situation with the guard shooting one prisoner more like the monte hall problem so that I can get my point across.
Let's say that there is a wall with three doors behind which you can hide. Now there are three prisoners who have to pick a door to hide behind, and there is a guard on the other side. Now the monte hall problem and the OPs problem says that if you hide behind a door, the guard is going to open a door - NOT your door though - and shoot the prisoner behind that door. Do you now want to switch doors? Yes you do, as has been shown a number of times in this thread.
But my point is that if you are just sitting on a bench and a guard comes up and shoots one of you, then no matter how far ahead he planned it, unless he first announces that he is not going to shoot you it is not analogous to the monte hall problem.
Let's say the guard knows what two people he is going to kill and who he is going to spare. Now he comes up and shoots one of the people he is going to kill.
Now my point is that this is analogous to the guard coming up to the wall with the doors, opening a door, and killing the prisoner behind it - but there's a 1/3 chance that the door he opens will be YOUR door and he'll kill you. Unless you somehow know that he is not going to shoot you first ahead of time, he IS going to shoot you first 1/3 of the time and therefore you have a 50% chance of survival once you know that someone else got shot first.

Do you understand what I'm saying?

Another way to put it is that if the guard comes up to prisoners A B C and shoots C, then according to your argument, both A and B can think: "oh dear I still only have a 1/3 chance of making it". But one of them is going to make it out, so there's a 100% chance that someone makes it. But if they both only have a 1/3 chance of making it, then that's as total chance of 2/3 that someone will make it. By contradiction, they can't both have a 1/3 chance of making it out, and since their situation is equal in all ways it has to be a 50% chance for each of them.

 

[Matthias99]

Now my point is that this is analogous to the guard coming up to the wall with the doors, opening a door, and killing the prisoner behind it - but there's a 1/3 chance that the door he opens will be YOUR door and he'll kill you. Unless you somehow know that he is not going to shoot you first ahead of time, he IS going to shoot you first 1/3 of the time and therefore you have a 50% chance of survival once you know that someone else got shot first.

Do you understand what I'm saying?

Sorry, you're right and I wasn't understanding the change in the scenario properly. The guard having the potential to reveal information about all the prisoners does change the situation; at that point you can say that, going forward, you have a 50% chance to survive (although your true 'overall' probability of survival is still 1/3).

This distinction is what tends to trip people up, and creates the 'paradox' in the Monty Hall problem. If you're only getting information about the other prisoners (or the doors you didn't pick), it doesn't necessarily give you any new insight into your own situation.

 

[skreet]

What if the guard shoots you first, are the other men doing math on death row too? :-D

 

[highwire]

Of course the chance of survival for you, corespondent (C) is 1/3. No question.

But what rhetorical element in this second statement misleads one to think it could be 1/2?

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.

It starts with the phrase "at least one" included in the sentence. If you break that down, there are actually TWO instances of a single "other", not ONE, (A) or (B), which is taken as a single possibility in this second statement to conclude that the chance is 1/2. The only good result for (C) is the third possibility, (A+B).

That first sentence was true, but did not inform, only mis-directing the reader to accept the false conclusion that follows.

So, now we have to correct the rest of the statement to make it true:
"This leaves the last spot on the electric chair to the second other prisoner, or you with prisoner (A), or you with prisoner (B). This is a one in three chance."

Makes it a bit more difficult to trust language, doesn't it?

 

[BrownTown]

Originally posted by: highwire
Makes it a bit more difficult to trust language, doesn't it?

Only if you have a fourth grade reading comprehension level like half the people in this thread...

 

[highwire]

Originally posted by: BrownTown

Originally posted by: highwire
Makes it a bit more difficult to trust language, doesn't it?

Only if you have a fourth grade reading comprehension level like half the people in this thread...

lmao

 

[pheonixstar]

Here is my take:
If you are in a cell and know that 2 out of 3 of you will die and that is your only information then you have a 1/3 chance of living(provided it is random)

If a guard walks by and says, "I just wanted to let you know, at least one of the other prisoners is going to die," then you can thank him for pointing out the obvious and your odds are still 1/3/

If the guard walks by and says, "Well we are going to kill only 2 of you. The warden really hates one of the other guys and so he determined he is to die for sure, and the one to die with him will be random." In this case you have new information, you knew each of them had a 66% chance of dying, but since one of them now has a 100% chance, your odds improve to 50%. The guard has given you new information so you can reevaluate the odds.

