Q.54 - A First Course in Probability

Question

Consider the following dice game: A pair of dice is rolled. If the sum is$7,$ then the game ends and you win $0.$ If the sum is not $7,$then you have the option of either stopping the game and receiving an amount equal to that sum or starting over again. For each value of$i, i = 2, ... , 12,$ find your expected return if you employ the strategy of stopping the first time that a value at least as large as $i$ appears. What value of$i$leads to the largest expected return? Hint: Let $Xi$ denote the return when you use the critical value $i.$ To compute $E\left[Xi\right]$, condition on the initial sum. 

Step-by-Step Solution

Answer

Largest expected return, $i=8$.

Step 1: Given information

Given in the question that, consider the following dice game: A pair of dice is rolled. If the sum is $7,$ then the game ends and you win $0$. If the sum is not$7,$ then you have the option of either stopping the game and receiving an amount equal to that sum or starting over again. For each value of$i, i = 2, ... , 12.$

Step 2: Explanation

Let $X_i$ denote the return when you use the critical value i.

$E\left[X_i\right]=i,P\left[X_i=i\right]$

$E\left[X_2\right]=2\frac{1}{36}$

$E\left[X_4\right]=4\frac{3}{36}$

$E\left[X_5\right]=5\frac{4}{36}$

$E\left[X_6\right]=6\frac{5}{36}$

$E\left[X_7\right]=7\frac{6}{36}$

$E\left[X_8\right]=8\frac{5}{36}$

$E\left[X_9\right]=9\frac{4}{36}$

$E\left[X_{11}\right]=11\frac{2}{36}$

$E\left[X_{12}\right]=12\frac{1}{36}$

Step 3: Expected return

$E\left[X\right]=E[X\mid$ sum is 7$]+E[X\mid$ sum is not 7$]$

$=0+E[X\mid$ sum is not 7$]$

$+E\left[X_7\right]+E\left[X_8\right]+E\left[X_9\right]+E\left[X_{10}\right]+E\left[X_{11}\right]+E\left[X_{12}\right]$

$=\frac{1}{36}(2+6+12+20+30+40+36+30+22+12)$

$=\frac{210}{36}$

If sum is not $7$ largest expected return,

$E\left[X_8\right]=8\frac{5}{36}=\frac{40}{36}$

If sum $7$largest expected return is $0$.

Therefore, largest expected return, $i=8$.

Step 4: Final answer

Largest expected return, $i=8$.