Raashan, Sylvia, and Ted play the following game. Each starts with $\$1$. A bell rings every $15$ seconds, at which time each of the players who currently have money simultaneously chooses one of the other two players independently and at random and gives $\$1$ to that player. What is the probability that after the bell has rung $2019$ times, each player will have $\$1$? (For example, Raashan and Ted may each decide to give $\$1$ to Sylvia, and Sylvia may decide to give her dollar to Ted, at which point Raashan will have $\$0$, Sylvia will have $\$2$, and Ted will have $\$1$, and that is the end of the first round of play. In the second round Rashaan has no money to give, but Sylvia and Ted might choose each other to give their $\$1$ to, and the holdings will be the same at the end of the second round.)
Source: AMC $2019$B Problem $22$
I am trying to solve the problem by recursion. Other solutions are already presented on the link above. However, I can't spot the error in my solution.
Let $a_n$ be the probability that after the bell has rung $n$ times, everyone will have one dollar each.
Let $b_n$ be the probability that after the bell has rung $n$ times, Raashan, Sylvia, and Ted will get $2,1,0$ respectively.
First we try to find $a_{n+1}$
- From $(1,1,1)$ there are $2^3=8$ possible ways to distribute it. It's easy to see that there are exactly two ways to get to $(1,1,1)$ again.
- From $(2,1,0)$ there are $4$ ways to distribute it. Only $1$ way to get $(1,1,1)$. Taking permutations into account, there are $3!=6$ ways.
So we have $$a_{n+1}=\frac{a_n}{4}+\frac{6b_n}{4}=\frac{a_n}{4}+\frac{3b_n}{2}$$ Next we try to find $b_{n+1}$.
- There are $2^3=8$ ways to distribute $(1,1,1)$. Only one results in $(2,1,0)$.
- There are $4$ ways to distribute $(2,1,0)$. Only one results in $(2,1,0)$
So we have $$b_{n+1}=\frac{a_n}{8}+\frac{b_n}{4}$$
Solving with the initial conditions of $a_0=1,b_0=0$ (There's no need to solve it given that options are simple suggests that there is most likely a pattern here). We get that $$a_n=2^{-2n-1}((1-\sqrt{3})^n + (1+\sqrt{3})^n)).$$ But $a_{2019}$ is not in the option.
The $a_{n+1}$ is correct. The $b_{n+1}$ isn't. $(2,1,0)$ can be reached from $(1,1,1),(2,1,0),(0,2,1), (2,0,1)$. So it should be $$b_{n+1}=\frac{a_n}{8} + \frac{3b_n}{4}$$ since there three out of four ($(2,1,0),(0,2,1), (2,0,1)$) are valid.