# Expectation value of repeated dice throws

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?

• The quick way would be to let $X$ be the random variable for the sum. Then we have $E[X] = \frac{1}{6}(1+2+3+4+5+(6+E[X]))$. Solve using simple algebra. As for evaluating your double sum... you can factor the $\frac{1}{6^n}$ out of the inner summation... leaving you with the product of two unrelated sums... the first of which is geometric, the second of which is just adding the whole numbers $1$ through $6$. Commented Mar 26 at 17:38

$$\mathbb{E}[Y] = 1\times P(1) + 2\times P(2) + 3\times P(3) + \cdots + (6+\mathbb{E}[Y]) \times P(6)$$
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 $$\sum\limits_{n=1}^{\infty} \sum\limits_{i=1}^{6} i\cdot \frac{1}{6^n} = \sum\limits_{i=1}^{6} i\cdot \sum\limits_{n=1}^{\infty} \frac{1}{6^n} = 21 \cdot \frac15= \frac{21}{5}$$.
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$$\sum_{n=1}^\infty6^{-n}=\sum_{n=1}^\infty\left({1\over6}\right)^n = {1\over5}$$
• I think that's wrong, wolframalpha says it's $\frac{21}{5} = 4.2$ Commented Mar 26 at 18:09