# Thread: Simple Rolling Die Question

1. ## Simple Rolling Die Question

Hi

Can someone tell me how would you do this question
A fair die is rolled several times until the first six is obtained
What is the average number of throws needed to get the first six?

now i think its six, but how do you calcuated the answer using geometric distribution?

P.S

2. ## Re: Simple Rolling Die Question

Hello, Paymemoney!

A fair die is rolled several times until the first six is obtained
What is the average number of throws needed to get the first six?

We are dealing with Expected Value.

. . $\displaystyle \begin{array}{cc}\text{Throws} & \text{Prob.} \\ \hline 1 & \frac{1}{6} \\ \\[-4mm] 2 & \frac{1}{6}(\frac{5}{6}) \\ \\[-4mm] 3 & \frac{1}{6}(\frac{5}{6})^2 \\ \\[-4mm] 4 & \frac{1}{6}(\frac{5}{6})^3 \\ \vdots & \vdots \end{array}$

$\displaystyle \begin{array}{cccccc}\text{We have:} &E &=& 1(\frac{1}{6}) + 2(\frac{1}{6})(\frac{5}{6}) + 3(\frac{1}{6})(\frac{5}{6})^2 + 4(\frac{1}{6})(\frac{5}{6})^3 + \hdots \\ \\[-3mm] \text{Multiply by }\frac{5}{6}\!: & \frac{5}{6}E &=& \qquad\;\; 1(\frac{1}{6}(\frac{5}{6}) \;+ 2(\frac{1}{6})(\frac{5}{6})^2 + 3(\frac{1}{6})(\frac{5}{6})^3 + \hdots \\ \\[-3mm] \text{Subtract:} & \frac{1}{6}E &=& \frac{1}{6} \;+\; \frac{1}{6}(\frac{5}{6}) \;+\; \frac{1}{6}(\frac{5}{6})^2 \;+\; \frac{1}{6}(\frac{5}{6})^3 \;+\; \hdots \qquad \end{array}$

$\displaystyle \text{We have: }\;\tfrac{1}{6}E \;=\;\tfrac{1}{6}\bigg[1 + \tfrac{5}{6} + (\tfrac{5}{6})^2 + (\tfrac{5}{6})^3 + \hdots\bigg]$

. . . . . . . $\displaystyle \tfrac{1}{6}E \;=\;\tfrac{1}{6}\left[\frac{1}{1-\frac{5}{6}}\right] \;=\;\tfrac{1}{6}\cdot\frac{1}{\frac{1}{6}} \;=\;\tfrac{1}{6}(6) \;=\;1$

. . . . . . . $\displaystyle \tfrac{1}{6}E \:=\:1 \quad\Rightarrow\quad E \:=\:6$

You were right!