Expectation value of repeated dice throws

Question

We throw a dice, if we throw a 6, then we throw again (any number of times). Let $X$ be the sum of all thrown numbers. Find $\mathbb{E} (X)$.

I know that if we were just throwing without repeating throws then it would be:

$\mathbb{E} (Y) = 1*P(1) + 2*P(2) + 3*P(3) + ... + 6*P(6) = \frac{7}{2}$

Now if we would get to throw only once after a six throw, then it's ($\mathbb{E} (Z)$):

Let $\mathbb{E}(Y')$ be the expectation value of the second throw after a six was thrown.

$\mathbb{E}(Y') = 1*P_{Y'}(1) + 2*P_{Y'}(2) + ... + 6*P_{Y'}(6)$
$ = 1*\frac{1}{6^2} + ... + 6*\frac{1}{6^2} = \frac{\frac{7}{2}}{6}$

$\mathbb{E} (Z) = \mathbb{E}(Y) + \mathbb{E}(Y')$

But how to calculate the repeated throws after a six is thrown?

Is it the sum up to infinity?

$$\mathbb{E}(X) = \sum_{n=1}^{\infty} \sum_{i=1}^{6} i\cdot \frac{1}{6^n}$$

How can I evaluate this double sum?

Answer 1

You can adapt your first expression to the "rethrow 6s" case with

$$\mathbb{E}[Y] = 1\times P(1) + 2\times P(2) + 3\times P(3) + \cdots + (6+\mathbb{E}[Y]) \times P(6) $$

which will give you $(1-P(6))\times \mathbb{E} [Y] = \frac72$ and so $\mathbb{E} [Y] = \frac{21}{5}$.

JMoravitz's comment says much the same thing, and also suggests you think about $\sum\limits_{n=1}^{\infty} \sum\limits_{i=1}^{6} i\cdot \frac{1}{6^n} = \sum\limits_{i=1}^{6} i\cdot \sum\limits_{n=1}^{\infty} \frac{1}{6^n} = 21 \cdot \frac15= \frac{21}{5}$.

Answer 2

To find the sum: $$ \sum^{\infty}_{n=1}\sum_{i=1}^6\frac{i}{6^n}= 21\sum_{n=1}^{\infty}6^{-n}={21\over5}=4.2 $$ Becuase the sum of$$\sum_{n=1}^\infty6^{-n}=\sum_{n=1}^\infty\left({1\over6}\right)^n = {1\over5}$$