Two teams A and B play series of five games... [closed]
Two teams A and B play series of five games. A tie is impossible for each game. When one of the teams wins three games the series of games are finished and no more games are played. What is the expected number of games played in this series if on average, team A wins 3 times more games than team B?
This problem appeared in my textbook and I'm confused about how to solve it.
Edit: It's 3/4 and 1/4. Sorry, I just woke up.
$\endgroup$2 Answers
$\begingroup$If A is "three times more likely to win a game than B then A must have probability of winning 3/4 and B 1/4
Writing "A" for a win by A an "B" for a win by B, the possible series are
AAA which has probability (3/4)^3= 27/64.
BBB which has probability (1/4)^3= 1/64.
BAAA, ABAA, AABA each of which has probability (3/4)^3(1/4)= 27/256.
ABBB, BABB, BBAB each of which has probability (3/4)(1/4)^3= 3/256.
There are 6 different ways B can win two games before A wins three BBAAA, BABAA, BAABA, ABBAA, ABABA, AABBA each of which has probability (3/4)^3(1/4)^2= 27/256.
There are 6 different ways A can win two games before B wins three AABBBL ABABB,, BAABA, BAABB, BABAB, BGAAB each of which has probability (3/4)^2(1/4)^3= 3/256.
$\endgroup$ 1 $\begingroup$Hints:
What is the expected number of games played in this series if on average, team A wins 3 times more games than team B?
First of all, I interpret this to mean that if $p = 0.75, q = 0.25$, then in each game, A's chance of winning is $p$ and B's chance of winning is $q$.
First, you need the general formula for Bernoulli trials:
Given $n$ trials of an independent event, with probability of success or failure $= p$ and $q$, respectively, then the chance of exactly $k$ successes in the $n$ trials is
$$\binom{n}{k}p^kq^{(n-k)}.$$
Next, you need to know that there are exactly 6 possibilities: either team A wins in either 3,4, or 5, games, or team B wins in either 3,4, or 5 games.
Next, you need to know that (for example) for team A to win in exactly $k$ games, two things must happen:
team A won exactly 2 games in the first $(k-1)$ games.
team A won the $k$-th game.
The over all computation will be
$\sum_{k=3}^5 k \times f(k) + \sum_{k=3}^5 k \times g(k).$
Here, $f(k)$ is the probability of team A winning in exactly $k$ games and $g(k)$ is the probability of team B winning in exactly $k$ games.
At this point, for the new student, it is simply a matter of forgoing elegance, and getting down in the mud. Later, with experience, you will see shortcuts that needn't concern you now.
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