TITLE: Thought experiment for dice game
QUESTION [6 upvotes]: Two players in a dice game to see who can roll a total of 60 first, taking turns each rolling 2 dice. 
For one player, one die is 4 sided and one is 6 sided. Therefore the average roll for this player will be a 6. 
The other player has two 6 sided dice. So this players average roll will be a 7. 
If they take turns, each rolling both dice, what % of the time will the first player (a 4-6 dice combo) get to 60 first and what % of the time will the second player (6-6 dice combo) get to 60 first. If they get to 60 on the same roll, the player who goes the furthest wins.

REPLY [7 votes]: Here is the exact answer, which I used Mathematica to compute. 
$P(player\ 1\ wins) = \frac{118598889714523902216022358617928917633636253614645787377890526452587094871641057161197}{778560366535929033488842048429259732340012411410736514701931793474234814991339289051136}$
$P(player\ 2\ wins) = \frac{858777828556435297558093986193070674512712644399583464104126713012616584348576395258605}{1038080488714572044651789397905679643120016548547648686269242391298979753321785718734848}$
$P(players\ tie) = \frac{63512421616314632416996800666111235287366697985612516983784929048741127433063741783941}{3114241466143716133955368193717038929360049645642946058807727173896939259965357156204544}$
Numerically, that's 
.15233101351178373415...
.82727479987591082853...
.02039418661230543732...

Here are the Mathematica rules I used to compute the probability of player 1 winnng. The other two cases are similar.

(sorry for the image) 

With these rules in place I then computed p[60,60]. I thought it was simpler to start the players at 60 and count down to 0, rather than the other way round.
The first rule just says if player 1 has reached 0 and he player 2 hasn't, then the probability of player 1 winning is 1. The second rule is similar for player 2.
The third rule handles the case where both players have crossed the finish line. Note that despite the game description, there aren't really turns involved—there are rounds. It doesn't matter which player goes first, so me might as well consider all four dice being rolled simultaneously.
The fourth and final rule is where all the action is. The recursive sum is over all possible outcomes of the four dice (two for each player). The constant $864$ is merely $4\cdot6\cdot 6\cdot 6$. The p[x,y] = thing in the center is a kind of memoization (caching). Without it, the simplistic recursion would take forever.
EDIT:
Here is a plot of $P(n)$, the probability that the weaker player wins, when playing to a total of $n$ (instead of $60$), for $n = 1\dots80$. To my surprise, it's not quite monotonic! But a little thought will explain why.