Originally posted by BLReidBLReid...you seem to have defeated your own argument...!
His 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.
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.
If 2 out of three were going to die, how can he have a 50/50 chance of survival? His chances of survival MUST be 1 in 3.
The guard is not allowed to name all possibilities, therefore, the ratio of outcomes is skewed. If the guard could name all possible scenarios, it would look something like this
A is spared B is named: 1/6
A is spared C is named: 1/6
B is spared A is named: 1/6
B is spared C is named: 1/6
C is spared A is named: 1/6
C is spared B is named: 1/6
The problem doesn't remove the possibilities that include A is known to die, only the possiblity that the guard can tell A that fact. In essence, the possibilities still exist, but they are sort of "covered up" by the limitation placed on the guards answer. Once subject B is known, using iamatiger's model, you must re-introduce the A is named possibilities into the overall odds to get an accurate picture of his survival chances. By removing the B is named possibilities, you are left with 2 options:
A is spared B is named
C is spared B is named
1 bullet / 2 people = 1/2
BLR - Hang on - the guard can't name A, so those options shouldn't be included in your calculations!
Options from A's point of view, given that the guard can't tell him whether he's going to die or not:-
A is spared, B is named: 1/6
A is spared, C is named: 1/6
B is spared, C is named: 1/3
C is spared, B is named: 1/3
Now look at what we know from the guard. B is not spared, he is named, so the only remaining options are:-
A is spared, B is named: 1/3
C is spared, B is named: 2/3 (doubling up the probabilities from above so they total one)
So, A has a 1/3 chance of being spared.
A definitely does NOT have a 50% chance of surving - we already know he has a 1/3 chance, and he KNOWS at least one of the others dies, so whether the guard says B or C is irrelevant as far as A is concerned - his chance of surviving is 1/3!!! End of.
C, on the other hand, should sleep more soundly in his bed if he overheard the guards conversation with A - he has a 2/3 chance of surviving...
🙂
How could C possibly have a better survival chance than A with the same knowledge? With respect, the logic is flawed. In the beginning there are 3 possibilities:
A lives, B lives, and C lives.
The guard eliminates the possibility that B lives.
Now there are 2 possibilities:
A lives and C lives.
That's 50/50.
I'm sorry, but...any formula that produces a different result is fundamentally flawed.
🙂
I think the argument you guys are having is over the way in which the question was asked.
If the guard picks one name at a time, and says that B will die, then the guard's next choice is 50/50 between A and C. However, the chance that A wasn't picked first is (2/3), and the chance that A isn't picked the second time is (1/2), so taken together you get (2/3)*(1/2)=(1/3).
If the guard picks two names at a time, then he's picking from this list:
1. A and B
2. B and C
3. A and C
The chances that A won't be picked is 1/3, which means that the chance he will be picked is 2/3. Once the guard says B is going to die, you can exclude item no. 3 from the list possibilities, but not from the list of choices, a subtle but important difference. The chance that one of item no. 1 or 3 could have been picked remains 2/3, but since we know item no. 3 wasn't picked, the 2/3 gets attirbuted to item no. 1 only. Therefore the chance that A survives is still 1/3.
One thing all of you seem to assume as granted (and though it may be implied it is not explicitly stated, and therefore not certain) is that the probabilities for choosing any of the prisoners are equal and are 1/3 for each of them.
There is however no way of verifying that and all your arguments are pointless. 🙄
BLR
With respect, the logic is not flawed. C is in a different boat to A. The guard could not possibly have said A as he is not allowed to tell A whether he dies or not. If C is eary-wigging at his cell door without the guards knowledge, he COULD have heard him say C (if he was due for execution). A, on the other hand, could not possibly have heard the guard say A. This is where the crux of the problem lies.
Look at it this way... the guard says to A, yes, one of the other two will die. I'm not telling you which one. A still has a 33% chance of surviving. Knowledge of which one dies doesn't improve his chances.
C's chances are improved, because the guard COULD have said C but didn't.
If the reasoning doesn't convince you, look at the various options in my earlier post from 20:17, and tell me which part of the maths you don't agree with, and I'll try and explain it further.
Originally posted by ilywrinYou have to assume the equal probability, or you really don't have a math puzzle, just a random question without enough information to solve it 🙂
One thing all of you seem to assume as granted (and though it may be implied it is not explicitly stated, and therefore not certain) is that the probabilities for choosing any of the prisoners are equal and are 1/3 for each of them.
There is however no way of verifying that and all your arguments are pointless. 🙄
Originally posted by PawnCurryThe problem is when you allow the limitations of the guard's answer to alter the overall probability of any individual being executed, except of course poor subject B, whose fate is known. To put it a different way, A asks the question and both B and C are eve's dropping...
BLR
With respect, the logic is not flawed. C is in a different boat to A. The guard could not possibly have said A as he is not allowed to tell A whether he dies or not. If C is eary-wigging at his cell door without the guards knowledge, he COULD have heard him say C (if he was due for execution). A, on the other hand, could not possibly have heard the ...[text shortened]... and tell me which part of the maths you don't agree with, and I'll try and explain it further.
Now B's chances have move from 1/3 to 1/1. He's out of the equation. C and A must equally absorb the difference. Each beginning with 1/3 and adding 1/6 (half of B's former status) they each now arrive at 1/2.
And no, we're not arguing. This is a friendly debate of perception. 🙂
BLR - at least we agree on one thing...yes this is a friendly debate, not an argument!
Have to agree with Palynka though, if C is listening, he knows his chances are 2/3. A and B can't just "absorb" the remaining chance of surviving equally - you need to look at all the possible outcomes, and how likely each outcome is, within the amount of information we have available.
Originally posted by PawnCurryWrong here. If C is listening, all he know is that B will die...that's what A knows as well. Both of them knew that A couldn't be named regardless. So, put yourself in C's position. He knows that the guard will say him or B. the guard says B. That doesn't rule out himself or A. His odds of survival are 1 in 2, just like A, operating on the same information.
BLR - at least we agree on one thing...yes this is a friendly debate, not an argument!
Have to agree with Palynka though, if C is listening, he knows his chances are 2/3. A and B can't just "absorb" the remaining chance of surviving equally - you need to look at all the possible outcomes, and how likely each outcome is, within the amount of information we have available.
Trust me, not wrong!!!
I've tried my hardest to explain so can only blame myself for not explaining it clearly enough. It is a tough one to get your head round... but the only way to do it is to ditch your usual assumptions, look at the possibilities, and do the maths.
Or you could try your own experiment with a friend, three cups to represent the cells, and two balls to represent who gets the chop. He can play prison guard, you can play prisoner A. Make sure you shut your eyes and don't cheat! Try it for at least 100 goes, preferably more.
Or hunt on the net for the Monty Hall problem. And then tell me if you'd switch boxes or not...!