deviation of video poker

Hello all,

i have some questions about video poker:

a) in WizardofOdds, this result is given for some kind of video poker when following perfect strategy :

Hand Payoff Probability Hit Frequency Variance
Royal Flush 800 0.000025 40,216.50 15.87
Straight Flush 50 0.000107 9,378.82 0.26
....
my question is how the Variance is calculated here? and

b) How to calculate Standard Deviation of a vp game when following perfect strategy ?

c) If i choose Double or nothing after each winning bet, then how to calculate the variance, assuming i follow the perfect strategy ?


Thank you in advance for your time.

(a):

If you have a payoff W with probability p, the monetary variance V is V = W^2 * p * (1-p).
For the Royal Flush W=800 and p = 1/40,216.50, yields a variance of V = 15.87.

(b): Standard deviation S is the square root of the variance.
For the Royal Flush, take the square root of 15.87, which is 3.98. Or equivalently (giving more precise results) take the square root of the variance formula, i.e.
S = W * sqrt(p * (1-p))

(c):
If you double up, you double the payoff W, but half the probability p. Hence, substitute W with 2*W and p with p/2 in the variance formula.
So the variance including double up is VD = 4*W^2 * p/2 * (1 - p/2). For the Royal Flush it is 31.83.

Observe that this is almost exactly double the variance without the double-up. In fact since p is much much smaller than 1 (and even more p/2), the (1 - p) factor or (1 - p/2) factor equates almost to 1 and could be neglected. Then for these rare events the variance including double-up is twice the variance without double-up.

Thank you for the answers and clear explanations. And i have a couple of more questions if it doesn't bother you too much :

a) Now we know that a Royal Flush takes place every 40216.50 times, EV = -0.9799(w*p+(-1)*(1-p)), and the variance is 15.87, but what does this actually mean? is it "for a big enough set of trials, there's 68.3% possibility that the mean value is between -0.9799-3.98 and -0.9799+3.98" ?

b) If i am allowed to double up for not sure how many times, i.e, once if i lose in comparison, or twice if i win in the first comparison, than how to calculate the variance? do i substitute the W~p with 4*W~p/4?

Thanks in advance again !

Variance in video poker doesn't really mean that much because the distribution is so skewed by the jackpots that it would take millions and millions of hands to converge to normal.

Case in point if you play 10/7 DB(100.1%) and Full Pay Pick-Em (99.95%) for one hour you're actually more likely to be ahead playing pick-em instead of DB even though it's a negative game with a much lower variance.

To get a sense of video poker it's better to run monte-carlo simulations.

Quote: Tortoise

Variance in video poker doesn't really mean that much because the distribution is so skewed by the jackpots that it would take millions and millions of hands to converge to normal.

Case in point if you play 10/7 DB(100.1%) and Full Pay Pick-Em (99.95%) for one hour you're actually more likely to be ahead playing pick-em instead of DB even though it's a negative game with a much lower variance.

To get a sense of video poker it's better to run monte-carlo simulations.

Well, variance still matters, but it's true (and important!) that the distribution is non-normal.

I asked some questions about this in a different thread recently and EnderMike and MustangSally did some really good simulations and shared some data that was very enlightening.

Quote: ilikevp

a) Now we know that a Royal Flush takes place every 40216.50 times, EV = -0.9799(w*p+(-1)*(1-p)), and the variance is 15.87, but what does this actually mean? is
it "for a big enough set of trials, there's 68.3% possibility that the mean value is between -0.9799-3.98 and -0.9799+3.98" ?

That is a over-simpification. You see your loss on any single game can never exceed your stake. You never end up with less than -1, and hence any mean result can never be below -1. Stating a loss of -0.9799-3.98 makes no sense.

The concept of "68% within one standard deviation" only applies to Gaussian probability distributions, i.e. a distribution of the famous "bell shape". You can calculate standard deviations for almost any other probability distributions, but then they won't have much to do with the 68% estimate.

In the case of the Royal Flush, the probability distribution is either hit a large amount or miss a small amount. That is very different from a Gaussian distribution. In fact you are 99.997% sure your result falls below the mean value of -0.9799. (For a Gaussian, this would be 50%)

Quote:

b) If i am allowed to double up for not sure how many times, i.e, once if i lose in comparison, or twice if i win in the first comparison, than how to calculate the variance? do i substitute the W~p with 4*W~p/4?

That depends on how often you want to double up. If you will always try it twice, then yes you substitude W by 4*W, and p by p/4. It will quadruple your variance.
If you *always* try the double up N times, you substitute W by 2^N * W, and p by p / N^2.

If you are "not sure how many times" to double up, then there is no real formula unless you specify exactly how you decide to try the next double up.

I am still confused about question a), need some time to get through this, but the answers do make it clearer, thank you all!

yes, i found that thread, although have some difficulties understanding the meanings of the plots. meanwhile This article seems to give more direct information, or at least can be a useful reference if your choice of vp is not listed.