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?

  • $\begingroup$ 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. $\endgroup$ – Bernard Freeman Aug 3 '11 at 11:08
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    $\begingroup$ Yes, that's wrong, since the other player will eventually have to go through the same "disadvantage" in order to win. $\endgroup$ – 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

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

and if it is your roll and you have already hit two and your opponent one

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

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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I'm not going to give you the full answer, but this should get you on track.

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 $P(D \le P)$.

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    $\begingroup$ $P(D\le P)$, 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. $\endgroup$ – joriki Aug 3 '11 at 11:48
  • $\begingroup$ Oops; and nice observation. $\endgroup$ – Peter Taylor Aug 3 '11 at 12:19
  • $\begingroup$ I'm sorry to have to ask for more, but I haven't been able to deduce much yet. I have not done an undergraduate course in probability theory yet so I was unfamiliar with "geometric distributions". I took a read of the Wikipedia article, but I am still quite confused, especially by some notation. For example, if X, Y and Z are distributions, how do we add such things? I believe this question can be solved with just a freshman's knowledge base, so could you please rephrase your hint for someone of that level? Sorry for the hassle and thanks for your help. $\endgroup$ – Bernard Freeman Aug 3 '11 at 12:41
  • $\begingroup$ @Bernard, I can't rephrase for a freshman level because I don't know what that is. I've never done an undergraduate probability course either; just A/S level statistics. Sorry. $\endgroup$ – Peter Taylor Aug 3 '11 at 13:06
  • $\begingroup$ Ok well I've probably been looking at this problem for too long tonight anyway, it's given me a headache. I'll read through the articles on Geometric distributions again tomorrow and I'll see if I can work through it with your hint. Thank you very much. $\endgroup$ – Bernard Freeman Aug 3 '11 at 13:18

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