Algorithm making of Die Roll for each pair who throws the same number + Mean and Variance

Question

Each member of a group of $n$ players rolls a die.

Find the Mean and Variance of the total score if any pair of the players who throw the same number of scores as that number.

Answer 1

I edited the question to make it more suitable, so hopefully, this question gets unflagged as an improper question!

There are many things wrong with your question. Remember, this is a math site and not a "code" site. If you want the mathematical answer to this question, then only ask this part: "Find the mean and variance of the total score if any pair of players who throw the same number scores that number." Remember to ask on Stack Overflow for "Create an algorithm and a code in R," but even then show what you've done.

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However, the approach for the mean and variance part should be along the lines of:

Let's declare variables for better understanding -

Say,

M$i$ = the score of players who actually throw a die 1, 2, . . . ,6

V$i$ = the number of people who throw those die (i) and 1xᵢⱼ (indicator function) and the indicator function can Iff be 1 if the jᵗʰ person throws $i$.

E = Expectation score.

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Based on those rules, we can form a simple series function for that:

V$i$ = $\sum_{j=1}^n = 1xᵢⱼ$

So,

Expectation

$E[Mᵢ][Vᵢ] = \frac{(Vᵢ)(Vᵢ-1)}{2}$, then $E[Mᵢ] = \frac{(n^2-n)}{12}$

• Expectation of 1xᵢⱼ = 1/6 (Die Roll: Probability), so 1/6 * (1 - 1/6) = $\frac{5}{36}$

Therefore,

• Expectation: => (n :: 2 [Pair]) * 1/6 = $\frac{n(n - 1)}{12}$

E[M] = $\frac{(n^2-n)}{12}$, in which M = $\sum_{i=1}^6 Mᵢ$ (Total Score)

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Variance

• Variance => V = (n :: 2) * 5/36 =

$\frac{5n(5n - 1)}{72}$

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