Suppose $A$ and $B$ roll a pair of dice in turn, with $A$ rolling first. Assume the rolls are independent. $A$ wants to obtain a sum of $6$ and $B$ a sum of $7$. The game ends when either player reaches the objective and that player is declared a winner.
(1) Find the expected number of rolls of the dice
(2)Find the variance of the number of rolls of the dice
My idea is as follows:
Let random variable $X$ means $A$ wins the game and $Y$ means $B$ wins the game. $N$ is the number of the rolls of the dice.
I can get the probability of the $A$ wins in the first turn , that is,
$$\mathbb{P}(X|N=1)=\frac{5}{36}$$
and the probability of the $A$ wins in the $2k+1$'s turn
$$\mathbb{P}(X|N=2K+1)=(\frac{31}{36})^k(\frac{5}{6})^k\frac{5}{36}$$
and the probability that $A$ wins that
$$\mathbb{P}(X)=\frac{5}{36}\sum_{k=0}^{\infty}(\frac{155}{216})^k=\frac{30}{61}$$
But the expectation of $N$ I am not sure. Is it to use the
$$\mathbb{E}(N)=\sum_{k=0}^{\infty}k\mathbb{P}(N=k)???$$
I try to use the law of the total of expectation to get a recursion.
Best Answer
I’m not sure that knowing the overall probability that $A$ wins helps you all that much here. Going with your approach, let $X$ be the r.v. that counts the number of rolls, $p_5=5/36$ the probability of rolling a five and $p_6=6/36$ the probability of rolling a six, and $q_i=1-p_i$. (I use $X$ instead of your $N$ because I use the standard notation of $N$ for a certain matrix later.) We then have $$\Pr(X=2k+1) = (q_5q_6)^kp_5 \\ \Pr(X=2k) = q_5(q_6q_5)^{k-1}p_6.$$ The expected number of rolls is then $$\mathbb E[X] = \sum_{k=0}^\infty(2k+1)(q_5q_6)^kp_5 + \sum_{k=1}^\infty (2k)q_5(q_6q_5)^{k-1}p_6.$$ With some painstaking work or the help of a computer we can evaluate these sums to get $$\mathbb E[X] = {402\over61} \approx 6.59.$$ Similarly, you can slog through the sums for $\mathbb E[X^2]-\mathbb E[X]^2$ to compute the variance.
There’s a much simpler way to compute the expected length of the game, though. Whenever B fails to roll a 6, we effectively start the game over. On the other hand, when A fails to roll a 5, we start a related game with the same victory conditions, but with $B$ going first. Let $x$ be the expected length of the first game and $y$ the expected length of the second. We then have the system $$x = 1+q_5y \\ y = 1+q_6x.$$ Solving gives $x=402/61$ as before.
Finally, you could model the game with an absorbing Markov chain. Its transition matrix is $$P = \left[\begin{array}{c|c}Q&R\\\hline\mathbf 0&I\end{array}\right] = \left[\begin{array}{cc|cc}0&q_5&p_5&0\\q_6&0&0&p_6\\\hline0&0&1&0\\0&0&0&1\end{array}\right]$$ with corresponding fundamental matrix $$N = (I-Q)^{-1} = \begin{bmatrix}{216\over61}&{186\over61}\\{180\over61}&{216\over61}\end{bmatrix}.$$ The expected game length is the first element of $\mathbf t = N\mathbf 1$, which agrees with the value computed above, and the variance in game length is the first element of $(2N-I)\mathbf t-\mathbf t\circ\mathbf t$. Here $\circ$ indicates the Hadamard (elementwise) product: $\mathbf t\circ\mathbf t$ is the vector with elements equal to the squares of the elements of $\mathbf t$.