# Probability question (2 people rolling dice: First to roll 1 or 2 wins, but two 6's = tie)

Suppose A and B are playing a game where they take turns rolling a standard die. If a 1 or 2 is rolled, the person who rolled that number wins. If a 3, 4, or 5 is rolled, the other person gets a turn. If two 6's a rolled in a row, the game ends in a tie. If A rolls first, what is the probability that A wins the game?

Here is what I have done:

If rolling a 6 means other player gets a roll (like 3,4,5), then the computation of this simpler case is: $\inline&space;Pr(A\;&space;wins,\;&space;no\;&space;tie)&space;=&space;\left&space;(&space;\frac{1}{3}&space;\right&space;)&space;+&space;\left&space;(&space;\frac{2}{3}&space;\right&space;)^{2}\left&space;(&space;\frac{1}{3}&space;\right&space;)&space;+&space;\left&space;(&space;\frac{2}{3}&space;\right&space;)^{4}\left&space;(&space;\frac{1}{3}&space;\right&space;)&space;+&space;...&space;=&space;\left&space;(&space;\frac{1}{3}&space;\right&space;)\left&space;(&space;\frac{1}{1-\frac{2}{3}^{2}}&space;\right&space;)&space;=&space;.6$

Denoting x as the event that a 1 or 2 is rolled, y that 3,4 or 5 is rolled; and z that 6 is rolled (where two 6's in a row means a tie): $\inline&space;P(tie)&space;=&space;P(zz)+P(yzz)+P(yyzz)+...&space;=&space;\frac{1}{36}&space;+&space;\frac{1}{2}\cdot&space;\frac{1}{36}+&space;(\frac{1}{2})^{2}\cdot&space;\frac{1}{36}+...&space;=&space;\frac{1}{36}\cdot&space;\frac{1}{1-(\frac{1}{2})}&space;=&space;\frac{1}{18}$

Therefore, (maybe): $\inline&space;Pr(A\;&space;wins)&space;=&space;\left&space;[1-&space;Pr(tie)&space;\right&space;]\cdot&space;Pr(A\;&space;wins,\;&space;no\;&space;tie)&space;=&space;\left&space;(&space;1-\frac{1}{18}&space;\right&space;)\times&space;0.6&space;=&space;\frac{17}{30}$ is the solution.

Whilst the result is intuitive (to me), I'm concerned that the above is 'biased' in some way by the fact that A rolls first, and there can't be a tie on the first roll. I have attempted to compute P(A wins) directly, e.g. using P(A Wins) = P(x)+P(yyx)+P(zyx)+P(yzx)+... but it was much too complicated to compute the limiting sum. Any help would be appreciated.

We need to introduce a state variable to tell us whether or not the prior roll was a $6$. to do that, let's have a few variables:

$p_0$ denotes the probability that the player whose turn it is will eventually win, given that the prior roll was not a $6$.

$p_1$ denotes the probability that the player whose turn it is will eventually win, given that the prior roll was a $6$.

$t_0$ denotes the probability that the game will end in a tie given that the prior roll was not a $6$.

$t_1$ denotes the probability that the game will end in a tie given that the prior roll was a $6$.

Thus, the answer you seek is $p_0$. We note that we have four variables, thus will want four equations.

To get the first equation, consider the first toss. We get $$p_0=\frac 26\times 1+\frac 36\times (1-p_0-t_0)+\frac 16\times (1-p_1-t_1)$$ So $$\boxed {9p_0=6-3t_0-p_1-t_1}$$

Also we get $$t_0=\frac 26\times 0+ \frac 36\times t_0+\frac 16\times t_1$$ so $$\boxed {3t_0=t_1}$$

Now consider a roll given that the prior roll was a $6$. We see that $$p_1=\frac 26\times 1 +\frac 36\times (1-p_0-t_0)+\frac 16\times 0$$ So $$\boxed{6p_1=5-3p_0-3t_0}$$ And $$t_1=\frac 26\times 0 +\frac 36\times t_0+\frac 16\times 1$$ So $$\boxed {6t_1=3t_0+1=t_1+1\implies t_1=\frac 15\implies t_0=\frac 1{15}}$$

We now rewrite the two equations in $p_1,p_0$ to get $$9p_0=6-\frac 25-p_1\implies 45p_0=28-5p_1$$ $$6p_1=5-3p_0-\frac 15\implies 30p_1=24-15p_0$$ From the first we see that $$5p_1=28-45p_0\implies 30p_1=168-270p_0$$ whence $$168-270p_0=24-15p_0$$ Which, implies $$\boxed {p_0=\frac {144}{255}=\frac {48}{85}}$$

A little additional effort yields $$p_1=\boxed{\frac {44}{85}}$$

Note I: I am a bit surprised that $p_1$ is so close to $p_0$, but it is hard to get intuition here. I don't think the method is conceptually flawed, but of course an arithmetic blunder is certainly possible. I'd check the calculation carefully.

Note II: Just to stress, $p_1$ is not the probability that the second player wins. Far from it. Indeed, at the start of the game three outcomes are possible: Tie, first player wins, second player wins. Call the probability that the second player wins $q_0$. We see that $$p_0+q_0+t_0=1$$ whence (assuming the earlier calculations were correct) $$1= \frac {48}{85}+q_0+\frac 1{15}\implies q_0=\frac {94}{255}$$ Thus, at the start, the probability that the first player wins is $0.564705882$, the probability that the second player wins is $0.368627451$, and the probability of a tie is $0.066666667$. Once again, this assumes that no blunder was made in the earlier computation.

• Thank you! A clever method of attempting this problem, but will take some time to look through it! – Anna Efron Feb 12 '17 at 14:42
• As a suggestion: I'd simulate it. Should be easy enough to run a million trials or so. Of course, the closed form solution is better, if it's correct. But the simulation should give a good sense of the numerical result and that might be helpful. To be sure, simulations can also contain flaws. – lulu Feb 12 '17 at 15:11