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Design a game where three coins are tossed and the player "wins" if he or she tosses either all three heads or all three tails. The question is if the "wager" is $1, then what is a fair winning payout for the player such that he or she would break about even playing this game over time?

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HINT: Only two of the possible outcomes, TTT and HHH, are wins. The other $n$ are losses, where I leave it to you to determine just what $n$ is. The $n+2$ outcomes are equally likely. If the winning payout is $p$, therefore, on average the player pays $n+2$ dollars and wins $2p$ dollars per $n+2$ games and has net winnings of $2p-(n+2)$ dollars. In a fair game the net winnings are $0$.

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The probability of $3$ heads in a row is $\left(\frac{1}{2}\right)^3$, as is the probability of $3$ tails in a row. So the probability of winning is $\frac{1}{4}$.

It is not precisely clear what "payout" means. We assume that you get your dollar back plus an amount $a$ that we will call the payout,

Then the random variable $X$, which is the amount you win, is $-1$ with probability $\frac{3}{4}$ and $a$ with probability $\frac{1}{4}$.

It follows that $E(X)=(-1)\left(\frac{3}{4}\right)+(a)\left(\frac{1}{4}\right)$. For a fair game, we need $E(X)=0$. Solve the equation $(-1)\left(\frac{3}{4}\right)+(a)\left(\frac{1}{4}\right) =0$ for $a$. We get $a=3$.

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