Ping Pong Game

Alice and Bob play table tennis, with Bob’s probability of winning any given point being $30\%$ . They play until someone reaches a score of $$21$$ . What, approximately, is the expected number of points played?

Answer

First note the minimum number of points played is $$21$$ (a $21\!-\!0$ game) and the maximum is $$41$$ (a $21\!-\!20$ game).

Let $p=\mathbb{P}(\text{Alice wins the game})$ . By conditioning on who wins,

\mathbb{E}[T] = p\,\mathbb{E}[T \mid \text{Alice wins}] + (1-p)\,\mathbb{E}[T \mid \text{Bob wins}].

Since Alice has a large edge and the target score is $$21$$ , $$p$$ is very close to $$1$$ . Hence, approximately,

\mathbb{E}[T]\approx \mathbb{E}[T \mid \text{Alice wins}].

If Alice wins, then $$T$$ is the number of games needed for Alice to obtain $$21$$ points, where each point is won by Alice with probability $$0.7$$ .

Hence, $T\mid(\text{Alice wins})$ is (approximately) a negative binomial RV, and the expected number of games to get $$21$$ points is

\mathbb{E}[T \mid \text{Alice wins}] \approx \frac{21}{0.7}=30.

Therefore, the expected number of points played is very close to $$30$$ (and in fact is a tad bit below $$30$$ ).

Notes and comments

Comment 1: $T\mid(\text{Alice wins})$ is not exactly negative binomial since its support is bounded above by $$41$$ . With that said, the EV is very close to $$30$$ (slightly below).