Richard and Vanessa are going to play a game of foosball. The first player to score 5 goals wins. Vanessa is better: at any given point sh is 60% likely to score the next goal. What is the probability that Richard wins?
For Richard to win in k games (k from 5 to 9; minimum of 5 games, maximum of 9, at 9 games at least 1 person has won 5 games) it is mandatory that Richard wins the last game and 4 of the first k-1 games. Vanessa wins k-5 of the first k-1 games. The probability of this is
(40/100)^5*(60/100)^(k-5).
There are k-1 choose 4 such configurations (Richard can win in any 4 of the first k-1 games), so there is a
(k-1 choose 4)*(40/100)^5*(60/100)^(k-5) probability that Richard wins in k games. We just sum this value for k from 5 to 9.
