A simple version of Expected Value of Iterative Dice Rolling

Probability Level 3

A 6-sided fair dice is rolled and the number is recorded. If the number is not a 6, then stop. Otherwise, roll the dice again and add the number to the previous 6. Continue this process until the dice show a non 6 value.

Here are two examples:

Example 1

Roll 1: the number 5 shown. Stop the process and the total is 5

Example 2

Roll 1: the number 6 shown. So subtotal is 6, rolled again.

Roll 2: the number 6 shown. So subtotal is 12, rolled again.

Roll 3: the number 6 shown. So subtotal is 18, rolled again.

Roll 4: the number 1 shown. So subtotal is 19. Stop the process and the total is 19

The expected total can be written as $\frac{a}{b}$ where $a$ and $b$ are coprime positive integers.

Find the value of $a+b$ .

Inspiration

The answer is 26.

1 solution

Brian Charlesworth
Oct 30, 2018

On the first roll there is a $1/6$ probability for each of the values $1$ through $6$ , but if the $6$ appears then one is essentially "rebooting" the process, so the expected value $E$ will satisfy the equation

$E = \dfrac{1}{6}(1 + 2 + 3 + 4 + 5) + \dfrac{1}{6}(6 + E) \Longrightarrow \dfrac{5}{6}E = \dfrac{21}{6} \Longrightarrow E = \dfrac{21}{5} \Longrightarrow a + b = 21 + 5 = \boxed{26}$ .

This is a beautiful line of reasoning, much more elegant than my method of setting up a summation(+1)

Aareyan Manzoor - 2 years, 7 months ago

A simple version of Expected Value of Iterative Dice Rolling

The answer is 26.

1 solution

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