Calculate the probability of winning a specific number of games if each individual game has a different probability of winning?

Question

Say I have to play 11 games. My goal is to win at least 6, and each game has a DIFFERENT probability of being won.

The probabilities of winning each game (as a %) are as follows: 24, 59, 3, 10, 85, 47, 14, 69, 92, 57, 63.

How can I calculate the probability that I will win at least 6 out of 11 of these games, in any particular combination of 6 wins?

This is something I am trying to implement in Java.

Answer 1

Rather than directly calculate the probability of winning "at least six games" I would calculate the probability of winning "five or fewer games" and subtract from 1. Since you have different probabilities of winning each game that will still be tedious!

Let's look at a simpler problem where you want to win 3 out of 5 games with probability of winng a, b, c, d, e.

I would calculate the probablity of winning 0, 1, or 2 games then subtract from 1.

The probability of not winning each of the games is 1- a, 1- b, 1- c, 1- e, and 1- e.

The probability of winning NO games is (1- a)(1- b)(1- c)(1- d)(1- e).

The probability of exactly one game is a(1- b)(1- c)(1- d)(1- e)+ (1- a)b(1- c)(1- d)(1- e)+ (1- a)(1- b)c(1- d)(1- e)+ (1- a)(1- b)(1- c)d(1- e)+ (1- a)(1- b)(1- c)(1- d)e.

The probability of exactly two wins is ab(1- c)(1- d)(1- e)+ a(1- b)c(1- d)(1- e)+ a(1-b)c(1- d)(1- e)+ a((1- b)(1- c)d(1- e)+ a(1- b)(1- c)(1- d)e+ (1- a)bc(1- d)(1- e)+ (1- a)b(1- c)d(1- e)+ (1- a)b(1- c)d(1- e)+ (1- a)b(1- c)d(1- e)+ (1- a)b(1- c)(1- d)e.

The probability of "two or less" is the sum of those:

(1- a)(1- b)(1- c)(1- d)(1- e)+ a(1- b)(1- c)(1- d)(1- e)+ (1- a)b(1- c)(1- d)(1- e)+ (1- a)(1- b)c(1- d)(1- e)+ (1- a)(1- b)(1- c)d(1- e)+ (1- a)(1- b)(1- c)(1- d)e+ ab(1- c)(1- d)(1- e)+ a(1- b)c(1- d)(1- e)+ a(1-b)c(1- d)(1- e)+ a((1- b)(1- c)d(1- e)+ a(1- b)(1- c)(1- d)e+ (1- a)bc(1- d)(1- e)+ (1- a)b(1- c)d(1- e)+ (1- a)b(1- c)d(1- e)+ (1- a)b(1- c)d(1- e)+ (1- a)b(1- c)(1- d)e.

The probability of "three or more" is 1 minus that: 1- (1- a)(1- b)(1- c)(1- d)(1- e)+ a(1- b)(1- c)(1- d)(1- e)+ (1- a)b(1- c)(1- d)(1- e)+ (1- a)(1- b)c(1- d)(1- e)+ (1- a)(1- b)(1- c)d(1- e)+ (1- a)(1- b)(1- c)(1- d)e+ ab(1- c)(1- d)(1- e)+ a(1- b)c(1- d)(1- e)+ a(1-b)c(1- d)(1- e)+ a((1- b)(1- c)d(1- e)+ a(1- b)(1- c)(1- d)e+ (1- a)bc(1- d)(1- e)+ (1- a)b(1- c)d(1- e)+ (1- a)b(1- c)d(1- e)+ (1- a)b(1- c)d(1- e)+ (1- a)b(1- c)(1- d)e.