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Suppose that your bank roll is \$75.00 and your goal is to win \$5. Your strategy is to spin the roulette wheel and wager \$5 on black (18 red numbers, 18 black numbers, and 2 green house numbers). If you lose, you will double your wager until you eventually win \$5 or are out of money. Find the probability that you lose your \$75 bankroll.

This is a geometric distribution so the probability distribution is:

$\Pr(X=x)=(18/38)(20/38)^x \text{ for } x=0,1,2, \cdots$ where

$X=$ number of rolls until you a win.

So I thought you evaluate the probability until the wager > 75 which in table below is x=4.

$$ \begin{array}{c|lcr} x & \text{Pr(X)} & \text{Wager}\\ \hline 0 & (18/38)(20/38)^0 & 5\\ 1 & (18/38)(20/38)^1 & 10\\ 2 & (18/38)(20/38)^2 & 20\\ 3 & (18/38)(20/38)^3 & 40\\ 4 & (18/38)(20/38)^4 & 80\\ \end{array} $$

The probability of a win after 4 losses is: $\Pr(X=4)=(18/38)(20/38)^4.$

I thought the probability of a loss is just: $1-(18/38)(20/38)^4$, but the solution is just $(20/38)^4$. Can someone explain why its just $(18/38)(20/38)^4$ and not $1-(18/38)(20/38)^4$? Thanks.

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up vote 2 down vote accepted

You lose precisely if the wheel goes against you $4$ times. You will have bet $5$, $10$, $20$, $40$, and your money is all gone.

Remark: We do not have a geometric random variable, since if $X$ has geometric distribution then $X$ can take on all positive integer values. The random variable $X$ here is our total winnings, which are $5$ if W, LW, LLW, or LLLW, and $-75$ otherwise.

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Thanks @AndréNicolas. So what does $1-(18/38)(20/38)^4$ represent then? – user1527227 Jul 19 '13 at 5:14
Misread your comment, starting again! The probability of not getting LLLLW is what you wrote in your comment, if we are allowed to stay for a fifth round and bet. The rules we are operating under make that impossible, since we go home after LLLL, – André Nicolas Jul 19 '13 at 5:29

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