# Expected duration of a die game

In a 2 person game, the player who first obtains a $6$ wins. I'm trying to determine the expected number of die rolls needed before a winner is determined. (One turn consists of two die rolls, assuming that neither player won in that turn).

$\textbf{My attempt:}$

This problem can be modeled by a geometric distribution with probability $p$, and by virtue of this the expectation, $E[X] = \frac{1}{p}$. Now since, the probability of getting a $6$ is $\frac{1}{6}$, then $E[X] = 6$. I'm wondering if this is correct, or am I missing something.

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If the first player throws a $6$, do you stop? If not, suppose they both throw $6$. Is it a tie, or do they keep playing? – Ross Millikan Mar 17 '14 at 17:25
The game seems to be played in turns. The first player throws the die. If he wins, the game stops there. Otherwise the second player gets his first chance and so on. That's what I've assumed while posting my answer. – Parth Thakkar Mar 17 '14 at 17:28
The $6$ is the correct expectation for the number of rolls, on the assumption the game ends immediately after the first $6$. The number of rolls has geometric distribution. – André Nicolas Mar 17 '14 at 17:36

We have to find the expected number of turns for the game to end. That is given by: $$\sum_{r=0}^\infty rP_r\;\;\;\;\;\;\;\text{(by definition)}$$ where $P_r$ is the probability for the game to end in the $r^\text{th}$ turn.

Now, let's see the probability for the game to end in the zeroth turn (that is, the players have just begun playing).

The game will end in the zeroth turn iff either the first player wins or the second player wins. Thus, the probability is $P_0 = \dfrac 16 + \left(\dfrac 56\right)\dfrac 16$

Similarly, $P_1 = \left(\dfrac 56\right)^2\dfrac 16 + \left(\dfrac 56\right)^3\dfrac 16$

It's easy to see that $P_r = \left(\dfrac 56\right)^{2r}\dfrac 16 + \left(\dfrac 56\right)^{2r+1}\dfrac 16 = \dfrac{275}{1296}\left(\dfrac{25}{36}\right)^{r-1}$

Thus, the expected number of turns of the game is:

\begin{align} &\sum_{r=0}^\infty r\dfrac{275}{1296}\left(\dfrac{25}{36}\right)^{r-1} \\\\ &=\dfrac{275}{1296} \sum_{r=0}^\infty r\left(\dfrac{25}{36}\right)^{r-1}\\\\ &=\dfrac{275}{1296}\dfrac{1}{\left(1-\dfrac{25}{36}\right)^2} = \boxed{\boxed{\dfrac{25}{11}}} \end{align}

where in the last step, I've used the fact that: $$\sum_{r=0}^\infty rx^{r-1} = \dfrac{1}{(1-x)^2} \forall |x| < 1$$

So, the expected number of turns turns out to be (sic) is $\dfrac{25}{11}$.

EDIT:

As suggested by Ross Millikan, it's better to start the counting from $1$. In that case, there's a slight change. Our $P_r$ becomes $\dfrac{11}{36} \left(\dfrac{25}{36}\right)^{r-1}$.

Hence, the 'new' expected number of turns is $\dfrac{36}{11}$.

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The game seems to end whenever either player throws $6$, not when both players throw $6$ – Ross Millikan Mar 17 '14 at 17:23
Yes, that's why the probability of the game ending in the $r^\text{th}$ turn is $\left(\dfrac 56\right)^{2r}\dfrac 16 + \left(\dfrac 56\right)^{2r+1}\dfrac 16$ and not: $P_r = \left(\dfrac 56\right)^{2r}\dfrac 16 \cdot \left(\dfrac 56\right)^{2r+1}\dfrac 16$ – Parth Thakkar Mar 17 '14 at 17:25
The game ends if either the first player puts a six, or the second does so. But for the second to do so, the first must put a non-six. – Parth Thakkar Mar 17 '14 at 17:26
It is a little odd to start counting turns from $0$ instead of $1$. As each turn is the last with probability about $\frac 13$, the expected number of turns should be around $3$. $25$ is way too high. – Ross Millikan Mar 17 '14 at 17:38
I'm wondering if this game was generalized to $k$ dice being thrown simultaneously and the ﬁrst player to obtain a total of k (or more) 6’s, accumulated over all his throws, wins the game. Would this game run for fewer turns? – Millardo Peacecraft Mar 17 '14 at 17:50