# Multinomial distribution for 4 side dice roll

Suppose I roll a 4-side dice 25 time to get a final sum between 25 and 100. How I calculated the distribution of probability for each sum between 25 and 100?

2. Do a recurrence starting with $f(i,0)=0$ except when $i=0$ in which case $f(0,0)=1$, and then use $f(i,j+1)=f(i-4,j)/4+f(i-3,j)/4+f(i-2,j)/4+f(i-1,j)/4$ to find $f(i,25)$
3. Expand $(x/4+x^2/4+x^3/4+x^4/4)^{25}$, for example by putting Expand[(x/4+x^2/4+x^3/4+x^4/4)^25] into Wolfram Alpha and press "=" then "show more terms" several times
4. Hope somebody has already done some recurrence calculations for you, which they have in rows 925-1000 of an OEIS table and then divide the results by $4^{25} = 1125899906842624$
• +1. For the particular case of a four-sided fair die, there exists an unusual shortcut: it's so effective that individual probabilities could easily be computed by hand. Noting that $$(x+x^2+x^3+x^4)/4= \frac{x}{4}(1+x)(1+x^2),$$ you could expand $(1+x)^{25}$ and $(1+x^2)^{25}$ via the Binomial Theorem and multiply the two polynomials. This gives the nice formula $$\Pr(X=k)=4^{-25}\sum_{i=0}^{25}\binom{25}{i}\binom{25}{k-25-2i}.$$ – whuber Aug 20 '17 at 19:24