On the Probabilities of the First Player Winning in a Two-Player Pile game where the winner is the first to accumulate n dollars By Shalosh B. Ekhad Consider the following simple 2-Player Pile game Players take turn tossing a die with, 2, faces marked as follows, with the given probabilities A face marked with , 1, dots , with probability, p A face marked with , 2, dots , with probability, 1 - p and they get the number of chips indicated by the die The player to be the FIRST to accumulate n chips is declared the winner. Theorem: The probability of the FIRST player winning is 1/2 + 1/2 a(n) where a(n) is a squence of numbers satisfying the linear recurrence a(n) = ( 2 3 2 2 3 2 3 2 2 n p - 4 n p - 8 n p + 16 n p + 6 p - 5 n p - 14 p + 2 n + 8 p - 2 2 5 2 4 5 2 3 4 5 ) a(n - 1)/(n %1) - (n p - 2 n p - 5 n p - 2 n p + 9 n p + 6 p 2 2 3 4 2 2 3 2 + 8 n p + 12 n p - 8 p - 10 n p - 42 n p - 22 p + 4 n + 45 n p 2 2 + 56 p - 16 n - 50 p + 16) a(n - 2)/(n %1) + (p - 1) ( 2 3 2 2 3 2 3 2 2 n p - 4 n p - 11 n p + 20 n p + 14 p + n p - 20 p - 2 n - 6 p + 6 4 (p - 1) (n - 4) (n p - 2 n - 2 p + 3) a(n - 4) ) a(n - 3)/(n %1) - ----------------------------------------------- n %1 %1 := n p - 2 n - 3 p + 5 Subject to the initial conditions 2 4 2 a(1) = 1, a(2) = 2 p - 2 p + 1, a(3) = 2 p - 2 p + 1, 6 5 4 3 2 a(4) = 2 p + 2 p - 2 p - 8 p + 10 p - 4 p + 1 and in Maple notation a(n) = (2*n^2*p^3-4*n^2*p^2-8*n*p^3+16*n*p^2+6*p^3-5*n*p-14*p^2+2*n+8*p-2)/n/(n *p-2*n-3*p+5)*a(n-1)-(n^2*p^5-2*n^2*p^4-5*n*p^5-2*n^2*p^3+9*n*p^4+6*p^5+8*n^2*p ^2+12*n*p^3-8*p^4-10*n^2*p-42*n*p^2-22*p^3+4*n^2+45*n*p+56*p^2-16*n-50*p+16)/n/ (n*p-2*n-3*p+5)*a(n-2)+(p-1)^2*(2*n^2*p^3-4*n^2*p^2-11*n*p^3+20*n*p^2+14*p^3+n* p-20*p^2-2*n-6*p+6)/n/(n*p-2*n-3*p+5)*a(n-3)-(p-1)^4*(n-4)*(n*p-2*n-2*p+3)/n/(n *p-2*n-3*p+5)*a(n-4) a(1) = 1, a(2) = 2*p^2-2*p+1, a(3) = 2*p^4-2*p^2+1, a(4) = 2*p^6+2*p^5-2*p^4-8* p^3+10*p^2-4*p+1