Trials until First Success: Challenging Probability Problem

On the average, how many times must a die be thrown until one gets a 6?

Let $p$ be the probability of a 6 on a given trial. Then the probabilities of success for the first time on each trial are (let $q = 1 - p$ ):

Trial	Probability of success on trial
1	$p$
2	$pq$
3	$pq ^2$
.	.
.	.
.	.

The sum of the probabilities is

\begin{align*} p+pq+pq^2+\ldots & = p\left( 1+q+q^2+\ldots \right) \\ \\ & = \frac{p}{1-q} \\ \\ & = \frac{p}{p} \\ \\ & = 1 \end{align*}

The mean number of trials, $m$ , is by definition,

m = p + 2pq + 3pq^2 + 4pq^3+ \ldots

Note that our usual trick for summing a geometric series works:

qm = pq + 2pq^2+3pq^3 + \ldots

Subtracting the second expression from the first gives

m-qm=p+pq+pq^2+\ldots

m\left( 1-q \right) = 1

Consequently,

mp=1

and

m=1/p

We see that $p=1/6$ , and so $m=6$ .

On the average, a die must be thrown 6 times until one gets a 6.