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There are $2$ volleyball teams A and B. They both have a $50 $% chance of winning a round. They play until one wins $3$ rounds. The first round is won by A. What is the probability of A winning.

I calculated by hand that it's $\frac{11}{16}$ but I'm wondering if there's a better more mathematical way of proving it.

I tried to use the Binomial Probability Theorem (example: https://www.cuemath.com/binomial-distribution-formula/ ) because this somewhat resembles the Bernoulli example question with coin throws but still I couldn't get $11/16$

So I'm looking for an explanation why the probability is $11/16$ using a math equation

Thank you

Aig
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3 Answers3

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Assuming that each game has the independent probability of $~50\%~$ of A winning, the standard shortcut here is to assume that all 5 games will be played, even if one of the two players wins before the 5th game. Note that playing all 5 games can not possibly change who won, when one of the players wins before the 5th game.

Then, A wins if he wins at least two of the four remaining games. Per the Binomial Distribution, the probability of exactly $~k~$ wins out of $~n~$ trials is $\displaystyle \binom{n}{k}p^kq^{(n-k)}.$

So, A's probability of winning is:

$$~\left\{ ~\binom{4}{2} (1/2)^2 \times [1 - (1/2)]^2 ~\right\} + \left\{ ~\binom{4}{3} (1/2)^3 \times [1 - (1/2)]^1 ~\right\} \\ + \left\{ ~\binom{4}{4} (1/2)^4 \times [1 - (1/2)]^0 ~\right\} $$

$$= \frac{1}{16} \times [6 + 4 + 1] = \frac{11}{16}.$$

user2661923
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To use the binomial distribution here, you need to realise two things

  • the match will need a maximum of $4$ more rounds

  • for $A$ to win, $B$ can be allowed to win at most $2$ matches

Then, writing in more general terms first, with $p$ being the probability for $A$ winning a round, and $q$ being that for $B$ winning a round,

P(A wins) $=\binom40q^0p^4 + \binom41q^1p^3 +\binom42q^2p^2$

If you compute the result with $p=q=\frac12$, you will get

ans $= \dfrac{1+4+6}{2^4} = \dfrac{11}{16}$

If you are wondering why we are seemingly allowing $A$ to win more rounds than needed to win the match, see here

true blue anil
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A won the first round, so A has to win two rounds to finish the game. You could use binomial probability to calculate the probability of A winning two rounds $C^{4}_{2}\frac{1}{2^{4}}$

Ok, agree this is wrong.

Ablet Imin
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    I disagree, see my answer. Specifically, while the assumption that all 5 games will be played is viable, your mistake is in thinking that A must win exactly 2 of the 4 remaining games, rather than at least 2 of the 4 remaining games. – user2661923 Mar 19 '24 at 18:32
  • This isn't right. Since $A$ won the first game, it should have more than 50% chance to win the tournament. – Alain Remillard Mar 19 '24 at 18:33
  • I agree my answer is wrong. – Ablet Imin Mar 19 '24 at 19:29