If we threw $30$ tetrahedral dice and summed the outcomes, how many values in the distribution would have a non-zero probability?
a) If we calculate the distribution of the sum of two thrown tetrahedral dice, how many values have a non-zero probability?
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$\begingroup$If we roll two tetrahedral dice, and let $S_2$ be the sum of the rolls, then $2\le S_2\le8$, and $S_2$ can be any integer in this range.
If we roll thirty tetrahedral dice, and let $S_{30}$ be the sum of the rolls, then $30\le S_{30}\le120$, and $S_{30}$ can be any integer in this range.
If we roll $n$ tetrahedral dice, then there are $4^n$ total outcomes. We can find out how many ways there are to get the rolls to add up to $k$ by using generating functions as follows:
Let $p(x)=x+x^2+x^3+x^4$. The number of ways for the $n$ rolls to add up to $k$ will be the coefficient of $x^k$ in $(p(x))^n$. The smallest degree term of $(p(x))^n$ is $x^n$. The highest degree term will be $x^{4n}$, and for every integer $m$ with $n\le m\le4n$, the coefficient of $x^m$ will be non-zero.
Hence if we roll $n$ tetrahedral dice, and let $S_n$ be the sum of the rolls, then $n\le S_n\le4n$, and $S_n$ can be any integer in this range.
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