My Math Forum Dice probability

 December 1st, 2009, 11:26 AM #1 Newbie   Joined: Dec 2009 Posts: 1 Thanks: 0 Dice probability I am trying to find a general solution to calculate a certain number with x number of dice. As in, what is the probability to roll a 10 with 3 6-sided dice? Or to roll a 17 with 4 6-sided dice? Could anyone please help me on this?
 December 1st, 2009, 11:49 AM #2 Member   Joined: Oct 2009 Posts: 64 Thanks: 0 Re: Dice probability It's very easy to write the recursion relation. Let C(N,k) be the number of ways to roll k dice so that the total is N. Then C(N,k+1) = C(N-1,k) + C(N-2,k) + ... + C(N-6,k) And C(N, 1) = 1 if 1<=N<=6 and 0 otherwise. That is, you look at the outcome of the first die, then count the number of ways to roll the remaining part of N on the remaining k dice. Add the counts for each possible roll of the first die, and you have the C(N,k+1) as desired. If you want to write a closed formula, you'll need to go a bit further.
 December 1st, 2009, 12:15 PM #3 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: Dice probability Wikipedia states that the probability of rolling k with n s-sided dice is: $F_{s,n}(k)=\frac{1}{s^n}\sum_{i=0}^{\left \lfloor \frac{k-n}{s} \right \rfloor} (-1)^i {n \choose i} {k-si-1 \choose n-1}$ So the probability of rolling 10 with 3 six-sided dice is $\frac{1}{6^3}\left [ {3\choose 0}{9\choose 2}-{3\choose 1}{3\choose 2}\right ]= \frac{27}{216} = \frac{9}{73}$. And the probability of rolling 17 with 4 six-sided dice is $\frac{1}{6^4}\left [ {4 \choose 0}{16 \choose 3}-{4 \choose 1}{10 \choose 3}+{4 \choose 2}{4 \choose 3} \right ]= \frac{560-480+24}{1296} = \frac{13}{162}$ If you want to solve the problem for all possible k at once, you can look instead at the coefficients of the expansion of (x+x²+x³+...+x^s)^n. For example, if you tell Wolfram Alpha to expand (x+x^2+x^3+x^4+x^5+x^6)^3, which represents three 6-sided dice, you get: $x^{18}+3 x^{17}+6 x^{16}+10 x^{15}+15 x^{14}+21 x^{13}+25 x^{12}+27 x^{11}+\\27 x^{10}+25 x^9+21 x^8+15 x^7+10 x^6+6 x^5+3 x^4+x^3$ which means that there is 1 way to get 18, 3 ways to get 17, 6 ways to get 16, 10 ways to get 15... etc.

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