I don't think the first example is illustrative at all. "Now consider the second case where we have a saw-tooth like region between them. Here also, the marbles will roll towards either ends with equal probability." Not if it's anything like the included illustration, in which case the marble is more likely to roll towards A than B. "Now if we tilt the whole profile towards the right, as shown in Figure 2, it is quite clear that both these cases will become biased towards B." Again, no, it's quite clear to me that the second case remains biased towards A.
No, because this paradox only applies to games where you chances of winning change over time.
I teach maths and have come across this before. I've always considered it just a really complex way of wrapping up something trivial and a bit stupid. Lets consider two much easier games, and take out the probability (it makes no difference that we do this).
Game A) At even time-steps, you give me $5. At odd time-steps, I give you $10.
Game B) I always give you $1, at each time-step.
Clearly both of these games are a lost cause for me. However, if we play ABABABA... then we alternate between you giving me $5, and me giving you $1, so I win!
Sure -- this is like playing roulette, (game A, a constant-losing-odds game,) while simultaneously watching the blackjack table (game B,) waiting for the deck to become favorable. As soon as the card count gets high enough, switch games. Once the blackjack game turns negative, switch back to roulette to pass the time.
This should be a winning strategy, against two losing games. The key is that the odds in game B vary, and you structure your gameplay such that you're mostly playing game B when the odds are better than even.
Oh, and you'd likely be kicked out of any casino in Vegas for doing this.
[+] [-] tedunangst|14 years ago|reply
[+] [-] gizzlon|14 years ago|reply
[+] [-] spektom|14 years ago|reply
[+] [-] CJefferson|14 years ago|reply
I teach maths and have come across this before. I've always considered it just a really complex way of wrapping up something trivial and a bit stupid. Lets consider two much easier games, and take out the probability (it makes no difference that we do this).
Game A) At even time-steps, you give me $5. At odd time-steps, I give you $10.
Game B) I always give you $1, at each time-step.
Clearly both of these games are a lost cause for me. However, if we play ABABABA... then we alternate between you giving me $5, and me giving you $1, so I win!
[+] [-] iclelland|14 years ago|reply
This should be a winning strategy, against two losing games. The key is that the odds in game B vary, and you structure your gameplay such that you're mostly playing game B when the odds are better than even.
Oh, and you'd likely be kicked out of any casino in Vegas for doing this.