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I need to create:

One 4 sided Die : Numbers add up to 14

One 6 sided Die: Numbers add up to 21

One 8 sided Die: Numbers add up to 28

Which combinations of numbers would result in my rolling the highest average number on each dice?

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3  
Didn't you already fix the averages by fixing the sums? –  Raskolnikov Sep 13 '12 at 18:52

3 Answers 3

The sum and the number of sides is all that matters in determining the average value of the dice.therefore no mattr how you label your die. The average through will give you $\frac{14}4+\frac{21}6+\frac{28}8$.

Proof that the average throw only depends on the sum of the sides: the average throw is simply the sum of each case multiplied with its probability. In a fair die the probability of each side is $\frac1n$ ($n$ being the number of sides) Therefore the average throw of an $n$ sided die is $\frac{a_1}{n}+\frac{a_2}{n}+\frac{a_2}{n}+\ldots+\frac{a_n}{n}$. ($a_1$ being the number on the first side, $a_2$ on the second side etc.)

Since $\frac{a_1}{n}+\frac{a_2}{n}+\frac{a_2}{n}+\ldots+\frac{a_n}{n}=\frac{a_1+a_1+\ldots +a_n}{n}$ The only thing that matters is the value of $a_1+a_1+\ldots +a_n $ which is already fixed.

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As Raskolnikov says, once you fix the total you have fixed the average for each die. Simple sets to give what you want are 2,3,4,5 for the four sided, 1,2,3,4,5,6 for the six sided (easily available) and 0,1,2,3,4,5,6,7 for the eight sided dice.

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probablty of multiples 2 or 3 while throughing a dice?

ans: 2\3

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