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I'm trying to figure out a general equation to determine the probability that the sum of $m$ dice will be greater than the sum of $n$ dice, where $m < n$. For example, if I roll $4$ dice, what's the probability that I will beat a roll of $6$ dice? Assume standard $6$ sided dice. I've been trying to adapt the answer I found here, but I think it's really a completely separate problem.

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You can compute the probability distribution of the sum of $k$ dice using generating series: the generating series is $$ \left(\frac{x+x^2+x^3+x^4+x^5+x^6}{6}\right)^k.$$ Given the two probability distributions for $k=n$ and $k=m$, it is easy to compute the probability that one sum beats the other.

I don't think you can obtain a more explicit formula. For large $m,n$, you can probably obtain a good estimate using the central limit theorem: roughly speaking, the sum of $k$ dice is distributed roughly $N(3.5k,(35/12)k)$. So the difference of a sum of $m$ dice and a sum of $n$ dice is distributed roughly $N(3.5(m-n),(35/12)(m+n))$. The probability that this is positive is roughly $\Pr[N(0,1)>\frac{3.5(n-m)}{\sqrt{(35/12)(n+m)}}]$, which you can look up in tables.

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  • $\begingroup$ I'm not sure I follow completely...I understand how to calculate the probability of rolling a specific number for a given number of dice, but I'm trying to figure out the probability of a hand of fewer dice beating a hand of more dice. Intuitively, the probability of 99 dice rolls beating 100 dice rolls is much higher than 1 dice roll beating 2 dice rolls, but I'm trying to figure out a way to formalize that intuition. $\endgroup$ – Archa3opt3ryx Feb 24 '15 at 5:07
  • $\begingroup$ If you can calculate the probability $P_k(x)$ that a roll of $k$ dice gives $x$, then the probability you're after is $\sum_{x>y} P_m(x) P_n(y)$. You can try to calculate this explicitly for some $n,m$ and see what happens. $\endgroup$ – Yuval Filmus Feb 24 '15 at 15:23

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