Expected value

#1
Could someone please explain the following?

A guy and a girl play a game where they roll a dice. If the guy rolls an even number, the girl pays him that amount. If he rolls an odd number he pays her that amount.

The expected value is supposedly $0.5. ie mean of -1+2-3+4-5+6

but if i split the values like so

girl: 1,3,5 total = 9 Average is 9/3 = 3

guy: 2,4,6 total = 12 Average = 12/3 = 4

dif = 1 and not 0.5

why can i not split up their winnings like the above?

Thanks for your help
 

Dason

Ambassador to the humans
#2
You can but you need to be more careful.

Let X be the amount won on any given roll.
\(E[X] = \sum XP(X)\)
\(= (-1)(1/6) + (2)(1/6) + (-3)(1/6) + (4)(1/6) + (-5)(1/6) + (6)(1/6) \)
\(= (1/6)(-1 +2 -3 +4 -5 +6) \)
\(= (1/6)(-1 -3 -5) + (1/6)(2 + 4 + 6) \) [Here we separate the outcomes into either losing or winning]
\(=(1/2)(-3) + (1/2)(4) \) [Here I bring 1/3 into each of the sums to get the averages you have in your calculations]

You need to account for the fact that you're losing $3 and that there is a probability of 1/2 of falling into either losing or winning. It's really that you didn't account for the probability of falling into losing or winning that messed you up here.

We can show something more general using a conditioning argument. Let \(I_W(X)\) be an indicator of whether or not you win money on a given roll. We can use the fact that
E(X) = E[E(X|Y)] and what I showed up above is really just an application of this where \(Y=I_W(X)\)
 
#3
Hey, thanks very much for replying to my post and clarifying my question.

Btw, sorry for the late reply but i hadnt seen your reply until now.