What is the probability of getting the sum of 14 with a modified deck of cards?
the q is equivalent to finding the probability of getting a sum of 14 on tossing four 10-sided dice firstly we have to find the # of favorable ways one way to do this by listing the patterns & their permutations systematically, e.g. 10-2-1-1: 4!/2! = 12 9-3-1-1 : 4!/2! = 12 9-2-2-2 : 4!/2! = 12 & so on but this is a tedious process and quite error prone however, it can be solved very neatly with a bit of algebra if we realize that the q is equivalent to finding the coefficient of x^14 in (x + x^2 + … + x^10)^4 = the coefficient of x^10 in (1 + x + ….x^9)^4 = coefficient of x^10 in (1-x^10)^4 / (1-x)^4 ………… [ sum of a G.P. ] = coefficient of x^10 in (1-x^10)^4 * (1-x)^(-4) = (coeff of x^0 in 1st expression)*(coeff of x^10 in 2nd) + (coeff. of x^10 in 1st expression)*(coeff of x^0 in 2nd) so # of favorable ways = (-4)C10 – 4C1 = 286 – 4 = 282 total # of ways = 10^4 reqd.probability = 2.82 x 10^-6 ——————— edit: —— i see that the equivalence on which the calcu
