Standard deviation for VP hands

Is the standard deviation for individual hands in video poker simply the square root? For example, if expectation over 8900 hands for straights is 100, is the SD of those 100 straights 10? Or is it more complex?

Yes, it is almost the square root, for sufficient "rare" events.

For an event occuring with probability p for N tries, expected number of events is N * p and standard deviation is sqrt(N * p * (1-p))

As for rare events 1-p is close to 1, you end up with sqrt(N*p), which is the square root of your observed events.

Quote: Caruso

Is the standard deviation for individual hands in video poker simply the square root?

it is the square root of variance. in this case it works out to be

Quote: Caruso

For example, if expectation over 8900 hands for straights is 100, is the SD of those 100 straights 10?
Or is it more complex?

to keep it simple
you want the binomial standard deviation for the number of successes in 8900 trials.
for JOB you have the mean (N*P)

vp tables shows the variance (sd) for your bets. the money part.
that will not work for what you want to know, if I understand your question

still square root of variance
N*P*Q


for a fair coin flip the square root of the ev would have too much error
100 flips
p=0.50
ev= 50
sqrt = 7.07

N*P*Q = 25
sd = 5


extra credit
for the SD of P in OP example
P = 1.1229%
SD of P = 0.1117%
correct?

Thanks, gentlemen.

In a current quarter million hand war of attrition with a sucky online JOB game I'm trying to establish probability parametres for sucky results, hence the question.