Suppose there's a game in which you have a 45% probability of winning. Let's make it a sedentary game, unlike tennis, so that you don't tire and the probability is always 45%.
Someone proposes a metagame: play a series of the original game, and if you win more games than your opponent, you win. But, there's a proviso: It must be an even number of games, and there's no tie-breaker, so again, it's not like tennis.
Your only strategy is to pick the actual, even, number of games. What's the choice of number of games that will minimize your probability of loss in the metagame?
Assuming you have to pick a nonzero number of games, I would have thought the answer is just 2. Given you're more likely to lose than win any individual game, I'd think your best chance of winning the metagame is to play as few original games as possible and hope variance works to your advantage.
But I did the math in Excel and found that the greatest chance of winning the metagame was actually by playing 10 games, which resulted in a ~26.16% chance of winning at least 6 of them.
Haven't had any coffee yet but I'm not sure I understand why it's the case that playing more games increases your chances of winning the metagame up to a point and then starts decreasing again. (Of course there is also a nonzero chance I did the math wrong.)
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Posted by tomarken
on 2020-10-27 08:01:51 |
Solution
