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I know it's basically asking for the expected value, however when I tried I was getting a positive number rather than a negative. I know I'm not doing it right because I have the answer key. The question says:

Someone plays roulette the following way: before each spin he rolls a die, and then he bets on red as many dollars as there were spots on the die. For example, if there were $4$ spots he bets $4$ dollars. If red comes up he gets the stake back plus an amount equal to the stake. If red does not come up he loses the stake. In the example above, if red comes up he gets the stake of $4$ dollars back plus an additional $4$ dollars. If red does not come up he loses his stake of $4$ dollars. The probability of red coming up is $\frac{18}{38}$. What is his expected gain on one spin?

I thought that I would try to use basic expected value rules and say $\sum_{x=1}^n xP(x)$ where I used the random variable $X$ as the variable for total gains/loses. I ended up with the answer just above $1$ dollar. I'm not sure what I'm doing wrong. Although this sounds like a homework problem, it is not, it's simply out of a probability text book and I'm studying for an exam.

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up vote 3 down vote accepted

Let the average value of the wager be $W$. We'll figure out what this is later.

100% of the time, he spends $W$ as his wager. With probability $p=18/38$, he wins $2W$, leading to a net gain of $W$. With $q=1-p$ probability, he wins nothing, leading to a net loss of $-W$.

Hence, his expectation on each spin is $W(18/38)-W(20/38) = -2W/38 = -W/19$.

Now, the average wager is the expectation of the roll of a die. This is an easy one; I'll just stipulate that it's $7/2=3.5$. The expectation is then $-7/38$.

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thank you thats a perfect explanation. –  TheHopefulActuary Nov 14 '12 at 19:15
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"100% of the time, he spends W as his wager" does not look right if W is the average amount of the wager (it is in fact 0% of the time here) but it does not affect the answer –  Henry Nov 14 '12 at 19:16
    
Fair point. It's more precise to say he spends $W$, which is a random variable, and then find the expectation of $W$. –  Muphrid Nov 14 '12 at 19:40
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