<s>[INST] 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.
 [/INST] 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.
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