Hold 'Em Standard Deviation.

[cardcounter0]

Okay, LOL DONKAMENTS.
AJs vs KJo, 3:1 favorite, all-in preflop - lose.
AQo vs ATs, 3:1 favorite, all-in preflop - lose.

So how many times can I push with a 3:1 favorite and lose before I exceed 1 SD?

[DWarrior]

isn't this a population proportion where p=.75?

I believe the formula for standard deviation is sqrt(p*(1-p)/n)

So if you run it 2 times, SD is 0.31. Losing twice seems more than 2 SDs away. But the SD will change depending on the number of trials.

PS, I don't think SD in the above example is very useful because this distribution is not normal, so you need N to be at least 30 before SD becomes a reasonable estimate.

[RonMexico]

I don't really see how standard deviation is relevant here. It's easier to just ask the probability of losing said number of successive 3:1 all-ins. Also, AQo isn't quite 3:1 against ATs.

[cardcounter0]

[ QUOTE ]
I believe the formula for standard deviation is sqrt(p*(1-p)/n)

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thanks. that is the formula I was looking for, used it, forgot it, found it again, used it, and after several years obviously forgot again.

[BruceZ]

[ QUOTE ]
isn't this a population proportion where p=.75?

I believe the formula for standard deviation is sqrt(p*(1-p)/n)

[/ QUOTE ]

That's the standard ERROR, which is the standard deviation of the success RATE or SD/sqrt(n). Note that this gets smaller as n increases. The standard deviation (SD) of the number of successes for n trials is sqrt[p*(1-p)*n]. Note that this increases as the sqrt(n). The standard deviation of a single trial is sqrt[p*(1-p)], where success = 1 and failure = 0.

[cardcounter0]

Dammit! you guys are going to make me dig up my stats books. [img]/images/graemlins/grin.gif[/img]

[fishyak]

Been a long time since my statistics classes, 30+ yrs. But I agree that SD deals with groups of #'s and unless N is big enough, the calculation for the value of the standard deviation becomes too unreliable. Particularly here where the two lone outcomes are rightfully suspected to be outside the probable outcomes.

Wouldn't it be more accurate to say that the odds of losing 2 3:1 shots in a row is 1/3 x 1/3 for 1/9. Since, by definition, +or- two standard deviations is the range that captures the true "average" about 90% of the time, that this outcome is close to the outer marker on using standard deviation to find the true "average."

Any effort to clean up my old terminology is appreciated.

[Copernicus]

[ QUOTE ]
Been a long time since my statistics classes, 30+ yrs. But I agree that SD deals with groups of #'s and unless N is big enough, the calculation for the value of the standard deviation becomes too unreliable. Particularly here where the two lone outcomes are rightfully suspected to be outside the probable outcomes.

Wouldn't it be more accurate to say that the odds of losing 2 3:1 shots in a row is 1/3 x 1/3 for 1/9. Since, by definition, +or- two standard deviations is the range that captures the true "average" about 90% of the time, that this outcome is close to the outer marker on using standard deviation to find the true "average."

Any effort to clean up my old terminology is appreciated.

[/ QUOTE ]

More important than terminology, losing two 3:1 shots in a row is 1/4 * 1/4, not 1/3 * 1/3.