If he walks by and says, well we selected the two that were to die and then randomly determined an order. For your information we have only killed one person so far and it was not you. Now he actually gave you some new information and you can thank him knowing that your odds are now 1/2.

It all depends on how "one of the others will die" is revealed to you. If it is something you did not already know, then the odds will change, otherwise 1/3.

 

[Special K]

Originally posted by: 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.

Can someone please tell me why my explanation, which I posted back towards the beginning of the thread, is wrong?

 

[BrownTown]

Yes, that is quite simple, you have P(B) = 2/3 when in fact P(B) = 1. This results from your poor reading comprehension not your math skills.

The first sentence of the second paragraph that should be labeled P(B) is "You know for sure that at least one of the other prisoners is going to die.". This statement provides no new information and is accurately represented as a probably of 1 due to the fact that it is already implied in the first paragraph. I am not sure where your inability to comprehend this comes from but I would suspect that a well educated person should understand the reasoning of this after completing elementary school. It might take a statistics course in high school or college to understand the rigorous reasoning behind this and the equation for conditional probability already stated. If I were to guess I would say that you are trying much to hard to complicate this problem when it is in fact very simple.

I know that this is something that many can fall into, so its probably just best to admit you are wrong. For example a test question I once had on a programming test was what is the next letter in the sequence "o-t-t-f-f-s", and of the 8 people in my class none of us got it right because we all tried to come up with complex algorithms to solve it instead of seeing the obvious answer.

 

[Special K]

Originally posted by: BrownTown
Yes, that is quite simple, you have P(B) = 2/3 when in fact P(B) = 1. This results from your poor reading comprehension not your math skills.

The first sentence of the second paragraph that should be labeled P(B) is "You know for sure that at least one of the other prisoners is going to die.". This statement provides no new information and is accurately represented as a probably of 1 due to the fact that it is already implied in the first paragraph. I am not sure where your inability to comprehend this comes from but I would suspect that a well educated person should understand the reasoning of this after completing elementary school. It might take a statistics course in high school or college to understand the rigorous reasoning behind this and the equation for conditional probability already stated. If I were to guess I would say that you are trying much to hard to complicate this problem when it is in fact very simple.

I know that this is something that many can fall into, so its probably just best to admit you are wrong. For example a test question I once had on a programming test was what is the next letter in the sequence "o-t-t-f-f-s", and of the 8 people in my class none of us got it right because we all tried to come up with complex algorithms to solve it instead of seeing the obvious answer.

I went through the problem again and saw this quote:

"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."

It seems I misread the question and interpreted it as scenario 2 instead of 1.

Nevertheless, I didn't need your smartass, elitist comments in your reply. FWIW I did take a probability class in college and received an A, so I really don't need to take your crap. It's not like I came in here criticizing everyone else's answers and proclaiming mine to be absolutely correct. I simply offered my interpretation of the problem, which turned out to be incorrect.

 

[BrownTown]

Originally posted by: Special KFWIW I did take a probability class in college and received an A, so I really don't need to take your crap. It's not like I came in here criticizing everyone else's answers and proclaiming mine to be absolutely correct. I simply offered my interpretation of the problem, which turned out to be incorrect.

Thats actually my point, someone who took a college statistics class is probably MORE likely to miss this question because they misinterpret it as a trick question instead of an obvious one. Like the problem I provided was actually done in a study and the results from fourth graders were compared to those of Duke PhD students and the fourth graders were more likely to get it right. All that extra knowledge is just getting in the way in this case.

 

[Nathelion]

lol this thread is STILL going? That's hilarious!

 

[mindless1]

You do not know that the last spot is either you or the other prisioner, the first spot could easily be you.

Someone probably mentioned this already but since the opening post wasn't edited to reflect this, it bears repeating.

 

[leekirlew]

LOL, I like this,

Your chances are 33% until the first is executed.

Then they are 50% if you were lucky enough to survive,

Depends on perspective and how you read the question, but the above is correct mathimatically

 

[TallBill]

The inmate asked the guard for a name for a future execution. Its not whack one, and then pick another. Both are pre-chosen.


So there are 4 possible results. You are A, other guys are B and C

A+B are to die, so guard will say B - 33%
A+C are to die, so guard will say C - 33%

B+C are to die, so guard will say B - 16.6%
B+C are to die, so guard will say C - 16.6%

Event 3 + 4 each have equal probability if B+C are picked to die.

