dice game

[cxuac]

rank the expected payoff of the following dice games:
1. Roll one dice, get paid the square of the number you get
2. Roll one dice twice, get the product of the two numbers
3. Roll three dices, get paid the square of the median of the three eg. 4,4,5 get paid 4^2=16

 

[soliton]

they are all the same, aren't they? assuming each roll is independent of each other

 

[Mike_B]

1,3,2

 

[dynamicD]

All 3 have expected value of 12.25 and yes they should independent of each other because they are dice.

 

[Brad Warren]

They are not the same

 

[CN Chen]

All 3 have expected value of 12.25 and yes they should independent of each other because they are dice.

Expected values are 15.1667, 12.25, and 14.1296 respectively, but intuitively look for the one more likely to get a product of two large numbers.

 

[PierreXu]

Let X, Y, Z be three i.i.d discrete uniform random variables. Let X* be the median of them. It's easy to see that E(X*)=E(X), Var(X*)<Var(X)
Expected payoff from option 1 = E(X^2)
Expected payoff from option 2 = E(XY)=E(X)E(Y)=[E(X)]^2
Expected payoff from option 3 = E(X*^2)
Due to Jensen's inequality: [E(X)]^2<E(X^2)=>option2<option1
Due to Jensen's inequality: E(X*^2)>[E(X*)]^2=[E(X)]^2=>option3>option2
Var(X*)=E[(X*)^2]-[E(X*)]^2=E[(X*)^2]-[E(X)]^2
Var(X)=E[(X)^2]-[E(X)]^2
Var(X*)<Var(X)=>E[(X*)^2]<E[(X)^2]=>option1>option3

=>option1>option3>option2

I'm not aware of an efficient way to calculate the exact value of option3 though. Maybe CN Chen can show us a good way to do this?

 

[jackcoke]

For Option 3...

Ways of getting a 1: 1, 1, [1 - 6] --> 6 ways
Ways of getting a 2: [1 - 2], 2, [2 - 6] --> 10 ways
Ways of getting a 3: [1 - 3], 3, [3 - 6] --> 12 ways
By symmetry,
Ways of getting a 4: 12 ways
Ways of getting a 5: 10 ways
Ways of getting a 6: 6 ways

Calculating expected payoff this way, I get a different value at 14.5...

 

[PierreXu]

For Option 3...

Ways of getting a 1: 1, 1, [1 - 6] --> 6 ways
Ways of getting a 2: [1 - 2], 2, [2 - 6] --> 10 ways
Ways of getting a 3: [1 - 3], 3, [3 - 6] --> 12 ways
By symmetry,
Ways of getting a 4: 12 ways
Ways of getting a 5: 10 ways
Ways of getting a 6: 6 ways

Calculating expected payoff this way, I get a different value at 14.5...

When you use brutal force method to solve this question, you need to distinguish between sequence 1, 1, 2 and sequence 1,2,1 as they count for two different "ways". The correct numbers using your method should be:
Ways of getting a 1: 1, 1, [1 - 6] --> 16 ways
Ways of getting a 2: [1 - 2], 2, [2 - 6] --> 40 ways
Ways of getting a 3: [1 - 3], 3, [3 - 6] --> 52 ways
By symmetry,
Ways of getting a 4: 52 ways
Ways of getting a 5: 40 ways
Ways of getting a 6: 16 ways
The answer is indeed 14.1296 as CN Chen suggested. I wonder if there's an easier way to get this number though.