Originally posted by BLReidI'm sorry for doubting, then. Please explain me what you mean by this sentence as I can't get it.
Palynka, the problem with factoring in the guard's answer as being specific to C's chances is that the guard is not allowed to consider A's chances. Therefore, anything he says relative to C is invalidated. All we learn from the guard is that B dies. Nothing else can be garnered from his statement.
Before any questions A's probability of surviving is 1/3 as is C's and B's.
Up to here we agree, right?
Then the guard tells all prisioners that IF A is to be executed he'll be the last to die.
The next day, B is executed first. Have A's chances improved to 50%?
No, he knew that B or C would die first and his chances remain at 1/3.
But C didn't know if he would survive and he had a 50% chance of dying in the first execution. For him to have had a 1/3 overall probability of surviving then he "needs" to have 2/3 in the second execution (1/2*2/3=1/3).
Originally posted by PalynkaAhh, you've hit directly on the misunderstanding...if C's chances of being executed in the first round are 50% (as he is one of 2) then they must again be 50% in the second round.
I'm sorry for doubting, then. Please explain me what you mean by this sentence as I can't get it.
Before any questions A's probability of surviving is 1/3 as is C's and B's.
Up to here we agree, right?
Then the guard tells all prisioners that IF A is to be executed he'll be the last to die.
The next day, B is executed first. Have A's chances ...[text shortened]... robability of surviving then he "needs" to have 2/3 in the second execution (1/2*2/3=1/3).
When you break down the original probabilities they look like this (forgive the redundancy, please)
A lives B named 1/6
A lives C named 1/6
B lives A named 1/6
B lives C named 1/6
C lives A named 1/6
C lives B named 1/6
next, you eliminate the "B lives" options, leaving the following:
A lives B named 1/4
A lives C named 1/4
C lives A named 1/4
C lives B named 1/4
We can also eliminate the "A" or "C" named possibilities, leaving
A lives B named 1/2
C lives B named 1/2
The point is, when you consider that the guard names B, you MUST re-introduce the "A named" possibilities to factor the remaining odds. The guard was not permitted to say that A was going to die, that doesn't eliminate the actual possibility that he wil die, so thescenarios in which he does die must be re-introduced.
The guard's statement clearly eliminates the B lives possibilities. We agree on that (this would be truly difficult if we didn't agree on that 😉). But that leaves us with only 2 possibilities remaining...namely A lives or C lives, both with B being named to die. From our original probability chart, we know that each of these is equally as likely to happen, and nothing has occured that would alter that fact.
Edit: The core of the problem is that A cannot be named. That is why this problem was created, to confuse people that forget that.You are wrong again and going in circles.
In your view, C would have a 50% chance of living in the first choice and 50% in the second, giving him a 25% initial chance of survival instead of 1/3.
You cannot reintroduce the possibility of A being named because you would be changing the problem completely. If someone else asked the question (and couldn't be named) then that someone else's probabilities would be 1/3 as well.
Anyway, I quit. I think you are trying too hard to prove your point and that is preventing you from seeing your mistake.
BLR
What's your answer to the Monty Hall problem?
######
Quiz show, you're in the final. There are three safes. One contains a prize. Chance of winning = 1 in 3. You choose box A. Monty opens one of the other safes to prove to you it's empty.
Do you
(a) Keep safe A
(b) Choose the remaining, unlocked safe
(c) Makes no difference as there are 2 boxes left, therefore 50/50 chance of winning
######
Given your reasoning, I would guess that you'd go for (c), makes no difference.
If I could prove to you that switching to the other safe would double your chances of winning, would you be prepared to believe that prisoner A has a 1/3 chance of surviving, prisoner C has 2/3?
If not, then I'm not contributing to this thread further, I'd have more chance of convincing you the moon is made of cheese!!!!!!!! (I know, where's the proof that it isn't....!)
🙂
Originally posted by PawnCurry(a) Keep safe A
BLR
What's your answer to the Monty Hall problem?
######
Quiz show, you're in the final. There are three safes. One contains a prize. Chance of winning = 1 in 3. You choose box A. Monty opens one of the other safes to prove to you it's empty.
Do you
(a) Keep safe A
(b) Choose the remaining, unlocked safe
(c) Makes no difference as ther ...[text shortened]... ng you the moon is made of cheese!!!!!!!! (I know, where's the proof that it isn't....!)
