Answer by I. J. Kennedy for Winning an unfair game
This is really a comment on Ross Millikan's answer, but too cumbersome to enter that way. I have verified the $ p= \frac{n-1}{2n-1}$ threshold, although it really should be called $\frac{n/2}{n+1}$...
View ArticleAnswer by Ross Millikan for Winning an unfair game
If you play two games, your chance is $ p^2$. If you play four, your chance is $p^4+4p^3(1-p)$, which becomes greater when $p\gt \frac 13 $. If you play six games, your chance is...
View ArticleAnswer by lynxlynxlynx for Winning an unfair game
Let's look at the number of plays vs the number of games he has to win:2: 2/2 + 1 = 24: 4/2 + 1 = 36: 6/2 + 1 = 4n: n/2 + 1Or in relative terms: 1, 3/4, 2/3, 1/2 + 1/n. So the more they would play, the...
View ArticleWinning an unfair game
I came across the following interesting problem today:A game consists of a sequence of plays; on each play either you or your opponent scores a point, you with probability 𝑝< 1/2, he with...
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