# Probability theory - Dice

Two guys are playing dice with each wagering $50. Player 1 chooses 2 as his lucky number, and Player 2 chooses 6. Every time their lucky number appears as a result, the player gets one point. The player who gets 3 points first wins$100

Suddenly, the game has to be stopped. Player 1 chalks up 2 points and Player 2 chalks 1 point by then. What is a fair way of splitting $100? Edit: the dice has 6 sides ## 2 Answers Denote the event that (by going on) the first result that belongs to the lucky numbers$2$and$6$is a$2$by$A$and that it is a$6$by$B$. Denote the event that player 1 wins by$W$. Then$P\left(W\right)=P\left(W\mid A\right)P\left(A\right)+P\left(W\mid B\right)P\left(B\right)=1.\frac{1}{2}+P\left(W\mid B\right).\frac{1}{2}=\frac{1}{2}\left(1+P\left(W|B\right)\right)$. Here$P\left(W|B\right)$stand for the probability that player 1 wins by going on if both players have$2$points, so it is evident that$P\left(W\mid B\right)=\frac{1}{2}$. This leads to$P\left(W\right)=\frac{3}{4}$. Player 1 should get$75$and 2 should get$25$This was exactly the type of problem that Pascal and Fermat discussed in length via a series of letters. Pascal thought it was easy. Fermat didn't get it. http://en.wikipedia.org/wiki/Problem_of_points Basically you count up the possible outcomes where one person wins and the number where the other person wins and then split the pot using that ratio. They will keep rolling until they get a 2 or 6. Both 2 and 6 have the same probability of happening... so we can ignore how many times they have to roll before getting one of them. If you imagine they kept rolling even once someone had one.. you get the following options [2,2] [2,6] [6,2] [6,6] In all cases except [6,6] player 1 wins. So player 1 has 3/4 chance.. so should get 3/4 of the pot... i.e. \$75 and player 2 gets \\$25

see @drhab's answer for the clever way to work it out :)

• Interesting link. Thank you. – drhab Oct 23 '13 at 8:31