Originally posted by Brother Edwin
I play a game where this is important, would someone be king enough to tell me the answer?
I roll three dice
am i more likely to roll at least two sixes, or that the total dice result will be 13?
i do the same with 4 dice, 5 dice, and so on, at what number of dice does it become more likely i will get at least two sixes?
Chance to roll a six on a dice is 1/6, so the chance to roll two sixes on two dice is 1/6 * 1/6 = 1/36 = 2/72. On three dice the chance to roll at least two sixes is 1/36 * 3 (because there are 3 ways of arranging your throw 66X 6X6 and X66).
The chance on one precise combination is 6*6*6=216. There are 15 combinations in wich the sum is 13 (661, 652, 643, 553, 544 and all reaarrangements of these). The chance to throw 13 on three dice is therefore 15/216=5/72. Thus the chance to throw 13 is greater on three dice.
For more dice the number of ways to throw 13 increases first, then decreases again. on 13 dice there is only one way to throw 13, on 12 there are 12, on 11 it's too much for for me to calculate without paper.
Perhaps you cn use this to clalculate it yourself:
The number of ways to arrange N things in a set of M (M>N) is M!/N!(M-N)! (N! means N* N-1 *N-2 *...*3*2*1)