# Difficult probability with a Die

Dirac and Pauli are playing a game with an ordinary six-sided die. Dirac’s target numbers are 1, 2, 3, and Pauli’s target numbers are 4, 5, 6. They take turns in rolling the die, with Dirac going ﬁrst. If the one whose turn it is rolls a target number which he has not previously rolled, he gets to roll again; if he rolls a target number which he has previously rolled, or a number which is not one of his target numbers, his turn ends. The ﬁrst player to have rolled all three of his target numbers (not necessarily all in the one turn) wins. What is the probability that Dirac wins?

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What have you tried? –  Raskolnikov Aug 3 '11 at 10:56
I haven't seen any effective methods to attack this question. To try to get some intuition as to what may be happening, I did a tree diagram, but i) the diagram becomes unwieldy within just 2 moves and ii) I realize it will never end, as there are scenarios where the players keep on rolling non-targets, though the probabilities of these scenarios decreases with each move. The first player to roll a target number disadvantages themselves since it becomes harder for them to roll a target number from then on, so Dirac's chances are probability less than 50%, but that may just be plain wrong. –  Bernard Freeman Aug 3 '11 at 11:08
Yes, that's wrong, since the other player will eventually have to go through the same "disadvantage" in order to win. –  joriki Aug 3 '11 at 11:49

If it is your roll and both players have already hit two of their targets, then write your probability of winning as $p_{2,2}$. If you make your remaining target then you win; if you don't then your chance of losing becomes $p_{2,2}$. So

$$p_{2,2} = \frac{1}{6} + \frac{5}{6}\left(1-p_{2,2}\right)$$

which you can solve for $p_{2,2}$.

Similarly if it is your roll you have already hit one and your opponent two, then your chance of winning is $p_{1,2}$ which (since you either hit or miss) satisfies

$$p_{1,2} = \frac{2}{6} p_{2,2} + \frac{4}{6}\left(1-p_{2,1}\right)$$

$$p_{2,1} = \frac{1}{6} + \frac{5}{6}\left(1-p_{1,2}\right)$$

which are a pair of simultaneous equations you can solve for $p_{1,2}$ and $p_{2,1}$. And the same techniques will work solving in turn single or simultaneous linear equations until you reach $p_{0,0}$, which is the probability at the start for the first person to roll.

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Each player starts with three target numbers. When they roll a target number, it ceases to be one. So if $X$ is a geometric distribution with parameter 3/6, $Y$ is a geometric distribution with parameter 2/6, and $Z$ is a geometric distribution with parameter 1/6, then the number of rolls Dirac must make is a discrete distribution $D = X + Y + Z - 2$ (the $-2$ corresponds to rolling again when he hits one, and is essentially irrelevant). Pauli's distribution $P$ is identical. What you're after is $P(D \le P)$.
$P(D\le P)$, to be precise, since Dirac goes first. In fact, since $P(D<P)=P(D>P)$ by symmetry, we only need to calculate $P(D=P)$, which is a single instead of a double summation. –  joriki Aug 3 '11 at 11:48