🙂
(b) Choose the remaining, unlocked safe
(c) Makes no difference as there are 2 boxes left, therefore 50/50 chance of winning
For option (b), I did, of course, mean LOCKed safe. Not that it matters within the spiri of the problem....
I await your response!!!!!!!!!!!!!!!!!!!!
First of all, to Palynka, I don't believe that I am mistaken, or I wouldn't maintain my point so tenuously. If I am in fact mistaken, I humbly await an explanaition as to why. All I have attempted to do if I have gone in circles is to try to restate the reasoning based on the objections that are thrown at me. Obviously, any reasoning that I have stated before can't be sinking in, or there wouldn't be an objection.
To PawnCurry, you are offering me exactly what I am asking for, a mathematical proof that the odds have been changed for safe A by showing one of the others to be empty. I don't believe it, and would certainly look at any such proof with scrutiny, but if it is true and accurate it should pass such scrutiny, shouldn't it?
And what have I posted thusfar to make you think that I would believe the moon to be made of cheese? As far as I can remember, I have not given you any cause to insult me.
Somehow seeing it helped it to make sense to me. If prisoner A survives 1/3 of the time, and we know prisoner B doesn't survive, then Prisoner C must survive 2/3 of the time. It's rather simple in this light. I don't know why I had to see it to understand it though...but it seems rather obvious now.
Hi, I haven't done probability theory in 4 years and I'm not going to state that I know the answer, so let me just give another problem statement since I'd like to know what everybody thinks of this.
There's 3 prisoners and 2 will die. The guard tells the 3 prisoners - two of you guys are going to and one will survive, and let me tell you now, prisoner A will die for certain!
What is the probability that B survive? I'd say 50%.
What is the probability that C survive? I'd say 50%.
Let me restate:
There are 3 prisoners and 2 will die. Through the grapevine it is determined that A will die for certain. What is the probablity that B or C will survive?
How does any of these problems differ from the original question? Would it change the probability of the outcome depending on which of the prisoners actually know that A will die for sure?
I am sure people will give mathematical arguments how this is the case, and the argument that the guard's answer only told C what he already knew is a very very good one... but as an engineer I must say I think the probabilty is down to 50%.
Given the information available, I'd say they both have a 50% chance.
The difference between this and the original poser, in the original the guard couldn't tell A if he was going to die, only which of the others would die. B and C weren't given any more information. Hence, if A was going to die, there would only be one possible answer he could give him (either B or C, dependent on which had been chosen). If A were going to live, the guard had a choice of telling him B or C would die (both MUST die if A lives)
Your new poser is subtly different. A and C both know the same information. Thinking logically (or trying to anyway!!!):-
Two of you guys are going to and one will survive
[At this point A, B and C all believe they have a 1/3 chance of surviving]
and let me tell you now, prisoner A will die for certain!
[OK, A is a bit down in the dumps, his chances have just plummetted to 0/3. B and C had an equal chance of surviving, so must still have an equal chance. The probabilities must add up to 1 for all possibilities, so each has a 1/2 chance of surviving]
Your new poser is different because the guard COULD tell someone he was going to die. In the original he couldn't.
If anyone can see a flaw in my logic, please point it out - I stand to be corrected!!!
🙂
To BLR - proof of the Monty Hall problem.
Assume there are three safes labelled A, B, C. There are only three possible outcomes each with P=1/3.
P(A) = P(B) = P(C) = 1/3 where P(A) is the probability of the prize being in safe A, etc
The three possible outcomes are shown below:-
A B C
1 O O X
2 O X O
3 X O O
where X shows the location of the prize.
Let's assume we choose safe A.
Option 1: Monty Hall opens safe B which is empty. You know the prize is in A or C. You have the option of keeping A or choosing C.
If you keep A, you lose. If you choose C, you win.
Option 2: Monty opens safe C which is empty. If you keep A you lose. If you choose B you win.
Option 3: Monty opens EITHER B or C (doesn't matter which). If you keep safe A you win. If you choose the remaining safe you lose.
Looking back at these options, in the first 2 cases, switching safes gave you the prize. Only in option 3 did keeping the safe result in a win.
Hence,
prob of winning if you choose the other safe = 2/3
prob of winning if you keep your first choice = 1/3
3 prisoners is a similar problem, where 2 choices do not have an equal chance of happening.
Phew! Do we get a prize for contributing to one of the longest threads on RHP?
🙂