I incidentally came across a probability sum , written as joke in one of the textbook of 3 rd standard . But when I tried it it is as really difficult .
Question : 3 person A,B,C throw a dice alternatively. The one who gets 2 or 4 first wins. Find the probabilities of their winnings.
On
Observe that A's odds on the first try are $\frac{1}{3}$.
B has $\frac{1}{3}*\frac{2}{3}$ (chance that B will get to roll) = $\frac{2}{9}$.
C has $\frac{1}{3}*\frac{7}{9}$ (chance that C will get to roll) = $1-(\frac{1}{3}+\frac{2}{9})$) = $\frac{7}{27}$.
A has $\frac{1}{3}*\frac{8}{27}$ ((chance that A will roll again, $1-(\frac{1}{3}$ + $\frac{2}{9}$ + $\frac{4}{27}$)) = $\frac{8}{81}$.
B has $\frac{1}{3}*\frac{16}{81}$(") = $\frac{16}{243}$, and so on.
As Aidan said, $\frac{n}{1-r}$ expresses the sum of the infinite series $n(r^0+r^1+r^2...)$ and since our ratio is $\frac{8}{27}$ we have A's probability = $\frac{9}{19}$, B = $\frac{6}{19}$, and C = $\frac{4}{19}$.
Hint: Use the formula for the sum of the infinite geometric series, $\frac{n}{1-r}$ where n is the starting value of the series and r is the ratio between successive terms.