# Probability of a player winning a snooker tournament

Two players have a snooker tournament, the first player to win 6 frames wins the match. (Best of 11)

Assume that the probability of player A winning each frame is P, regardless of who starts. If A does not win then his opponent does. (No Draws)

If P = 0.4 what is the probability that A wins the tournament?

This one stumped me, so any help from you math experts would be really appreciated!

• What are the possible events that lead to a victory for $A$? What are their associated probabilities? Commented Apr 8, 2015 at 15:40
• A possible event that leads to a victory for $A$ is $AAAAAA$ with which I mean $A$ won the first $6$ game in a row, other events are $AABAAAA$ with which I mean, $A$ won the first $2$ games, then $B$ won a game, then $A$ won $4$ games. Can you list all of those events? What are the associated probabilities of these events happening? If the answer to one of both questions is no, please let me know. Commented Apr 8, 2015 at 15:43
• Correct me if I'm wrong here but there should be 30 ways in which A can win 6 times. Using permutations P(6,2) = 30 ? Commented Apr 8, 2015 at 15:59
• @Ordered we should be using combinations, not permutations. Furthermore, it is out of the total 11 possible games, 6 of those being wins. There are thus $\binom{11}{6} = 462$ ways of winning exactly six out of 11 games. Commented Apr 8, 2015 at 16:01
• The events of interest are the events corresponding to the final scores $6-0$, $6-1$, $6-2$, $6-3$, $6-4$, $6-5$ with on the left the final score of player $A$ and on the right the final score of player $B$. The probability that $6-0$ happens, can only happen in one way. Player $A$ wins all games. The probability of that happening is $0.4^6$ Commented Apr 8, 2015 at 16:48

Hint: You can assume they in fact play all $11$ games. A has to win at least $6$ of them. You have a binomial distribution. Can you compute the chance that A wins exactly $6$ of $11$? Now just add the chances of $7, 8, 9, 10, 11$