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Old 12-15-2005, 10:13 PM
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Default Re: Does a decreased winrate have an exponential effect on variance?

By variance you actually mean "swings." Variance is the square of standard deviation; it is NOT a function of winrate.

Let's say you wanted to calculate what your worst possible downswing could be.

Let's assume a 3 SD downswing for "worst possible".
The equation you want to minimize is:

Winnings = Winrate_per_100hands*(N) - 3*SD_per_100hands*SQRT(N)

This is how much you're up (down if negative) after N blocks of 100 hands.
We take the derivative with respect to N (abbreviate Winrate_per_100hands as WR and SD_per_100hands as SD):

D(winnings)/Dn = WR - (3/2)SD/[SQRT(N)].

We set this derivative equal to 0 and solve for N:

N= [1.5SD/WR]^2

Now plug this back into your Winnings equation:

Winnings = 2.25SD^2/WR - 4.5*SD^2/WR = -2.25SD^2/WR

So...we see that your worst possible downswing is proportional to the square of your SD and inversely proportional to your WR. So halving your winrate doubles your worst possible downswing. Going from 2bb/100 to 0.2bb/100 multiplies your worst possible downswing by a factor of 10.

Hope this helps.
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