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?
These are binomial distributions probability with n trials of probability p. The prob of x successes is given by the TI83 or 84 calculators by
binompdf(n,p,x)
The calculator can also give the cumulative probability of x successes or fewer by
binomcdf(n,p,x)
Consequently, its easier to turn the problem around. The probability of winning more than half is the same as the opponent losing less than half. The following then gives the probability of losing the metagame with 2n games:
binomcdf(2x,0.55,x-1)
A table shows the minimum occurs when x=5 (10 games) and the probability of opponent winning is about 0.2616
and so your losing probability is 0.7384
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Posted by Jer
on 2020-10-27 12:51:08 |
Implementing by TI calculator
