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Q:

How many different ways are there to win a 6-sided die with a specific value?

I’m not very good at probability so let’s say I roll a 6 sided dice and it shows a ‘3’ (of those 6 numbers there are 2 with a 3) and I want to calculate how many different ways there are to win with that die. How would I go about this?

A:

The number of possible distributions of outcomes on a die with $n$ faces is given by the binomial coefficient:
$$\binom{n}{k} = \frac{n!}{k! (n-k)!}$$
This gives us $6^3$ distributions that can be rolled.

A:

Let $p_k$ be the probability of rolling $k$ in a single roll. The probability of rolling a $3$ is $2/6$, and if we pick a distribution of numbers $\{p_k\}$ for a die with $n$ sides, then our calculation is:
$$\sum_{k=0}^n p_k = 1$$
$$\sum_{k=0}^n p_k k = 0$$
$$\sum_{k=0}^n p_k k^2 = 1$$
$$\sum_{k=0}^n p_k k^3 = 0$$
$$\sum_{k=0}^n p_k k^4 = 1$$
So, $p_k$ is found by solving for it in the following polynomial:
\begin{align}
1 &= p_0 + p_1 k + p_2 k^2 + p_3 k^3 + p_4 k^4 \\
0 &= p_0 + p_1 k + p_2 k^2 + p_3 k^3 + p_4 k^4 \\
1 &= p_0 + 2p_1 k + 3p_2 k^2 + 4p_3 k^3 + 5p_4 k^4 \\
0 &= 2p_1 + 3
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