Simple Rolling Die Question

**Paymemoney** · September 11th 2011, 03:59 PM

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

**Soroban** · September 11th 2011, 07:10 PM

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.

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

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

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

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

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

You were right!

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