Two teams A and B compete in a "best-of-5" competition. This means
they play each other until one team has won 3 games. Suppose that for any
of the games, the probability that A beats B is x . What is the probability
that A wins the "best-of-5" competition?
can someone help me with this problem
Best Answer
We need to assume a little more, that the results of the games are independent, like the results of tossing repeatedly a coin that has probability $x$ of landing heads.
I will assume you know something about the Binomial distribution, so will recycle results from there.
Imagine the rules are changed so that no matter what happens, the full $5$ games are played. Then Team A wins the series with the original rules if and only if A wins $3$ or more games under the modified rules. This probability, by a standard formula, is $$\binom{5}{3}x^3 (1-x)^2+\binom{5}{4}x^4(1-x)^1 +\binom{5}{5}x^5(1-x)^0.$$
Another way: Or else we can do a cases analysis. Team A can win by (i) winning $3$ in a row or (ii) by winning exactly $2$ of the first $3$ games, and winning the fourth or (iii) winning exactly $2$ of the first $4$ games, and winning the fifth.
The probability of (i) is $x^3$.
For (ii), the loss can come in any one of $3$ places. The probability of the pattern WWLW is $x^3(1-x)$. The other two patterns WLWW and LWWW have the same probability, for a total of $3x^3(1-x)$.
For (iii), use the same reasoning. There are $\binom{4}{2}=6$ patterns, each of which has probability $x^3(1-x)^2$. So the probability is $6x^3(1-x)^2$.
Add up.