## An unbiased single $4$-sided die is thrown and its value, $T$, is noted.$T$ unbiased $6$-sided dice are thrown and their scores are added together. The sum, $C$, is noted.$C$ unbiased $8$-sided dice are thrown and their scores are added together. The sum, $O$, is noted.$O$ unbiased $12$-sided dice are thrown and their scores are added together. The sum, $D$, is noted.$D$ unbiased $20$-sided dice are thrown and their scores are added together. The sum, $I$, is noted.

Find the variance of $I$, and give your answer rounded to $4$ decimal places.

### The solution to this problem will make use of the formula for the variance of a sum of independent random variables: the variance of a sum is the sum of the variances.

Firstly, we must figure out the mean and variance for a single roll of a fair n-sided die.

For a fair n-sided die, where the faces are numbered 1, 2, …, n,

the expected value or mean (E[X]) is (1 + 2 + … + n)/n = (n + 1)/2.

and the variance (Var[X]) is E[X^2] – (E[X])^2, where E[X^2] is (1^2 + 2^2 + … + n^2)/n = (n(n + 1)(2n + 1))/6.

Now we derive the expressions for T, C, O, D, and I.

1) For T (a single 4-sided die):

E[T] = (4 + 1)/2 = 2.5,

Var[T] = (4*(4 + 1)*(2*4 + 1))/6 – (2.5)^2 = 1.25.

2) For C (T 6-sided dice):

E[C] = T * E[6-sided die] = T * (6 + 1)/2 = 3.5T,

Var[C] = T * Var[6-sided die] = T * ((6*(6 + 1)*(2*6 + 1))/6 – (3.5)^2) = 35/12*T.

3) For O (C 8-sided dice):

E[O] = C * E[8-sided die] = C * (8 + 1)/2 = 4.5C,

Var[O] = C * Var[8-sided die] = C * ((8*(8 + 1)*(2*8 + 1))/6 – (4.5)^2) = 21/2*C.

4) For D (O 12-sided dice):

E[D] = O * E[12-sided die] = O * (12 + 1)/2 = 6.5O,

Var[D] = O * Var[12-sided die] = O * ((12*(12 + 1)*(2*12 + 1))/6 – (6.5)^2) = 455/12*O.

5) For I (D 20-sided dice):

E[I] = D * E[20-sided die] = D * (20 + 1)/2 = 10.5D,

Var[I] = D * Var[20-sided die] = D * ((20*(20 + 1)*(2*20 + 1))/6 – (10.5)^2) = 3850/3*D.

Now, note that the variances at each stage are determined by the expected value from the previous stage.

Therefore, substituting for T, C, O, and D in Var[I], we have

Var[I] = 3850/3 * (455/12 * (21/2 * (35/12 * 1.25))).

Evaluating this gives approximately 302018.161, which to four decimal places is 302018.1611.

So the variance of I is 302018.1611 to four decimal places.

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