Two people want to investigate who is best at playing Scrabble. They decide to settle this by playing a series of games. Assume that there is always a winner for each game, and that winning or losing a game is independent from the outcome of the previous game. At the start they both have $0$ points. After each game, the winner increases his/her number of points by $1$, the loser stays put at his/her number of points before the game. We can define this investigation as a stochastic process by defining $X_n$, for $n = 0, 1, 2,...$, as the absolute value of the difference in points after $n$ games. Note that this implies $X_0 = 0$.

Question: Assume that the two people are equally good at playing Scrabble and that they can play infinitely many games. Given an integer $K > 0$, what is the expected number of games before the difference in points is $K$?

My answer: If both players, $A$ and $B$, are equally good at playing scrabble, then $$P(A\text{ wins}) = P(B\text{ wins}) = \frac12$$ This question is then analogous to a Gambler's Ruin scenario, where both players start with $K$ chips and the game ends when either we reach state $0$ or $2K$. Hence, asking the expected number of games before the difference in points is $K$ is the same as asking what the expected number of games is before either player is bankrupted in the Gambler's Ruin scenario. Thus $p_{ij}=p_{ji}=\frac12$. Letting $Y_n=\# \text{chips of }A$, we have the new state space $$ S=\{0,1,\dots,2K\}. $$ We are then interested in $E(T_i)$, where $T_i$ is the time to absorption (to state $0$ or $2K$) given $Y_0=i$, which can be written as $$ T_i=\min\{n\geq 0 \,:\, Y_n=0 \text{ or } Y_n=2K\mid Y_0=i \}. $$ In our case, we have $i=K$. It was shown that, with $p=q=\frac12$ $$ E(T_i)=i(2K-i). $$ Hence, we have that the expected number of games before point difference is $K$ is $$E(T_K)=K(K+K-K)=K^2$$ Is this correct? If not, what is the correct approach?

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    $\begingroup$ $K^2$ is indeed correct when $p=q=\frac12$ $\endgroup$ – Henry Apr 13 at 17:56
  • $\begingroup$ And is my approach correct? $\endgroup$ – sam wolfe Apr 13 at 18:24
  • $\begingroup$ Clearly not, because if one person wins all the time it will take $K$ games for the score difference to be $K$ $\endgroup$ – Ross Millikan Apr 15 at 2:45
  • $\begingroup$ The question is regarding expectation, so I don't follow your comment. Can you please tell or hint at what's the correct approach? $\endgroup$ – sam wolfe Apr 15 at 12:18

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