# Expectation of trials until success

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## Example 1 - Fair coin

### Problem

What is the average number of tosses we need until reaching a head, given $P(head) = p = 0.5$

### Solution

The probability of tossing only once and getting a head $P(n = 1) = 0.5$

The probability of tossing twice to get a head $P(n = 2) = P(T, H) = (\frac{1}{2})^2$

The probability of tossing 3 times to get a head $P(n = 3) = P(T, T, H) = (\frac{1}{2})^3$

The probability of tossing $t$ times to get a head $P(n = t) = P(T_1, T_2, \dots, T_{t-1}, H) = (\frac{1}{2})^{t}$

The expectation of number of tosses until head

\begin{align} \newcommand{\E}[1]{\text{E}[#1]} \newcommand{\blueseries}{\color{blue}{\left[ \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \dots \right]}} \E{n} &= \frac{1}{2} + 2\times\frac{1}{4} + 3\times\frac{1}{2^3} + \dots \\ &= \blueseries + \left[ \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \dots \right] \end{align}

Note that the infinite sum of the geometric series

\begin{align} S &= \sum\limits_{k=0}^{\infty} a r^k = \frac{a}{1 - r} \\ \blueseries &= \frac{1}{2 (1 - \frac{1}{2})} \\ &= 1 \end{align}

The expectation above can be written as the infinite sums of several geometric series

\begin{align} \newcommand{\bluesum}{\color{silver}{\sum\limits_{k=0}^{\infty} \frac{1}{2} \left( \frac{1}{2} \right) ^2}} \E{n} &= \bluesum + \frac{1}{2} \bluesum + \frac{1}{4} \bluesum + \dots \\ &= 1 + \frac{1}{2} + \frac{1}{4} + \dots \\ &= 1 + S \\ &= 2 \end{align}