perplexus.info :: Probability : Metagame: Set and Match

Metagame: Set and Match (Posted on 2020-10-27)

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?

Submitted by Charlie

Rating: 3.0000 (1 votes)

Solution: (Hide)

p games P(metawin) 0.45 2 0.6975 0.2025 0.45 4 0.6090 0.2415 0.45 6 0.5585 0.2553 0.45 8 0.5230 0.2604 0.45 10 0.4956 0.2616 0.45 12 0.4731 0.2607

The next to last number in each row is the probability you would have won if ties counted in your favor. But alas, ties go to your opponent's favor.
Your best chance of success is with 10 games, with 26.16% probability of success. With fewer games there'd be more ties, favoring your opponent; with more games, your lack of skill (or the house advantage) kicks in more.

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
No Subject	Mattie Jones	2021-11-17 06:22:50
re(2): Solution	tomarken	2020-10-27 14:54:43
Implementing by Desmos	Jer	2020-10-27 12:59:09
Implementing by TI calculator	Jer	2020-10-27 12:51:08
re: Solution	Steve Herman	2020-10-27 10:40:46
Solution	tomarken	2020-10-27 08:01:51

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