| We can model this game with a system of equations. Let pn be the
probability of a player winning if they are n points ahead. Then p6 is
1, because the player has won. If a player is five points ahead, there
is a 3 in 6 chance they will roll a 1, 2, or 3 in their favor, and one
chance each of rolling a 1, 2, or 3 against them, so p5 is 3/6 plus
(p4+p3+p2)/6. Similarly, we can write each of the other probabilities
in terms of some constants plus other probabilities. Also, let qn be
the probability of a player winning if they are n points behind. Then
the equations are:
p6 = 1
p5 = 3/6 + (p4+p3+p2)/6
p4 = 2/6 + (p5+p3+p2+p1)/6
p3 = 1/6 + (p5+p4+p2+p1+p0)/6
p2 = (p5+p4+p3+p1+p0+q1)/6
p1 = (p4+p3+p2+p0+q1+q2)/6
p0 = (p3+p2+p1+q1+q2+q3)/6
q1 = (p2+p1+p0+q2+q3+q4)/6
q2 = (p1+p0+q1+q3+q4+q5)/6
q3 = 0/6 + (p0+q1+q2+q4+q5)/6
q4 = 0/6 + (q1+q2+q3+q5)/6
q5 = 0/6 + (q2+q3+q4)/6
q6 = 0
Since p6 and q6 are constants, the remaining equations are a system of
11 equations with 11 variables. We could reduce this further. For
example, by symmetry, we expect p0 to be 1/2, and we could make quick
substitutions for q1 through q5 by replacing them with (1-p1) through
(1-p5), again by symmetry. But this is the age of computers, so let's
just stick the equations in an array and tell a computer to figure it
out. I multiplied all the equations by six, to make all the
coefficients integers, and wrote them in the array below, with the
constant terms moved to a vector on the right.
p5 p4 p3 p2 p1 p0 q1 q2 q3 q4 q5 =
[-6 1 1 1 0 0 0 0 0 0 0] -3
[ 1 -6 1 1 1 0 0 0 0 0 0] -2
[ 1 1 -6 1 1 1 0 0 0 0 0] -1
[ 1 1 1 -6 1 1 1 0 0 0 0] 0
[ 0 1 1 1 -6 1 1 1 0 0 0] 0
[ 0 0 1 1 1 -6 1 1 1 0 0] 0
[ 0 0 0 1 1 1 -6 1 1 1 0] 0
[ 0 0 0 0 1 1 1 -6 1 1 1] 0
[ 0 0 0 0 0 1 1 1 -6 1 1] 0
[ 0 0 0 0 0 0 1 1 1 -6 1] 0
[ 0 0 0 0 0 0 0 1 1 1 -6] 0
My calculator produces these numbers for the variables:
p5 = .864442127217
p4 = .803962460898
p3 = .732533889469
p2 = .650156412932
p1 = .576642335768 (X's probability of winning after gaining 1)
p0 = .500000000002 (which gives us an estimate of numerical error)
q1 = .423357664236
q2 = .349843587072
q3 = .267466110533 (X's probability of winning after losing 3)
q4 = .196037539105
q5 = .135557872785
-- edp
|
| P.S., exact answers are p5 = 829/959, p4 = 771/959, p3 = 1405/1918, p2
= 1247/1918, p1 = 79/137.
-- edp
|
| Happily, the probabilities are actually rationals and MAPLE cranks them out
in that form, so precision is not an issue. The vector of solutions is
829 771 1405 1247 79 58 671 513 188 130
---, ---, ----, ----, ---, 1/2, ---, ----, ----, ---, ---
959 959 1918 1918 137 137 1918 1918 959 959
|