AA

BB

ABAA

ABBB

BAAA

BABB

ABABAA

ABABBB

ABBAAA

ABBABB

BAABAA

BAABBB

BABAAA

BABABB

......

*Possible scenarios*

Answer: Let $N$ be the number of rounds. $N$ cannot be odd, otherwise the difference of rounds won by the two players could not be 2.

Therefore $N$ is even, say $N=2k$. Each player has to win exactly one of $2r-1$-th and $2r$-th round for $1\le r\le k-1$ and then one player has to win the last two games.

In other words, the pattern should be $[..][..]...[..](..)$, the first $k-1$ $[..]$'s should be AB or BA, and the last $(..)$ is AA or BB.

Summing up, we know

$P(N=2k-1)=0, P(N=2k)=(1/2)^k$ for $k=1,2,\ldots$

The game most likely ended in 2 rounds, but the expected number of rounds is $\sum_{k=1}^\infty\frac{2k}{2^k}=4$.

## No comments:

## Post a Comment