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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

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  • $\begingroup$ Here's a plot of the probabilities of winning a best of $1$, best of $3$ and best of $5$, all plotted together over $x$; this gives an impression how much the better team benefits from the longer competition. The difference between best of $3$ and best of $1$ is more pronounced than the one between best of $5$ and best of $3$. $\endgroup$
    – joriki
    Feb 21, 2013 at 1:46

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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.

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  • $\begingroup$ ok so we want to add them all together and get x^3+3x^3(1-x)+6x^3(1-x)^2 simplifying we get x^3(1+3(1-x)+6(1-x)^2) $\endgroup$
    – dotdotdot
    Feb 21, 2013 at 2:00
  • $\begingroup$ i understand x^3 but what does winning 3 games and loosing one have to do with (1-x) is it the probability that B beats A? $\endgroup$
    – dotdotdot
    Feb 21, 2013 at 2:04
  • $\begingroup$ By W I meant A wins, by L that it loses. If two events $E$ and $F$ are independent, you get probability they both happen is the product of the probabilities. So for example the probability of WW is $(x)(x)$. The probability of WWL is $(x)(x)(1-x)$. So the probability of WWLW is $(x)(x)(1-x)(x)=x^3(1-x)$. $\endgroup$
    – André Nicolas
    Feb 21, 2013 at 2:11
  • $\begingroup$ ok i see and how did you deduce that for the x^3(1-x)^2 there were (4C3) or 6 patterns without writing out all the events? $\endgroup$
    – dotdotdot
    Feb 21, 2013 at 2:19
  • $\begingroup$ It was $\binom{4}{2}$, not $\binom{4}{3}$. We are considering words of length $4$, that have exactly $2$ W's and $2$ L's. How many such words are there? We need to choose the two places where the $W$ will go. $\endgroup$
    – André Nicolas
    Feb 21, 2013 at 2:34

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