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Suppose $A$, $B$, and $C$ roll a die in turns. What is the probability that $A$ is the second person to roll a $6$ for the first time?

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  • $\begingroup$ Perhaps you could model this as a Bernoulli process and set $T=$ throw number where the second occurance of 6 occurs. Then you would have to consider it $\mod 3$ ... $\endgroup$ – Matti P. Aug 16 at 6:05
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Avoiding explicit infinite sums:

Let's try a simpler question: two people X and Y are playing, with X rolling first. What is the probability $p_x$ that X is the first to get a $6$?

Considering what might happen on the first roll (either X rolls a $6$ and is first to do so or X does not roll a $6$ with Y becoming the first to roll next), you have $p_x = \frac16 + \frac56 p_y$ and $p_y = \frac56 p_x$. You can solve these simultaneous equations to get $p_x=\frac6{11}$ and $p_y= \frac5{11}$

Now let's try your question with $p_a, p_b, p_c$ being the respective probabilities of being second to roll a $6$ in the three-player game. You want to find $p_a$

Again considering what might happen on the first roll, you have $p_a = \frac56 p_c$ and $p_b=\frac16 p_x + \frac56 p_a$ and $p_c=\frac16 p_y + \frac56 p_b$. You can solve these, using the earlier results, to get $p_a=\frac{300}{1001}$, $p_b=\frac{341}{1001}$, $p_c=\frac{360}{1001}$

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So consider A about to succeed at the event. For some positive integer $k$, everybody has rolled $k$ times. A has gotten no sixes (which can happen in $5^k$ ways), one of the other players has already rolled one or more sixes (the player can be chosen in 2 ways and their rolls in $6^k-5^k$ ways), and the third player has not yet rolled a 6 (which can happen in $5^k$ ways). With that set up, A then needs to roll a 6. The probability of all of this is $$\frac{1}{6}\sum_{k=1}^\infty\frac{5^{2k}\cdot2(6^k-5^k)}{6^{3k}}=\frac{300}{1001}\approx0.2997$$

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Since you only consider the cases where players roll a $6$ or not, you may write each turn as a tripple $(Y,Y,N)$, meaning A rolled a 6, B rolled a 6 but C did not. Your game then becomes a sequence of tripples of the previous form, i.e., $$(Y,Y,N), (Y,N,Y),...,(Y,N,N),...$$

Now the probability that A rolls the second 6 for the first time (call this event E) translates to any finite sequence of the form $$(N,N,N), (N,N,N),...,(N,Y,N),...,(Y,N,N)$$ or $$(N,N,N), (N,N,Y),...,(N,N,N),...,(Y,N,N)$$ where all but one tripple have the form $(N,N,N)$. You can write this now via a geometric random variable $X_A\sim G(1/6)$, namely $$P[E] = \sum_{n=2}^{\infty} P[E|Number\;of\;rounds = n]P[Number\;of\;rounds = n]$$ $$\sum_{n=2}^{\infty} P[E|X_A = n]P[X_A = n]$$

Now, assuming independence of the players and the rolls, we can describe the number of $6$ which player B, and analogously C, rolls in $n$ rounds by a binomial $B_n\sim Binom(n,1/6)$, or $C_n\sim Binom(n,1/6)$. The term $P[E|X_A = n]$ has then the form $$P[E|X_A = n] = P[B_n = 1]P[C_n = 0]+P[B_n = 0]P[C_n = 1] = 2P[B_n = 1]P[C_n = 0].$$

From here, you put everything together in $\sum_{n=2}^{\infty} P[E|X_A = n]P[X_A = n]$ and should arrive at $$P[E] = \dfrac{2(1/6)^2}{(1-1/6)}\sum_{n=1}^{\infty}n(1-1/6)^{(3n-1)}=\dfrac{2(1/6)^2}{(1-1/6)}\dfrac{(1/6)^2}{(1-(1/6)^3)^2}.$$

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  • $\begingroup$ I understood the question like that. A rolling the second $6$ in total but her first 6. My bad, I guess... $\endgroup$ – Jfischer Aug 16 at 7:24
  • $\begingroup$ It should work, though, if I just change the $P[B_n = 1]P[C_n = 0]$ to $\sum_{k=1}^{n} P[B_n = k]P[C_n = 0]$. I'll try it later and then adjust the post.Thanks for the remark! $\endgroup$ – Jfischer Aug 16 at 7:27

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