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In the bet on a “single number” at roulette, there is chance $\dfrac 1 {38}$ of winning. The bet

pays $35$ to $1$; you can take this to mean that if you win the bet, your net gain is $35$,

and if you lose the bet then you lose $1$ (that is, your net gain is $-1$).

Suppose you bet repeatedly on a single number; assume that the bets are independent

each other. Find the expected value of your net gain from $38$ bets?

$35 (1); -1(37)$

Let $X_1; X_2; \dots X_n$ be independent and identically distributed (i.i.d.)

random variables, and let $S_n = X_1 + X_2 + \dots + X_n$. Then

$E(S_n) = nE(X_1)= 38 \left(35 \cdot \dfrac 1 {38}+(-1)\cdot \dfrac {37} {38}\right)= 38 \cdot\left(\dfrac {-2} {38}\right)=38(-0.0526)= -1.9988$ USD???

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On a single play, your expectation is $$\frac{1}{38}(35)+\frac{37}{38}(-1)=-\frac{2}{38}.$$

For multiple plays, your analysis is correct. Let random variable $X_i$ be your net gain in the $i$-th bet. Then your total gain $Y$ is $X_1+X_2+\cdots+X_n$.

By the linearity of expectation, $$E(Y)=E(X_1+X_2+\cdots+X_n)=E(X_1)+E(X_2)+\cdots+E(X_n)=-\frac{2}{38}n.$$ In particular, if $n=38$ then $E(Y)=-2$.

Remark: Your calculation was basically correct. The small departure from the correct $-2$ is probably due to writing down an intermediate calculator result, and then reentering the number, perhaps not to full accuracy. It is a good thing to learn to use the memory feature of your calculator. It will save on roundoff errors, and on time. It will also help cut down on keying errors, which are a significant source of mistakes.

The independence of your bets plays no role in the calculation, as long as the roulette wheel itself is fair. We have precisely the same expectation if we bet on our favourite number, say $13$, all $38$ times. And remember, on average, the more you play, the more you lose.

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Thanks Andre Nicolas! – statistics-student13 May 4 '13 at 0:54
You are welcome. Have you reached an understanding in your other question (the very persistent dice thrower)? – André Nicolas May 4 '13 at 0:56
Yes after a long while by symmetry! – statistics-student13 May 4 '13 at 1:04
The point is that $16000$ and $160000$ are below the mean. The probability that $A\gt 16000$ is $1$ for all practical purposes. And the probability that $B\gt 160000$ is enormously closer to $1$ still. – André Nicolas May 4 '13 at 1:09
Now I get it why B is more closer! Thanks! – statistics-student13 May 4 '13 at 1:21

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