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A roulette from a casino contain 18 red cases, 18 black cases and 1 green cases. A player shows up with $10$dollars. He decides to bet $1$dollar on red 10 consecutives times. If it's red, he wins a dollar and if not he loses his dollar. Let $S$ be the the amount of money the player has after 10 bets. Find the value $S$ can have, then calculate $P(S<3|S\leq18)$.

So the value S can take is every natural numbers $0\leq S\leq20$

but after that I'm confused on how to proceed. I guess we could do it using a density function but i don't know how.

Any help to point me in the right direction would be greatly appreciated.

Thank you.

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  • $\begingroup$ How should we interpretate $P(S < 3|S\leq 18)$ ? Probability that the player has less than 3 Dollars after 10 bets given that he has less than 19 Dollars after 10 bets ? $\endgroup$
    – callculus42
    Commented Nov 11, 2015 at 2:39
  • $\begingroup$ That is what i understand from the question yes $\endgroup$
    – spexel
    Commented Nov 11, 2015 at 2:41
  • $\begingroup$ Let $p$ be the probability of the win in a single bet. Then $S\sim 2Bin(10,p)$ which you can approximate with a normal $N$ of same mean and variance. Then $P(S<3|S\le 18)\approx \frac {P(N\le 2.5)}{P(N\le 18.5)}$. $\endgroup$
    – A.S.
    Commented Nov 11, 2015 at 3:11

2 Answers 2

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Some hints: First we can use the converse probability:

$P(S\leq 18)=1-P(S>18)=1-P(S=20)-P(S=19)$

In case of $S=20$ the player has to win 10 times. $P(S=19)$ is zero, beause it is not possible.

$P(S<3)=P(S=0)+P(S=1)+P(S=2)$

For $S=0$ the player has to loose 10 times.

$S=1$ is not possible.

For $S=2$ the player has to win once.

Bayes theorem:

$P(S<3|S\leq 18)=\frac{P(S<3 \cap S\leq 18)}{P(S\leq 18)}$

$P(S<3 \cap S\leq 18)=P(S<3)$

Therefore $P(S<3|S\leq 18)=\frac{P(S<3)}{P(S\leq 18)}$

A calculation example:

$P(S=16)=\binom{10}{8}\cdot \left(\frac{18}{37}\right)^8\cdot \left(\frac{19}{37}\right)^2$

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  • $\begingroup$ Here's the answer i got: $\frac{P(S< 3)}{P(S\leq 18)}=\frac{(\frac{19}{37})^{10}+(\binom{10}{1}(\frac{18}{37})(\frac{19}{37})^9)}{1-(\frac{18}{37})^{10}}\approx \frac{0.000015401}{0.99925748}\approx 0.000015516$ correct? $\endgroup$
    – spexel
    Commented Nov 11, 2015 at 18:31
  • $\begingroup$ In general your expression is right. I have the same value for the denominator. The value for the numerator calculated by w.a. is different from yours. See here (You have to mark the whole link and then copy it): wolframalpha.com/input/?i=%28%2819%2F37%29%5E10%2B10*%2819%2F37%29%5E9*18%2F37%29 $\endgroup$
    – callculus42
    Commented Nov 11, 2015 at 18:50
  • $\begingroup$ this is what i get : wolframalpha.com/input/… $\endgroup$
    – spexel
    Commented Nov 11, 2015 at 18:57
  • $\begingroup$ @spexel I have the same final result, $1.34\%$ $\endgroup$
    – callculus42
    Commented Nov 11, 2015 at 19:00
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The value of $S$ cannot take every natural number in that range; though the range itself is okay.

In ten games the count of wins will be somehow distributed.   $W$ is the count of wins in $10$ games with success probability $p$.   What is $p$, and what is the distribution of $W$.

$$\mathsf P(W=k) = \underline{\qquad}\quad[k\in\{0,1,2,3,4,5,6,7,8,9,10\}]$$

Assume the player started with $\$10$, and in each game bets one dollar and gets back two on a win.   So the amount of money $S$ taken away will be $S= \underline{\quad}$ (somehow related to $W$), and thus have distribution:

$$\mathsf P(S=s) = \underline{\qquad}\quad[s\in\{\underline{\qquad}\}]$$

Then by definition of conditional probability:

$$\begin{align}\mathsf P(S< 3\mid S\leq 18) & = \dfrac{\mathsf P(S< 3 \cap S\leq 18)}{\mathsf P(S\leq 18)} \\[1ex] & = \dfrac{\mathsf P(S< 3)}{1-\mathsf P(S> 18)} \\[1ex] & = \underline{\dfrac{(\qquad)}{(\qquad)}}\end{align}$$

Fill in the blanks and complete.

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