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- Homework Statement:
- There is a game where one person rolls a 20-sided dice and another person rolls three 6-sided dice. They both roll their dice and whoever gets a sum of numbers wins the game. Which person would you rather be? Is it a fair game?

- Relevant Equations:
- Expectation and variance

Hi,

I was taking a look at the following question.

I think the general concept is about looking at expectation and variance for both cases and then make a decision from there.

So for the person with the 20-sided die, the expectation and variance are given by:

[tex] E(\text{20 sided die}) = \frac{n + 1}{2} = \frac{21}{2} = 10.5 [/tex]

[tex] Var(\text{20 sided die}) = \frac{n^2 - 1}{12} = \frac{(20)^2 - 1}{12} = \frac{399}{12} [/tex]

For the person with the three 6-sided dice:

[tex] E(\text{three 6 sided die}) = E(X_1 + X_2 + X_3) = 3 \cdot E(\text{one 6 sided die}) = 3 \cdot \frac{7}{2} = 10.5 [/tex]

[tex] Var(\text{three 6 sided die}) = Var(X_1 + X_2 + X_3) = 3 \cdot \frac{n^2 - 1}{12} = 3 \cdot \frac{(6)^2 - 1}{12} = \frac{35}{4} [/tex]

Thus, both of the distributions have the same expected value/mean. Therefore, perhaps we would rather be the person who chooses the lower variance side, which is the three 6-sided die. Does that seem reasonable?

Also, what does the 'is the game fair' part mean? Is that basically asking if the mean and variance are the same? If so, then based on the calculations above, I would say the game isn't fair...

Any help would be greatly appreciated.

I was taking a look at the following question.

**Question:**There is a game where one person rolls a 20-sided dice and another person rolls three 6-sided dice. They both roll their dice and whoever gets a higher number (for 20-sided die) OR sum of numbers (for three 6-sided dice) wins the game. Which person would you rather be? Is it a fair game?*Assume the dice are all fair***Attempt:**I think the general concept is about looking at expectation and variance for both cases and then make a decision from there.

So for the person with the 20-sided die, the expectation and variance are given by:

[tex] E(\text{20 sided die}) = \frac{n + 1}{2} = \frac{21}{2} = 10.5 [/tex]

[tex] Var(\text{20 sided die}) = \frac{n^2 - 1}{12} = \frac{(20)^2 - 1}{12} = \frac{399}{12} [/tex]

For the person with the three 6-sided dice:

[tex] E(\text{three 6 sided die}) = E(X_1 + X_2 + X_3) = 3 \cdot E(\text{one 6 sided die}) = 3 \cdot \frac{7}{2} = 10.5 [/tex]

[tex] Var(\text{three 6 sided die}) = Var(X_1 + X_2 + X_3) = 3 \cdot \frac{n^2 - 1}{12} = 3 \cdot \frac{(6)^2 - 1}{12} = \frac{35}{4} [/tex]

Thus, both of the distributions have the same expected value/mean. Therefore, perhaps we would rather be the person who chooses the lower variance side, which is the three 6-sided die. Does that seem reasonable?

Also, what does the 'is the game fair' part mean? Is that basically asking if the mean and variance are the same? If so, then based on the calculations above, I would say the game isn't fair...

Any help would be greatly appreciated.

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