[BruceZ]

[ QUOTE ]
Can somebody calculate the variance for these two situations for me?

I'm a little rusty on my stats, and I think I'm going about it the wrong way

Situation 1:
Getting 1.5:1 on a coin flip

Situation 2:
Getting 6:1 on a single number, on a the roll of a die

Please show all steps.

FWIW: I have the EVs calculated as:

1:
(1.5x-1x)/2 = 0.25x = E(x)

[/ QUOTE ]

E(x^2) = [(1.5x)^2 + (-1x)^2]/2 = (13/8)*x^2

var(x) = E(x^2) - [E(x)]^2 = (13/8)*x^2 - (0.25x)^2 = (25/16)*x^2


[ QUOTE ]
2:
(6x-5x)/6 = 0.167x = E(x)

[/ QUOTE ]

E(x^2) = (1/6)*(6x)^2 + (5/6)*(-1x)^2 = (41/6)*x^2

var(x) = E(x^2) - [E(x)]^2 = (41/6)*x^2 - (x/6)^2 = (245/36)*x^2