One more attempt at basic logic here.
You start with three possibilities: A lives, B lives, or C lives.
They are all equally probable.
You eliminate 1 possiblity, namely: B lives
That leaves you with two equally probable possibilities: A lives and C lives. Two and only two equally probable possibilities are always 1/2.
Therefore, the prisoner's survival probability is 1/2 after knowing that B does not live.
BLR
Originally posted by BLReidNice try BLR, but no dice!!!
One more attempt at basic logic here.
You start with three possibilities: A lives, B lives, or C lives.
They are all equally probable.
You eliminate 1 possiblity, namely: B lives
That leaves you with two equally probable possibilities: A ...[text shortened]... rvival probability is 1/2 after knowing that B does not live.
BLR
The guard could speak to each of them and tell them which other prisoner dies, does that mean they EACH have a 50% chance of surviving?
If the prisoners to die have ALREADY been decided, A has a 1/3 chance of surviving, C has a 2/3 chance (mathematical proof by iamatiger on Page 1 of thread.
If they shoot B, THEN decide who to kill... A and C both have a 50% chance of surviving.
Dave
Originally posted by PalynkaNever exclude A as a possible answer. You begin with 3 possibilities. You gain knowledge that eliminates one possibility. That leaves 2...no more or less.
No, because you eliminate probability of B living while excluding A as a possible answer.
Edit: There are 1.000.000 prisioners. A says that he knows that at least 999.998 of the other prisioners will die and asks the guard to name them. Do his chances leap from 1/1.000.000 to 1/2?
True, the answer couldn't be A, but that doesn't exclude A from dying.
So it remains that 2 out of 3 will die. A knows that 1 will not be him (B). Therefore, the second to die must be either himself or C. Ther is no other option. His survival odds are 50/50...(caveat) from his perspective, which is how the original problem was posed. It doesn't really matter that the deaths are pre-ordained. It only matters what his calculation of his survival odds are based on what he knows.
Originally posted by BLReidNo, it doesn't. That's why his probabilities are not 0 but 1/3.
True, the answer couldn't be A, but that doesn't exclude A from dying.
Edit to answer your edit: There are two options when you buy a lottery ticket, you win or you lose. Do you have a 50% chance of winning?
Originally posted by THUDandBLUNDERHis chances haven't increased at all. His chances were 50/50 to begin with, since 2 out of the three were going to die anyway.
There are 3 prisoners in a prison. Let’s call them A, B, and C. Tomorrow, 2 of them will be executed, but the prisoners don’t know which of them have been chosen. Prisoner A reasons that his chances of survival are 1/3 (a third). He ...[text shortened]... nces of survival have increased from 1/3 to 1/2. Is he right?
.
ex: He knows going into it that either B or C will definitely die. Therefore, it is simply left to either him or the other survivor of the first round as to who gets the second bullet.
Originally posted by iamatiger
I agree - this is the way I would put it.
Before A asks the guard there are 3 possibilities, each with the probability stated below:
A spared : 1/3
B spared : 1/3
C spared : 1/3
After A asks the guard, but before the guard answers there are the following possibilities of who is spared and what the guard might say.
A spared & Guard names B : 1/6
A spared & Guard names C : 1/6
B spared & Guard names C : 1/3
C spared & Guard names B : 1/3
After the guard names B we know it was either the first or last possibility that happened, but we have to normalise those two remaining probabilities so they add up to 1
A spared & Guard names B: (1/6)/(1/6 + 1/3) = 1/3
C spared & Guard names B: (1/3)/(1/6 + 1/3) = 2/3
Originally posted by Polynikes
It seems to me that his chances do not increase from 1/3 to 1/2.
Reason being, there is a 100% chance that B or C will be named. His question,
“tell me the name of one person (B or C) who is to be executed.”
tells him absolutely nothing at all. Of course, at least one of them is to die. The guard's answer only tells us what we already know.
Lets rephrase the discussion.
Prisoner: Will I die tomorrow?
Guard: I cant tell you that.
Prisoner: Ok, fair enough. Does one of them die tomorrow?
Guard: Yes, of course fool. If two of you are to die, it's obvious that at least one of them will be chosen.
Prisoner: What is the name of one prisoner who dies, not including me?
Guard: No matter which name I say, you still dont know who dies with him, and no math or reasoning can possibly increase your odds of survival. Now lights out.