It is interesting to contemplate that the almost fair solution favors bob:
Bob counts the numbers of apples in 1st room and accepts it unless it has zero apples in it, in which case he rejects it. If he hasn’t rejected room 1 he counts the apples in 2 and if it is more than in 1 he accepts it else he rejects it.
For all possible numbers of apples in rooms EXCEPT one room has zero apples, Bob has 50% chance of getting it right. But for all possible number of apples in rooms where one room has zero apples in it, Bob has 5⁄6 chance of winning and only 1⁄6 chance of losing.
I think in some important sense this is the telling limit of why Coscott is right and how Alice can force a tie, but not win, if she knows Bob’s strategy. If Alilce knew Bob was using this strategy, she would never put zero apples in any room, and she and Bob would tie, i.e. Alice was able to force him arbitrarily close to 50:50.
And the strategy to work relies upon the asymmetry in the problem, that you can go arbitrarily high in apples but you can’t go arbitrarily low. Initially I was thinking Coscott’s solution must be wrong, that it must be equivocating somehow on the fact that Alice can choose ANY number of apples. But I think it is right, but that every strategy Bob uses to win can be defeated by Alice if she knows what his strategy is. I think without proof, that is :)
I think in some important sense this is the telling limit of why Coscott is right
Right about what? The hint I give at the beginning of the solution? My solution?
Watch your quantifiers. The strategy you propose for Bob can be responded to by Alice never putting 0 apples in any room. This strategy shows that Bob can force a tie, but this is not an example of Bob doing better than a tie.
Right about it not being a fair game. My first thought was that it really is a fair game and that by comparing only the cases where fixed numbers a, b, and c are distributed you get the slight advantage for Bob that you claimed. That if you considered ALL possibilities you would have not advantage for Bob.
Then I thought you have a vanishingly small advantage for Bob if you consider Alice using ALL numbers, including very very VERY high numbers, where the probability of ever taking the first room becomes vanishingly small.
And then by thinking of my strategy, of only picking the first room when you were absolutely sure it was correct, i.e. it had in it as low a number of apples as a room can have, I convinced myself that there really is a net advantage to Bob, and that Alice can defeat that advantage if she knows Bob’s strategy, but Alice can’t find a way to win herself.
So yes, I’m aware that Alice can defeat my 0 apple strategy if she knows about it, just as you are aware that Alice can defeat your 2^-n strategy if she knows about that.
So yes, I’m aware that Alice can defeat my 0 apple strategy if she knows about it, just as you are aware that Alice can defeat your 2^-n strategy if she knows about that.
What? I do not believe Alice can defeat my strategy. She can get arbitrarily close to 50%, but she cannot reach it.
It is interesting to contemplate that the almost fair solution favors bob:
Bob counts the numbers of apples in 1st room and accepts it unless it has zero apples in it, in which case he rejects it.
If he hasn’t rejected room 1 he counts the apples in 2 and if it is more than in 1 he accepts it else he rejects it.
For all possible numbers of apples in rooms EXCEPT one room has zero apples, Bob has 50% chance of getting it right. But for all possible number of apples in rooms where one room has zero apples in it, Bob has 5⁄6 chance of winning and only 1⁄6 chance of losing.
I think in some important sense this is the telling limit of why Coscott is right and how Alice can force a tie, but not win, if she knows Bob’s strategy. If Alilce knew Bob was using this strategy, she would never put zero apples in any room, and she and Bob would tie, i.e. Alice was able to force him arbitrarily close to 50:50.
And the strategy to work relies upon the asymmetry in the problem, that you can go arbitrarily high in apples but you can’t go arbitrarily low. Initially I was thinking Coscott’s solution must be wrong, that it must be equivocating somehow on the fact that Alice can choose ANY number of apples. But I think it is right, but that every strategy Bob uses to win can be defeated by Alice if she knows what his strategy is. I think without proof, that is :)
Right about what? The hint I give at the beginning of the solution? My solution?
Watch your quantifiers. The strategy you propose for Bob can be responded to by Alice never putting 0 apples in any room. This strategy shows that Bob can force a tie, but this is not an example of Bob doing better than a tie.
Right about it not being a fair game. My first thought was that it really is a fair game and that by comparing only the cases where fixed numbers a, b, and c are distributed you get the slight advantage for Bob that you claimed. That if you considered ALL possibilities you would have not advantage for Bob.
Then I thought you have a vanishingly small advantage for Bob if you consider Alice using ALL numbers, including very very VERY high numbers, where the probability of ever taking the first room becomes vanishingly small.
And then by thinking of my strategy, of only picking the first room when you were absolutely sure it was correct, i.e. it had in it as low a number of apples as a room can have, I convinced myself that there really is a net advantage to Bob, and that Alice can defeat that advantage if she knows Bob’s strategy, but Alice can’t find a way to win herself.
So yes, I’m aware that Alice can defeat my 0 apple strategy if she knows about it, just as you are aware that Alice can defeat your 2^-n strategy if she knows about that.
What? I do not believe Alice can defeat my strategy. She can get arbitrarily close to 50%, but she cannot reach it.