# Expected value of the number of trials until it occurs with probability p

Hello, guys! It’s me! Warotan!

Japanese version.

This time let us calculate the expected value of the number of trials until it occors with probability $$p$$!
There are so many ways to do it, but here we use summation of inifinite series.
I write this article in case we need it but can’t remember and panic

## Summation of infinite geometric series

First, let us review the formula.

We set the geometric ratio $$r < 1$$ for converge and consider the series

$$$$S = \sum_{i = 1}^{\infty}r^i$$$$

$$$$S = \frac{r}{1-r}$$$$

Below are the specific calculation to find this, please skip it if you don’t need it.

Starting from the partial sum

$$$$S_{n} = \sum_{i = 1}^{n}r^i$$$$

and multiple $$r$$ on it,

$$\begin{eqnarray} S_{n} &=& r+r^2+r^3+\cdots + r^n \\ rS_{n} &=& \,\,\,\,\,\,\,\,\,\,\, r^2+r^3+\cdots + r^n+r^{n+1} \end{eqnarray}$$

Subtract the lower one from the upper one, we get

$$\begin{eqnarray} \left(1-r\right)S_{n} &=& r-r^{n+1} \\ &=& r\left(1-r^n\right) \\ S_{n} &=& \frac{r\left(1-r^n\right)}{1-r}. \end{eqnarray}$$

Letting $$n \to \infty$$, we finally reach the answer as $$r < 1$$.

## Expected value of the number of trials with probability $$p$$

The main part.

We consider an event with probability $$p$$ and calculate the expected value of the trials until it occurs.

Setting the value $$E$$, the answer is

$$$$E = \frac{1}{p}.$$$$

We will derive this below.
The number can be from 1 time to $$\infty$$ times.
According to the definition of an expected value, we can write $$E$$ as

$$\begin{eqnarray} E &=& 1\times p+2\times\left(1-p\right)p+3\times\left(1-p\right)^2 p+\cdots \\ &=& \sum_{k = 1}^{\infty}k\times\left(1-p\right)^{k-1}p \end{eqnarray}$$

and dividing this into the part $$k = 1$$ and others,

$$\begin{eqnarray} E &=& p+\sum_{k = 2}^{\infty}k\left(1-p\right)^{k-1}p \end{eqnarray}$$

Rewriting the index of the second term from $$k \geq 2$$ to $$k \geq 1$$ and do some breaking up,

$$\begin{eqnarray} E &=& p+\sum_{k = 1}^{\infty}\left(k+1\right)\left(1-p\right)^{k} p \\ &=& p+\sum_{k = 1}^{\infty}k\left(1-p\right)^{k} p+\sum_{k = 1}^{\infty}\left(1-p\right)^{k} p \\ &=& p+\left(1-p\right)\sum_{k = 1}^{\infty}k\left(1-p\right)^{k-1} p+\sum_{k = 1}^{\infty}\left(1-p\right)^{k} p \end{eqnarray}$$

Here, the summation of the second term of the right hand side is exactly the same of the original form $$E$$,

$$\begin{eqnarray} E &=& p+\left(1-p\right)E+\sum_{k = 1}^{\infty}\left(1-p\right)^{k} p, \end{eqnarray}$$

and the third one is just an infinite geometric series and easily calculated from the formula above,

$$\begin{eqnarray} \sum_{k = 1}^{\infty}\left(1-p\right)^{k} p &=& p\times\frac{1-p}{1-\left(1-p\right)} \\ &=& p\times\frac{1-p}{p} \\ &=& 1-p \end{eqnarray}$$

Finally we get

$$\begin{eqnarray} E &=& p+\left(1-p\right)E+\left(1-p\right) \\ &=& 1+\left(1-p\right)E, \end{eqnarray}$$

and

$$$$E = \frac{1}{p}$$$$

However, it’s not so hard to see why.
I feel like an event with probability $$p = 1/n$$ happen if we do it $$n$$ times and $$p = 2/n$$ needs $$n/2$$ times.

## Summary

Now, that’s all what I wanted to write.

This time, we calculated the expected value of the number of trials until it occurs with probability $$p$$.

If you have any comment, please let me know from the e-mail address below or the CONTACT on the menu bar.
tsunetthi(at)gmail.com

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