debitncredit
debitncredit
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April 24th, 2015 at 10:36:46 AM permalink
Just realized that this would be a better category to post under.

I play bonus poker. A royal shows up, on average, about little over 40,000 hands. What's the standard deviation for the number of hands for getting a royal?

For example, if it's been 60,000 hands since my last royal, I'd say that's variance, and it'd probably fall within one standard deviation. However, if it's been 500,000 hands since my last royal, that's likely be at the tail end of the distribution.

Just wondering. Some people on another forum had no concept of statistics, and said things like, "Theory is often different from the reality." Umm.... no. That's not how statistics works.

Before anyone says something like, "Each hand is independent, so there's no 'due to hit.'" Of course it's never due, but you absolutely can get a distribution and stddev of number of hands between royals. If you run a bootstrapping simulation for # of hands to hit royals, you'll have a normal distribution with a mean of 40,233.
dwheatley
dwheatley
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April 24th, 2015 at 11:33:21 AM permalink
Getting a royal within a certain number of hands has the negative binomial distribution, with failure p=39999/40000 to make the math easy. We set r=1, the number of royals we want.

The mean is p*r / 1-p = 39999 (we fail 39999 times on average, and get it once)

The variance is known to be: p*r / (1-p)^2 = 39999*40000
so the standard deviation is essentially 40000 hands. Convenient.

You can casually say that within 80,000 hands is within one standard deviation, but this implies that the curve looks normal. You can use a different calculation to find the probability of not getting a royal for 500,000 hands.
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teliot
teliot
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April 24th, 2015 at 11:48:14 AM permalink
Quote: dwheatley

Getting a royal within a certain number of hands has the negative binomial distribution, with failure p=39999/40000 to make the math easy. We set r=1, the number of royals we want.

The mean is p*r / 1-p = 39999 (we fail 39999 times on average, and get it once)

The variance is known to be: p*r / (1-p)^2 = 39999*40000
so the standard deviation is essentially 40000 hands. Convenient.

You can casually say that within 80,000 hands is within one standard deviation, but this implies that the curve looks normal. You can use a different calculation to find the probability of not getting a royal for 500,000 hands.

I just posted this same thing in his other thread on this topic. I think a more accurate description is that of a "geometric distribution."

http://en.wikipedia.org/wiki/Geometric_distribution

The geometric distribution is the discrete form of the exponential distribution.
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Seriouslyfunny
Seriouslyfunny
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April 24th, 2015 at 11:54:17 AM permalink
I have to say, the knowledge here just flows! Nice.
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