Exercise Problem 25, Chapter 4, Blitzstein and Hwang, Intro to Probability

probability solution-verification expected-value self-learning probability-distributions

Solution 1:

I have used a different method.

Taking win probability of Calvin as $p$
let $x$ = additional games needed from start (= tied) for Calvin to win
$y$ = additional games needed if Calvin leads by $1$ game
$z$ = additional games needed if Calvin trails by $1$ game,

then $x = 1 +py +qz, y = 1+qx, z = 1 + px$

Solving, $x = \frac{2}{2p^2-2p+1}= \frac{2}{1-2pq},$

and by result symmetry, expected # of games needed by Hobbes to win is also $\;\frac{2}{1-2pq}$

Thus for one or the other to win, $\Bbb{E}[N] = \frac{2}{1-2pq}$

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