MJ: CVCX and Standard Deviation

[MJ]

On CVCX my Win Rate is $24.00/Hr. Does that mean the statistic of 7.832 SD/Hand represents +/- 1 one sigma of my win rate? In other words 68% of my hands played will fall within the range of $24/Hr +/- $7.83? Is this correct? If so then 95% of my hands will earn $24/Hr +/- $15.86(2 sigmas)? There is no Avg Earning/Hand given so I dont know how to interpret the SD in this case.

If that is correct I can apply the same idea to SD per hour I suppose. But something tells me I am adding apples and oranges in the above example because I was never given the Mean earning per Hand.

Lastly I don't understand how 1 SD is MORE likely to occur then 2 SD. If 1 SD falls within a range that occurs 68% of the time, and 2 SD is falls within a range occuring 95% of the time, then 2 SD clearly occurs MORE OFTEN then 1 SD. Oddly enough it seems as though the further we veer away from the mean the more likely an event will occur. Where am I going wrong? Thanks for any answers.

-MJ

[Don Schlesinger]

> On CVCX my Win Rate is $24.00/Hr. Does that
> mean the statistic of 7.832 SD/Hand
> represents +/- 1 one sigma of my win rate?

What is your unit size? The 7.832 is expressed in units, while the hourly win rate is expressed in dollars. You can't compare the two the way you're doing. What's more, the win rate is per hour (100 hands?), while the SD is per hand, so if you play 100 hands per hour, you'd have to multiply the per-hand SD by 10 (square root of 100) to get the hourly SD.

> In other words 68% of my hands played will
> fall within the range of $24/Hr +/- $7.83?

Don't we wish!! No, not at all. The hourly SD is usually anywhere from 10 to 20 times the hourly win rate.

> Is this correct?

Nope.

> If so then 95% of my hands
> will earn $24/Hr +/- $15.86(2 sigmas)? There
> is no Avg Earning/Hand given so I dont know
> how to interpret the SD in this case.

See above.

> If that is correct I can apply the same idea
> to SD per hour I suppose. But something
> tells me I am adding apples and oranges in
> the above example because I was never given
> the Mean earning per Hand.

If you've set CVCX to play 100 hands per hour, the per-hand win rate is 1/100 of the hourly win rate.

> Lastly I don't understand how 1 SD is MORE
> likely to occur then 2 SD.

It isn't.

> If 1 SD falls
> within a range that occurs 68% of the time,
> and 2 SD is falls within a range occuring
> 95% of the time, then 2 SD clearly occurs
> MORE OFTEN then 1 SD.

You have it backwards. The result of your play falls withing a one-SD bandwidth around the mean 68.3% of the time. If you widen that bandwidth to two SDs, then the result you obtain will fall within that widened range 95.4% of the time.

> Oddly enough it seems
> as though the further we veer away from the
> mean the more likely an event will occur.

No, not at all. See above.

> Where am I going wrong? Thanks for any
> answers.

Hope I've helped.

Don