So there is a 50% chance that he'll say B, and a 50% chance that he'll say C, but still a 66% chance that you will die.


I WIN!!!

 

[mh47g]

It makes much sense to me, and I suck at math

3 people.

1 is dead already.

2 of you left.

1 is going to die.

50% it's going to be you, suckah.

 

[Nathelion]

Except that's not the way it works, as has been explained 15+ times in this thread already. In the OPs problem (at least in the revised version), the chance of you surviving is still 1/3.

 

[imported_Finny]

I think this problem is just being silly with it's wording.

Alright, so we know that 2/3 guys are gonna die from the start, right? One today, one tomorrow. The guard tells you that one of the other guys is gonna die, right? Great but... You already knew that! This is the sole source of confusion, as people are seeing this and saying "Oh, new situation" when, in fact, it presents zero news- Up until one of the prisoners has been executed, you've got a 1/3 chance of survival.

IF the guard had said that one of the other prisoners would be killed FIRST, THEN you could say that your chance of survival was 50%. However, this is never stated, so we have to conclude that there is a chance that you'll die in the first pick (Where there's a 66% chance of survival), then survive the 2nd execution (back at 50%), and 2/3 x 1/2 = 1/3 = 33%.

To break it down:

Survive the first day: 2/3 or 66% (He either picks one of the other guys, or you)
Survive the 2nd day: 1/2 or 50% (He picks the one guy or you)
Survive BOTH days: 1/3 or 33% (He picked the other guys each day, leaving you alive)

The 50% only comes into play if you survive the first execution.

 

[Vee]

The OP stated that he knows that the chance of survival is 1/3.
Neither did he ask for correct approaches to calculate the odds of survival. Implication being that he already knows how to do it correctly.
What he did ask for was what was wrong with the reasoning that seemed to imply there is a 50% chance of survival.

I cannot see that anyone has really answered that question in an explicit and straightforward manner, so here goes:

The false reasoning is founded on the fact that we know at least one of the other two prisoners will die.
The problem here is the compound "one of the other". This is a grouping statement.
If we look at this in terms of dices, the three possible outcomes of the selection are:
A and B dies
A and C dies
B and C dies
So this would be the proper faces on a 3-faced dice.
From the observation that in every case either B or C dies or both, the false 50% reasoning correctly concludes that we are guaranteed that one of the others dies. From this 100% he derives two groups of outcomes:
one of the other dies and A dies
one of the other dies and the last other one also dies ( = A survives)
Then he assigns 50% chance to each of these outcomes. Just right out of the blue! - 50%. Can he really do that? - No!

Look what happens here. He groups together
A and B dies and A and C dies as a compound statement: A and one of the other dies
then keeps the original B and C dies as the equivalent both of the others dies
and assumes that we now have a 2-faced dice and thus a 50% survival chance for A.

A and one of the other dies
both of the others dies

Yes, we are guaranteed in any case that one of the others dies. Does that mean we have two cases? Each with a 50% chance? No we don't. We merely have a generalized ("one of the other") statement/label that happen to be true for two different possible cases. Above we have 2 groups of cases. Not 2 individual equivalent cases.
Actually we still have the same 3 cases and the same 3-faced dice with these three sides:

A and one of the other dies (this new sticker now hides original A and B dies label)
A and one of the other dies (this new sticker now hides original A and C dies label)
both of the others dies -> A survives (this new sticker now hides original B and C dies label)

It's still a three faced dice with only 1/3 chance for survival of A. That we can come up with an outcome statement that is true for several cases, and thus group them together, doesn't change the odds. Obviously we can group cases anyway we please, but pasting on group labels on multiple faces of the dice can never reduce the number of faces. It's a very common pitfall and this is in no way special or any kind of paradox whatsoever. The way you group together the various outcomes of a random event never affects the outcome. I'm sure the OP is well aware of this and the reason he is confused this time is merely a fluke due to the wording. Beware of grouping things when trying to analyze probability problems. You can group things together but the groups then don't have equal chances. In this case:

2/3 chance one of the other dies and A dies
1/3 chance one of the other dies and the last other one also dies ( = A survives)

The two groups do not divide the 1/1 = 3/3 = 100% probability of "one of the other dies" equally, 50/50. And our guarantee that one of the others dies is thus totally irrelevant and chance of survival is still only 1/3.