# Trials until First Success

## Geometric Distribution

The geometric distribution describes the number of Bernoulli trials until the first success. According to the distribution, the mean number of trials until success is $$\frac{1}{p}$$, where $$p$$ is the probability of success. From the geometric distribution, we can expect to toss a die $$\frac{1}{\frac{1}{6}} = 6$$ times before landing a $$6$$.

## Another Approach

Let $$X$$ be the number of trials until the first success (including the success). Given a probability of success of $$\frac{1}{6}$$, we can write $$E[X]$$ as:

$\begin{split} E[X] &= \sum_{i=0}^{\infty} (i + 1)\left( \frac{5}{6} \right)^i \left( \frac{1}{6} \right) \\ &= \left( \frac{1}{6} \right) \left[ \sum_{i=0}^{\infty} i\left( \frac{5}{6} \right)^i + \sum_{i=0}^{\infty} \left( \frac{5}{6} \right)^i \right] \end{split}$

Since $$|\frac{5}{6}| < 1$$, the second infinite geometric series converges, and has the value $$\frac{1}{1 - \frac{5}{6}} = 6$$. Manipulating the second series gives the following:

$\begin{split} \sum_{i=0}^{\infty} i\left( \frac{5}{6} \right)^i &= \sum_{n=1}^{\infty} \sum_{m=1}^{n} \left( \frac{5}{6} \right)^n \\ &= \sum_{m=1}^{\infty} \sum_{n=m}^{\infty} \left( \frac{5}{6} \right)^n \\ &= 6 \sum_{m=1}^{\infty} \left( \frac{5}{6} \right)^m \\ &= 30 \end{split}$

Substituting back into the original equation for $$E[X]$$, we get $$E[X] = \frac{1}{6} \left( 30 + 6 \right) = 6$$.