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I must be missing something here. > if we follow this strategy we’ll win a little bit less than one-third of the time. That is not possible. If N players use
by lisper 4y ago
I must be missing something here.
> if we follow this strategy we’ll win a little bit less than one-third of the time.
That is not possible. If N players use the same strategy then by symmetry each of them must win on average 1 time in N. So against two other players we should win exactly one third of the time, not "a little bit less" than one third of the time.
Also, if all players use the same strategy then it doesn't matter what the strategy actually is. The only requirement is that it have some probabilistic element, otherwise all players would choose the same number every time.
- deleted 4y ago[deleted]
- Jyaif 4y agosometimes all 3 players will choose the same number
- ZetaZero 4y agoI believe this strategy will work against any other strategy, not just mirror strategy. Of course, against mirror strat, 1/N is correct.
- andrewla 4y agoIt seems like there are pathological counterexamples. If you're player three, and player one chooses 1 always, and player two chooses 2 always, then the player using this strategy can never win. Player 2 wins with probability .45 (only wins when player 1 gets "knocked out"), and player 1 with probability 1-.45, and player 3 never wins.
- alexmolas 4y ago> If N players use the same strategy then by symmetry each of them must win on average 1 time in N. The point here is that there's an option of tying, if all players choose the same number. So P_1_win + P_2_win + P_3_win + P_tie = 1, means P_i_win < 1/3. > Also, if all players use the same strategy then it doesn't matter what the strategy actually is. The only requirement is that it have some probabilistic element, otherwise all players would choose the same number every time. A player choose its next move using some probabilities P, ie: it's going to choose randomly number 1 the 45% of times, 2 the 25% of times, etc. So there's a random element in the game, which makes players following the same strategy to not choose the same numbers every time.
- lisper 4y agoAh. I didn't realize that ties were counted as non-wins. I thought ties would just be ignored.
- feoren 4y ago> Also, if all players use the same strategy then it doesn't matter what the strategy actually is. But it's not very interesting to add "all players use the same strategy" as an assumption, because yes, obviously, if every player randomly picks from the set {33, 247, 17855433344} each has a 29.6% chance of winning. But that's a different game. In fact the author doesn't assume everyone will use the same strategy; rather, the author simply asks what optimal play is, when you don't know the others' strategies. The fact that all players choose the same strategy (if they're playing optimally) is a consequence of this. Optimal play has to work regardless of the strategy your opponents pick; assuming they'll play optimally too is just a shortcut to figuring out what optimal play is. For example, let's say you knew the other two players' strategy is to pick from this distribution. If you simply always pick 1, you win any time neither of the others picked 1, which happens with probability (1-0.456)^2 ~ 0.296 -- the same probability of winning as the optimal strategy! So why bother with the distribution? Precisely because you don't know whether your opponents are playing optimally or not. If another player gets the same bright idea as you, you counteract each other and the 3rd player wins 54% of the time (the rest are ties). An optimal strategy has to be optimal even if opponents are not using it; otherwise it's not really optimal (one may not exist). So all players using the same strategy is not an assumption, but a consequence, and therefore much more interesting.