First to Six

Solution

Let \(X\) be the number of \(6\)'s a player gets on one turn. Since he rolls \(k\) dice on that turn, \[ X \sim \mathrm{Binomial}\!\left(k,\frac16\right). \] Each player keeps taking turns, adding these \(6\)'s up, until his cumulative total first reaches \(k\). A very clean way to think about this is to ignore the turn structure for a moment. For one fixed player, imagine listing all the dice he will ever roll: after \(r\) turns he has rolled exactly \(rk\) dice in total, so the total number of \(6\)'s he has seen by then is \[ S_r \sim \mathrm{Binomial}\!\left(rk,\frac16\right). \] He has not yet won after \(r\) of his own turns exactly when \(S_r < k\). So define \[ a_r=\Pr(S_r2r)=a_r^2. \] After \(2r+1\) total turns, player 1 has had \(r+1\) turns and player 2 has had \(r\) turns. The game is still going exactly when player 1 is still below \(k\) after \(r+1\) turns and player 2 is still below \(k\) after \(r\) turns, so \[ \Pr(T>2r+1)=a_{r+1}a_r. \] For any positive-integer-valued random variable \(T\), \[ \mathbb E[T]=\sum_{n\ge 0}\Pr(T>n). \] Applying that here gives \[ \mathbb E[T] = \sum_{r=0}^{\infty}\Bigl(\Pr(T>2r)+\Pr(T>2r+1)\Bigr) = \sum_{r=0}^{\infty}\left(a_r^2+a_ra_{r+1}\right), \] where \[ a_r=\sum_{j=0}^{k-1}\binom{rk}{j}\left(\frac16\right)^j\left(\frac56\right)^{rk-j}. \] So the expected number of turns needed to finish the game is \[ \mathbb{E}[T] = \sum_{r=0}^{\infty} \left[ \left( \sum_{j=0}^{k-1}\binom{rk}{j}\left(\frac{1}{6}\right)^j\left(\frac{5}{6}\right)^{rk-j} \right)^2 + \right. \] \[ \left. \left( \sum_{j=0}^{k-1}\binom{rk}{j}\left(\frac{1}{6}\right)^j\left(\frac{5}{6}\right)^{rk-j} \right) \left( \sum_{j=0}^{k-1}\binom{(r+1)k}{j}\left(\frac{1}{6}\right)^j\left(\frac{5}{6}\right)^{(r+1)k-j} \right) \right]. \]