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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?

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There are several approaches:

  1. Use the normal distribution as an approximation as described in whuber's link to the earlier question, or use some of the other suggestions there. The mean will be 62.5 while the variance will be 31.25.
  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 $4^{25} = 1125899906842624$
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    $\begingroup$ +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 $$\Pr(X=k)=4^{-25}\sum_{i=0}^{25}\binom{25}{i}\binom{25}{k-25-2i}.$$ $\endgroup$ – whuber Aug 20 '17 at 19:24

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