Probability Questions! Alex, Bret, and Chloe repeatedly take turns tossing a fair die. Alex begins; Bret always follows Alex; Chloe always follows Bret; Alex always follows Chloe, and so on. Find the probability that Chloe will be the first one to toss a six.
 A: A natural approach involves summation of series. We look at the problem another way.
Let $p$ be the required probability. The event "C is first to get a $6$" can happen in two ways:
(i) In the first $3$ tosses, the results are not $6$, not $6$, $6$. The probability of this is $\left(\frac{5}{6}\right)^2\cdot\frac{1}{6}$.
(ii) In the first $3$ tosses, the results are not $6$, not $6$, not $6$, but C ultimately is first to get a $6$. The probability of this is $\left(\frac{5}{6}\right)^3p$. 
It  follows that
$$p=\left(\frac{5}{6}\right)^2\cdot\frac{1}{6}+\left(\frac{5}{6}\right)^3p.$$
This is a linear equation in $p$. solve. 
A: The third player in line wins if the first player fails to roll a six ($5/6$) and then the second player in line wins.  So $p_3 = \frac{5}{6}p_2$.  Similarly, $p_2 = \frac{5}{6}p_1$.  Finally, the first player in line wins if he rolls a six immediately ($1/6$) or if he fails to roll a six ($5/6$) and then the third player in line wins:
$$
p_3 = \frac{5}{6}p_2 = \frac{5}{6}\left(\frac{5}{6}p_1\right)=\frac{5}{6}\left(\frac{5}{6}\left(\frac{1}{6}+\frac{5}{6}p_3\right)\right)=\frac{25}{216}+\frac{125}{216}p_3.
$$
Solving this gives
$$
p_3=\frac{25}{216-125}=\frac{25}{91};
$$
then
$$
p_1=\frac{1}{6}+\frac{5}{6}\cdot\frac{25}{91}=\frac{91+125}{6\cdot 91}=\frac{36}{91}
$$
and
$$
p_2=\frac{5}{6}\cdot\frac{36}{91}=\frac{30}{91}.
$$
These satisfy $p_1+p_2+p_3=1$, as they must.
A: Let $S$ be the event that a six is thrown, and $\bar{S}$ the event that a six is not thrown.
Then the patterns with which Chloe wins are $(\bar{S} \bar{S} \bar{S}) ^k\bar{S}\bar{S} S $, hence the probability that she wins is given by 
$\sum_{k=0}^\infty ({ 5 \over 6})^{3k+2} {1 \over 6} = { 5^2\over 6^3} { 1\over 1-({5 \over 6})^3}$.
