Calculate the expected time spend playing the game. Hint: Use Wald's Equation.

Question

• Alice and Bob play each other in a checkers tournament, where the first player to win four games wins the match.
• The players are evenly matched, so the probability that each player wins each game is $1$ $2$, independent of all other games.
• The number of minutes for each game is uniformly distributed over the integers in the range $\left[\,{30, 60}\,\right]$, again independent of other games.
• What is the expected time they spend playing the match ?.

I am trying to use Wald's equation in this but very new to this thing. Don't know how to proceed.

Answer 1

For convenience, we can assume that Alice and Bob play on for a total of $\ 7\ $ games, even after one of them has already won. Then they will play some number, $\ L\ $, of live games until one of them has won the match, followed by $\ 7-L\ $ "dead" games. Wald's equation tells you that the expected number of games needed to determine who won the match is $\ 45\,\mathbb{E}\left(L\right)$ minutes.

In this problem, the identity $\ \mathbb{E}\left(L\right)=\sum_\limits{n=1}^7\Pr\left(L\le n\right)\ $ is a handy way of calculating $\mathbb{E}\left(L\right)\ $ . The probability that Alice has won by the end of game $\ n\ $, for $\ n=4\mbox{–}7\ $, is just the probability of four or more successes in $\ n\ $ Bernoulli trials, or $\ \sum_\limits{j=4}^n \frac{1}{2^n}{n\choose j}\ $. Likewise, the probability of Bob's winning by the end of the $\ n^\mathrm{th}\ $ game is the same. Thus $\ \Pr\left(L\le n\right)=2\sum_\limits{j=4}^n \frac{1}{2^n}{n\choose j}=\sum_\limits{j=4}^n \frac{1}{2^{n-1}}{n\choose j}\ $. This should be enough for you to find the answer you're looking for.

Addendum: If you don't want to assume that Bob and Alice continue to play each other after the match has been decided, then the probability that Alice wins the match after exactly $\ n\ $ games is the probability that she wins $\ 3\ $ out of the first $\ n-1\ $, and then wins the $\ n^\mathrm{th}\ $—that is $\ \frac{1}{2^{n-1}}{n-1 \choose 3}\cdot\frac{1}{2}\ $. The probability that Bob wins after exactly $\ n\ $ matches is the same. Thus, the probability that they take $\ n\ $ or fewer games to decide the match is $\ 2\sum_\limits{j=4}^7\ \frac{1}{2^n}{n-1 \choose 3}=\sum_\limits{j=4}^7\ \frac{1}{2^{n-1}}{n-1 \choose 3}\ $, which is easily checked to give the same value for $\ \Pr\left(L\le n\right)\ $ as before.