# Expected value and standard deviation of roulette

Suppose you are playing multiple independent rounds of roulette where you bet $$x$$ dollars on black then with probability $$18/38$$ you will double that bet (net gain of $$x$$ dollars) and with probability $$20/38$$ you will lose that money.

If you decide to bet 50 dollars on each round, let J be the amount of money you have after playing 20 rounds of roulette. Find $$E(J)$$ and $$SD(J)$$.

The answers are $$E(J)=947.37$$ and $$SD(J)=223.3$$

I am not sure how to approach this problem; initially I tried to think of it as an indicator problem where each indicator $$=1$$ when the $$i$$th round lands black on the roulette, but did not arrive at the correct solution. Any help is appreciated!

• Hint: can you work out the answers for 2 instead of 20 rounds? Now use your results to work them out for 3 rounds ... and iterate. Alternatively, you may view all 20 rounds as independent so use what you know about expectations and variances of sums of random variables. For perspective, note that you need to start this process with at least \$1000 in order to bet on 20 rounds, so the expectation of \$947.37 reflects an expected loss of about 5.3% of your capital. That proportion is, of course, 5.3% = 2/38. – whuber Oct 27 at 20:07
• I was able to work out the expectation that way but am having trouble figuring out the variance, I am unsure of how to calculate it for even just two trials – user301139 Oct 27 at 20:21
• Have you learned that the variance of the sum of independent variables is the sum of their variances? – whuber Oct 27 at 20:29
• yes I have, so it would be 20 times the variance of each round? – user301139 Oct 27 at 20:34