Dice Game: Probability of rolling three 6's before my opponent

Question

Let's say I'm playing a dice game with a friend. We each get two dice, and we roll all of our dice at the same time. You get one point for each 6 you have. First to score at least three points wins. If both of us get to 3 (or more) points on the same trial, it is a tie.

1. What is the probability that I win? lose? tie?
2. What is the expected number of rolls required for either of us to win?
3. If I replace one of my dice with an unfair dice which rolls 6 2/6 times, how does that affect the above answers?

More info: I'm actually trying to design a slot game (5 reels). If a particular symbol appears on reels 1 or 2, the player gets a point, if the symbol appears on reels 4 or 5 the computer gets a point. First to three points wins. In order to balance the payout of the machine, I need to know the number of trials expected before the game ends, and how often the player wins vs the computer wins or ties.

Answer 1

This is an example of an absorbing Markov chain, where the states are Player 1 wins, Player 2 wins, tie, and then any combination of Player 1 having 0/1/2 points and Player 2 having 0/1/2 points. We can construct a Markov matrix where the jth entry in ith row is the probability of going from state i to state j after one two-die roll by both players. We can determine these probabilities pretty easily, since the player scores are independent of each other - e.g. the probability of going from P1 having 1 point and P2 having 0 points to P1 having 2 points and P2 having 2 points is simply the product of the probabilities of P1 getting 1 point and P2 getting 2 points.

Once we have the matrix, we can apply the formulas for Markov matrices to answer your questions - by raising the Markov matrix to an infinite power (or approximating the result of this by raising it to a high power) we obtain a matrix which gives us the expected distributions of the final state given a starting state, and we can also obtain the expected number of steps needed to reach an absorbing state (in this case, either player reaching 3 points) from a starting state by summing the state's corresponding row of the fundamental matrix. The exact formulas can be found on the wiki pages for Markov chains and absorbing Markov chains.

Here's a download link to the Excel sheet I used to actually runs these calculations - if you change the value in Q1 to the probability of rolling a 6 on any one die in this example, it will generate for you the probabilities of either player winning or a tie, as well as the expected number of rolls needed to end the game. For p = 1/6, either player has a 46.6% chance of winning, and there is a ~6.8% chance of a tie, and games on average will take ~6.69 rolls. For p = 1/3 the probability of a win is ~42.4%, of a tie is ~15.2%, and the expected number of rolls is ~3.61.