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A roulette wheel has 21 red, 21 black numbers and 4 zeros, A player places $20 bets on red numbers. If the roulette wheel spins a red number the player gets his $20 back and wins another $20. If the roulette does not spin a red number the player loses his bet. the player sits down with $40 and keeps playing until he has no money. How many spins can he expect to be able to play until he has no money?

This problem has got me in a bit of a pickle. I know I have to use the binomial theory but I am getting confused on what you would consider the number of successful trials.

Eg. to go bust after 2 spins you would have to land on a colour other then red twice. This probability is 23/44 and each event is independent.

For 4 spins there are 2 ways to go bust. For 6, there are 4 ways to go bust and so one and so forth.

I am not sure if I am heading the right path but the probability of going bust after 4 attempts say is:

Pr(x=4) = 2!/(2-4)!*4! * (21/44)^4!(23/44)^(4-2)...

Am I heading down the right path?


marked as duplicate by JMoravitz, Claude Leibovici, user91500, kingW3, George Law Aug 18 '17 at 11:40

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  • 2
    $\begingroup$ The picture gets truncated on my screen. Please type up the problem. MathJax will help a lot with your final expression. $\endgroup$ – Ross Millikan Aug 17 '17 at 14:51
  • $\begingroup$ Here's a link to a quick MathJax tutorial, but even typing it with ascii is better than relying on a picture some of the times. As for your question, I assume (but because of your picture not containing the final lines cannot be certain) that you are asking for the probability of a roulette player going bust in exactly $k$ turns given that he always bets on red, correctly betting on red pays out $2:1$, he always bets $\$20$ and never stops, there are both 0 and 00 spaces, and starts with $\$40$ $\endgroup$ – JMoravitz Aug 17 '17 at 15:06
  • $\begingroup$ It is not clear where the expression you wrote for $Pr(x=4)$ came from (assuming my above interpretation to the problem is correct) but recognize that to go bust in exactly $4$ turns, you must win once and lose three times and further the first two games cannot simultaneously be losses. This should have been then $\frac{2\cdot 21\cdot 23^3}{44^4}$. In general, counting the number of ways to lose in exactly $k$ turns can be done in a similar manner to counting catalan numbers, specifically look at technique used in second proof. $\endgroup$ – JMoravitz Aug 17 '17 at 15:22
  • $\begingroup$ I added the typed question :). @JMoravitz I used Pr(x=4) to indicate loss after 4 spins. $\endgroup$ – MagicMagician Aug 18 '17 at 1:56
  • $\begingroup$ dollar signs are used to initiate math mode for mathjax. Use the word dollars instead until you learn how to mathjax $\endgroup$ – JMoravitz Aug 18 '17 at 2:01

Hint: when counting how many sequences of wins and losses he can take to go bust after exactly $2k$ turns, pretend that before the $2k$ turns happened he began at $\$0$ and won twice to begin. The first and last games then will have been a win and a loss respectively, but the middle $2k$ games will have a special property/pattern about them.

They form a Dyck word and can be counted using Catalan numbers. There should have been $1$ way to lose in two turns, $2$ ways to lose in four turns, $5$ (not four) ways to lose in six turns, etc...

Using ('s for wins and )'s for losses, red for the nonexistent two wins at the beginning and an underline to highlight the specific pattern, we have the following possibilities to end in two, four, six turns:









These patterns in the underline are precisely what Catalan numbers count.

The pdf is then $Pr(X=k)=\begin{cases}0&\text{if}~k~\text{is odd}\\C_{k/2}\frac{21^{k/2-1}25^{k/2+1}}{46^{k}}&\text{if}~k~\text{is even}\end{cases}$ where $C_n$ is the n'th Catalan number, $C_n=\binom{2n}{n}/(n+1)$


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