1. ## Finding expected value

A casino patron will continue to make $5 bets on red in roulette until she has won 4 of these bets. On each bet, she will either win$5 with probability $\frac {18}{38}$ or lose $5 with probability $\frac {20}{38}$. If W is her final winnings and X is the number of bets she makes, then, since she would have won 4 bets and lost (X-4) bets, it follows that $W=20-5(X-4)=40-5X$ Hence $E[W]=40-5E[X]=40-5[4/(\frac {9}{19})]=-\frac{20}{9}$ but why is $E[X]=4/(\frac{9}{19})$? 2. Originally Posted by noob mathematician A casino patron will continue to make$5 bets on red in roulette until she has won 4 of these bets.
On each bet, she will either win $5 with probability $\frac {18}{38}$ or lose$5 with probability $\frac {20}{38}$.
If W is her final winnings and X is the number of bets she makes, then, since she would have won 4 bets and lost (X-4) bets, it follows that
$W=20-5(X-4)=40-5X$
Hence $E[W]=40-5E[X]=40-5[4/(\frac {9}{19})]=-\frac{20}{9}$
but why is $E[X]=4/(\frac{9}{19})$?
The number of failures is k = X - 4.

k follows a negative binomial distribution: Negative binomial distribution - Wikipedia, the free encyclopedia

p = 18/38 = 9/19 and the number of successes is r = 4.

$E(k) = r \cdot \left( \frac{1 - p}{p} \right)$.

Therefore $E(X) = E(k + 4) = E(k) + 4 = 4 \cdot \left( \frac{1 - \frac{9}{19}}{\frac{9}{19}} \right) + 4 = \, ....$