# Suppose that we roll a fair die until a 6

Suppose that we roll a fair die until a 6 comes up.
a) What is the probability that we roll the die n times?
b) What is the expected number of times we roll the die?
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Wendy Boykin
Step 1
a) If we roll a fair die, then we have 1 chance in 6 of rolling a six and 5 chances in 6 of not rolling a 6.
$P\left(6\right)=\frac{1}{6}$

We roll the die until a 6 is rolled.
Let X be the number of rolls of the die.
When we roll the die n times, then the first $n-1$ rolls cannot be a 6, while the n-th roll has to be a 6:

Note: The variable X has a geometric distribution with $p=\frac{1}{6}$.
Step 2
b) The expected value of a random variable X with a geometric distribution is the reciprocal of the constant probability of success $p=P\left(6\right)$
$E\left(x\right)=\frac{1}{p}=\frac{1}{P\left(6\right)}=\frac{1}{\frac{1}{6}}=6$

Heather Fulton
a) We would have to roll $n-1$ numbers that are not 6 followed by a 6, so the probability is ${\left(\frac{5}{6}\right)}^{n-1}\cdot \left(\frac{1}{6}\right)$
b) $E\left(N\right)=\sum _{n=1}^{\mathrm{\infty }}n\cdot {\left(\frac{5}{6}\right)}^{n-1}\left(\frac{1}{6}\right)=\left(\frac{1}{6}\right){\left(\frac{1}{1-\left(\frac{5}{6}\right)}\right)}^{2}=6$.
This is a nice answer since after 6 rolls we would expect to have rolled exactly one 6.

nick1337

a. If we roll the die n times (assuming $n\ge 1$ and the dice is 6-sided and fair), then we must roll n-1 "not 6" rolls followed by 1 "6" roll. The probability of that is:
$\left(\frac{5}{6}{\right)}^{n-1}\left(\frac{1}{6}{\right)}^{1}$
b. We could roll the die any number of times from 1 to infinite. Consider a random variable R which is the number of rolls to roll a 6. We could determine the expected number in two ways:
i. We could say that R is a geometric RV with a chance of success of $\frac{1}{6}$, so $E\left(R\right)=\frac{1}{\left(\frac{1}{6}\right)}=6$
ii. We could show i. explicitly, using the formula for expected value:
$E\left(R\right)=\sum _{s\in S}R\left(s\right)p\left(s\right)$
$=\sum _{i=1}^{\mathrm{\infty }}\left(i\right)\left(\left(\frac{5}{6}{\right)}^{i-1}\left(\frac{1}{6}{\right)}^{1}\right)$
$=\frac{1}{6}\left(\sum _{i=1}^{\mathrm{\infty }}\left(i\right)\left(\left(\frac{5}{6}{\right)}^{i-1}\right)$
This summation takes the form
$\sum _{i=1}^{\mathrm{\infty }}\left(i\right)\left({r}^{i-1}\right)$, where $|r|<1$
$\sum _{i=1}^{\mathrm{\infty }}\left(i\right)\left({r}^{i-1}\right)=\sum _{i=1}^{\mathrm{\infty }}\frac{d}{dr}\left({r}^{i}\right)$
$=\frac{d}{dr}\sum _{i=1}^{\mathrm{\infty }}\left({r}^{i}\right)$
$=\frac{d}{dr}\frac{1}{1-r}$ (since $|r|<1$)
$=\frac{1}{\left(1-r{\right)}^{2}}$
Let $r=\frac{5}{6}\left(|r|=\frac{5}{6}<1$.) Then,
$E\left(R\right)=\frac{1}{6}\left(\frac{1}{\left(1-\frac{5}{6}{\right)}^{2}}\right)$
$⇒E\left(R\right)=6$
Either way, we get the same answer: 6 rolls.