# Gambler's ruin vs point difference

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?

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