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$\begin{array}{l}{\text { In each game played one is equally likely to either win or lose 1. Let } S \text { be your }} \\ {\text { cumulative winnings if you use the strategy that quits playing if you win the first }} \\ {\text { game, and plays two more games and then quits if you lose the first game. }} \\ {\text { (a) Use Wald's equation to determine } E[S] \text { . }} \end{array}$

Let $X_i$ be the amount won in game $i$. $$E[S] =E\bigg[\sum_{i=1}^N X_i\bigg]$$ Applying Walds theorem, $$=E[N]E[X]$$ And since this is a fair game where the player starts with 0 dollars, we know that $E[X]=0$. So, $E[S]=0$

Is this correct?

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    $\begingroup$ Suppose $E[S]\neq0$, then this strategy (or its reverse) has a positive expected value on a fair game which is impossible So, in short, yes, it is true. $\endgroup$ – Stan Tendijck Apr 22 at 23:40
  • $\begingroup$ @StanTendijck Thank you for verifying. Could you also take a look and verify the following if you are able? link to question $\endgroup$ – Jac Frall Apr 23 at 0:08

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