Statement of the problem : Fermat and Pascal play a coin flipping game. Each person contributes 50 francs to the pot. A fair coin is tossed repeatedly. A "heads" is a point for Fermat; a "tails" is a point for Pascal. The first to 6 points wins the pot. Play is halted with Fermat at 4 points and Pascal at 3 points. What is the fair way to divvy up the pot?

Here are the solutions of Fermat and Pascal, which we discussed in class.

Fermat's solution:

List all 16 equally likely ways to complete the game to completion:

FFFF
FFFP
FFPF
FFPP
FPFF
FPFP
FPPF
    FPPP
    PPPP
    PPPF
    PPFP
PPFF
    PFPP
PFPF
PFFP
PFFF

Note that 11/16 are F wins; 5/16 are P wins.  So F gets 11/16 of the
pot; P gets 5/11.

Pascal's solution:

         0        1        2        3       4       5   6
0  0.00000  0.00000  0.00000  0.00000  0.0000  0.0000 NaN
1  1.56250  3.12500  6.25000 12.50000 25.0000 50.0000 100
2  6.25000 10.93750 18.75000 31.25000 50.0000 75.0000 100
3 14.45313 22.65625 34.37500 50.00000 68.7500 87.5000 100
4 25.39063 36.32813 50.00000 65.62500 81.2500 93.7500 100
5 37.69531 50.00000 63.67188 77.34375 89.0625 96.8750 100
6 50.00000 62.30469 74.60938 85.54688 93.7500 98.4375 100


R code to compute Pascal's solution:

w <- matrix(numeric(49),ncol=7)
w[,7] <- c(NaN,rep(100,6))
for (i in 2:7) {
     for (j in 6:1) {
           w[i,j] <- .5*w[i,j+1] + .5*w[i-1,j]
     }
}

Here is the simulation we developed in class using R; it seems to work:

sum(replicate(100000,sum(sample(coin,4,T))) < 3)/100000
[1] 0.68925  # pretty close to .6875.


> 11/16
[1] 0.6875