Q. 4.18 - A First Course in Probability

Question

A casino patron will continue to make $\backslash (5$ bets on red in roulette until she has won $4$ of these bets.

1. What is the probability that she places a total of $9$ bets?
2. What are her expected winnings when she stops? 

Remark: On each bet, she will either win $\backslash )5$ with probability $\frac{18}{38}$ or lose $\$5$ with probability $\frac{20}{38}$.

Step-by-Step Solution

Answer

1. The probability she places all the bet is $\left(\begin{array}{c}8\\ 3\end{array}\right){\left(\frac{9}{19}\right)}^3{\left(\frac{10}{19}\right)}^5\left(\frac{9}{19}\right)$
2. The expected winnings when she stop is $-\frac{20}{9}$
   

Step 1: Given Information (Part a)

A casino patron will continue to make $\$5$bets on red in roulette until she has won $4$ of these bets.

Step 2 : Explanation (Part a)

We have to find the probability that she places a total of bets.

Either she can $\$5$ with probability $\frac{18}{38}$$=\frac{9}{19}$ or she can lose $\$5$ with probability $\frac{10}{38}=\frac{10}{19}$

The patron will continue until she has won all  the $4$ bets.

So out of $9$ bets, she will win the last bet.

So out of remaining $8$ bets, she has to win $3$ bets.

Therefore, the probability she places all the bets is:

$\left(\begin{array}{c}8\\ 3\end{array}\right){\left(\frac{9}{19}\right)}^3{\left(\frac{10}{19}\right)}^5\left(\frac{9}{19}\right)$

Step 3: Final Answer (Part a)

The probability she places all the bets is: $\left(\begin{array}{c}8\\ 3\end{array}\right){\left(\frac{9}{19}\right)}^3{\left(\frac{10}{19}\right)}^5\left(\frac{9}{19}\right)$

Step 4: Given Information (Part b)

A casino patron will continue to make $\$5$bets on red in roulette until she has won $4$of these bets.

Step 5: Explanation (Part b)

If $W$ is his final winnings and $x$ is the number of bets he makes, then since he would have won $4$ bets and lost $x-4$ bets, it follows that

$W=20-5(x-4)=40-5x$

Hence,

$E\left(w\right)=40-5E\left[x\right]$

width="108" style="max-width: none;" $=40-5\left(\frac{4}{{\displaystyle \frac{9}{19}}}\right)$

$=40-5\left(\frac{19\times 4}{9}\right)$

$=-\frac{20}{9}$

Therefore, the expected winnings when she stop is $-\frac{20}{9}$

Step 6: Final Answer (Part b)

The expected winnings when she stop is $-\frac{20}{9}$