A probability problem by Soumava Pal

Let $P(X)$ be the probability that player $X$ wins.
We can think in this way:
'If $A$ got a non-six, start a new game in which $B$ rolls first, then $C$ , then $A$ .'
So $P(B)=\frac 56 P(A)$
Similarly, $P(C)=\frac 56 P(B)$
After some calculations of ratios,
$\frac ab=\frac{36}{91},\frac cd=\frac{30}{91},\frac ef=\frac{25}{91}$
Answer $=36+91+30+91+25+91=364$

If Player $X$ wins we write $X$ , else we write $X'$ .

Also we note that the probability of getting a 6 is $\frac{1}{6}$ while that of getting any of the other 5 numbers is $\frac{5}{6}$ .

So $A$ wins if any of the following events happen,

$A, A'B'C'A, A'B'C'A'B'C'A, A'B'C'A'B'C'A'B'C'A, ...$

Thus the total probability of $A$ winning is $\frac{1}{6}+\frac{5.5.5.1}{6.6.6.6}+\frac{5.5.5.5.5.5.1}{6.6.6.6.6.6.6}+...=\frac{1}{6}+(\frac{5}{6})^3\frac{1}{6}+((\frac{5}{6})^3)^2\frac{1}{6}+...=\frac{\frac{1}{6}}{1-(\frac{5}{6})^3}=\frac{36}{91}$ (by the formula for the sum of an infinite geometric progression)

Similarly $B$ wins if any of the following events happen,

$A'B, A'B'C'A'B, A'B'C'A'B'C'A'B, A'B'C'A'B'C'A'B'C'A'B, ...$

and the total probability of $B$ winning the game will come out to be $\frac{30}{91}$ .

Similarly the total probability of $C$ winning the game turns out to be $\frac{25}{91}$ .

So we get $36+91+30+91+25+91=364$

The first three rolls give $6^3$ possible outcomes.

In the $5^3$ cases where there is no six, nobody wins. However, the probabilities in a second (third, fourth, ...) round are the same as in the first round. Therefore we only need to consider the $6^3 - 5^3$ cases in which there is actually a winner.

The probabilities are of the form $P(i) = \frac{5^{i-1}\cdot 6^{3-i}}{6^3 - 5^3}.$ The numerator and denominator are coprime. If we add them all together, we get $\sum (\text{numerators} + \text{denominators}) = (6^3 - 5^3) + 3\cdot (6^3-5^3) = 4\cdot (6^3 - 5^3) = 364.$

Generalization: With $n$ players and a die that has $k$ possible outcomes, the probabilities will be $P(i) = \frac{(k-1)^{i-1}\cdot k^{n-i}}{k^n-(k-1)^n},\ \ \ \ i = 1, \dots, n$ with coprime numerators and denominators; adding them all gives a total of $(n+1)(k^n-(k-1)^n)$ .

Hack:

The sample space is not $6^3 = 216$ . If all three throw non-sixes, the turn passes back to A, and we're effectively in an identical situation as the start. Thus the sample space is only when some of them throw a six in the first throw, which is $6^3 - 5^3 = 91$ . Thus we know that $P(A), P(B), P(C)$ have denominator 91 before simplifying. Additionally, they add up to 1, so their numerators sum up to 91. Adding all of them together, this gives 364; since there is no smaller number in the choices, this must be the answer.

$P(A) = \frac 16 + \frac 56 \times \frac 56 \times \frac 56 P(A) \Rightarrow P(A)=\frac {36}{91}$

$P(B) = \frac 56 \times \frac 16 + \frac 56 \times \frac 56 \times \frac 56 P(B) \Rightarrow P(B)=\frac {30}{91}$

$P(C) = \frac 56 \times \frac 56 \times \frac 16 + \frac 56 \times \frac 56 \times \frac 56 P(C) \Rightarrow P(C)=\frac {25}{91}$