Calculating standard deviation for fixed odds

[joeg]

Hi

I was wondering if there is an equation to calculate the standard deviation for fixed odds events, for example if I was to get odds of 2-1 on a event with a 50% chance, how would I calculate the standard deviation here?

Thanks for any replies

Joe.

[BruceZ]

[ QUOTE ]
Hi

I was wondering if there is an equation to calculate the standard deviation for fixed odds events, for example if I was to get odds of 2-1 on a event with a 50% chance, how would I calculate the standard deviation here?

Thanks for any replies

Joe.

[/ QUOTE ]

Half of the time you win 2 bets, and half of the time you lose 1 bet, so your average or EV is (2-1)/2 = 0.5 bets. That means that the half of the time when you win 2 you will be 1.50 above the average, and the other half of the time when you lose 1 you will be 1.50 below the average. The variance is the average (or expected value) of the squared difference from the average. This is (1.50^2 + 1.50^2)/2 = 2.25.

We have used the definition of the variance of x which is

var(x) = E[x - E(x)]^2

where E(x) means expected value. Another equivalent formula you can use is:

var(x) = E(x^2) - [E(x)]^2.

In this case E(x^2) = [2^2 + (-1)^2 ]/2 = 2.50, and E(x)^2 = 0.50^2 = 0.25. So the variance is 2.50 - 0.25 = 2.25.

[BruceZ]

I just noticed that you asked for the standard deviation. This is just the square root of the variance or sqrt(2.25) = 1.5 bets.

[joeg]

Hi Bruce

Thanks for the detailed explenation, I think I have got that now. Just one more quick example to check that I fully understand it with more than 2 outcomes, suppose for example I was playing dice, where I was paid 8-1 when I hit a 6 and lost when I missed, my E.V. here would 0.5 and 5 times out of 6 I would be 1.5 times below average and the other 1 time I would be 7.5 above average, so I would work out my variance as follows :

(5*(-1.5^2) + 7.5^2)/6 = 11.25

and my standard deviation would be 3.35, is that correct?

Thanks again for your help

Joe.

[BruceZ]

[ QUOTE ]
Hi Bruce

Thanks for the detailed explenation, I think I have got that now. Just one more quick example to check that I fully understand it with more than 2 outcomes, suppose for example I was playing dice, where I was paid 8-1 when I hit a 6 and lost when I missed, my E.V. here would 0.5 and 5 times out of 6 I would be 1.5 times below average and the other 1 time I would be 7.5 above average, so I would work out my variance as follows :

(5*(-1.5^2) + 7.5^2)/6 = 11.25

and my standard deviation would be 3.35, is that correct?

Thanks again for your help

Joe.

[/ QUOTE ]

That's correct.