You have a basketball hoop and someone says that you can play one of two games.
Game1 : You get one shot to make the hoop.
Game2 : You get three shots and you have to make two of three shots.
If p is the probability of making a particular shot, for which values of p should you pick one game or the other?

Probability of winning the Game 1: p

Probability of winning the Game 2:

P(game2) = P(make 1 and 2) + P(make 1 and 3) + P(make 2 and 3) + P(make 1, 2 and 3)
= p*p*(1-p) + p*(1-p)*p + (1-p)*p*p + p*p*p
= 3p^2 - 2p^3

Coming back to game selection, Play Game 1 if:

P(game1) > P(game2)
=> p > 3p^2 - 2p^3
=> 1 > 3p - 2p^2       // Since p > 0
=> (2p - 1)(p - 1) > 0

Both terms must be positive, or both must be negative, But we know p < 1, so p – 1 < 0. This means both terms must be negative i.e.

=> 2p - 1 < 0
=> 2p < 1
=> p < 0.5

So, we should play Game 1 if 0 < p < 0.5 and Game 2 if 0.5 < p < 1. If p = 0, 0.5, or 1 then P(Game 1) = P(Game 2), so it doesn’t matter which game we play

December 28th, 2018|Categories: Puzzles|
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