Statistical Analysis of the Number of Spins needed to play a Snakes And Ladders Game (w/o Snakes and Ladders) With a fair k-faced die, terminating as soon as the total meets or exceeds n\ By Shalosh B. Ekhad Suppose that you keep tossing a fair k-faced die marked with 1 through k dot\ s, and at each step you advance according to the number of dots shown by the die. You keep playi\ ng until the first time your total meets or exceeds n Let X(n) be the random variable, number of rolls until finishing. Of course \ it depens on k. Since the expected size of a single donation is k/2 + 1/2 n The expectation of X(n) is roughly, --------- k/2 + 1/2 More precisely, up to exponentially small error, the Expectation of X(n) is 2 n 2 (k - 1) ----- + --------- k + 1 3 (k + 1) and in Maple notation it is 2/(k+1)*n+2/3*(k-1)/(k+1) The variance of X(n), up to exponentially small error is 2 2 (k - 1) 2 (k - 1) n ---------- + ----------- 2 2 9 (k + 1) 3 (k + 1) and in Maple notation it is 2/9*(k-1)^2/(k+1)^2+2/3*(k-1)/(k+1)^2*n The , 3, -th moment about the mean of of X(n), up to exponentially small error is 2 2 (k - 1) (k - 7) (7 k - 1) 2 (k - 1) n --------------------------- + ------------ 3 3 135 (k + 1) 3 (k + 1) and in Maple notation it is 2/135*(k-1)*(k-7)*(7*k-1)/(k+1)^3+2/3*(k-1)^2/(k+1)^3*n The limit of the scaled , 3, -th moment as n goes to infinity is 0 The , 4, -th moment about the mean of of X(n), up to exponentially small error is 2 2 2 2 (13 k - 110 k + 13) (k - 1) 2 (k - 1) (13 k - 30 k + 13) n ------------------------------- + ------------------------------- 4 4 135 (k + 1) 15 (k + 1) 2 2 4 (k - 1) n + ------------- 4 3 (k + 1) and in Maple notation it is 2/135*(13*k^2-110*k+13)*(k-1)^2/(k+1)^4+2/15*(k-1)*(13*k^2-30*k+13)/(k+1)^4*n+4 /3*(k-1)^2/(k+1)^4*n^2 The limit of the scaled , 4, -th moment as n goes to infinity is 3 The , 5, -th moment about the mean of of X(n), up to exponentially small error is 4 3 2 2 (k - 1) (145 k - 3892 k + 8790 k - 3892 k + 145) ----------------------------------------------------- 5 1701 (k + 1) 2 2 3 2 2 (139 k - 530 k + 139) (k - 1) n 40 (k - 1) n + ----------------------------------- + -------------- 5 5 81 (k + 1) 9 (k + 1) and in Maple notation it is 2/1701*(k-1)*(145*k^4-3892*k^3+8790*k^2-3892*k+145)/(k+1)^5+2/81*(139*k^2-530*k +139)*(k-1)^2/(k+1)^5*n+40/9*(k-1)^3/(k+1)^5*n^2 The limit of the scaled , 5, -th moment as n goes to infinity is 0 The , 6, -th moment about the mean of of X(n), up to exponentially small error is 4 3 2 2 2 (841 k - 146860 k + 459222 k - 146860 k + 841) (k - 1) ------------------------------------------------------------ 6 25515 (k + 1) 4 3 2 2 (k - 1) (2321 k - 18284 k + 32358 k - 18284 k + 2321) n + ------------------------------------------------------------ 6 567 (k + 1) 2 2 2 3 3 4 (13 k - 30 k + 13) (k - 1) n 40 (k - 1) n + --------------------------------- + -------------- 6 6 3 (k + 1) 9 (k + 1) and in Maple notation it is 2/25515*(841*k^4-146860*k^3+459222*k^2-146860*k+841)*(k-1)^2/(k+1)^6+2/567*(k-1 )*(2321*k^4-18284*k^3+32358*k^2-18284*k+2321)/(k+1)^6*n+4/3*(13*k^2-30*k+13)*(k -1)^2/(k+1)^6*n^2+40/9*(k-1)^3/(k+1)^6*n^3 The limit of the scaled , 6, -th moment as n goes to infinity is 15 ----------------------- This ends this article that took, 12.878, seconds to produce.