Standard Deviation Answers
Byron,
You asked what is the standard deviation (SD) of this sidebet. That's easy to calculate. As I mentioned in my last post, the winning probability is W = 0.09676, and a win pays 9:1. Since the SB has no pushes, that means the losing probability is L = 1 - W = 0.90324, and each loss "pays" -1:1. Finally, we saw that the overall EV of this SB is -3.24%, so that means the average (actually, the "mean", represented by the Greek letter "mu") outcome is -0.0324.
As any elementary statistics textbook (or, fo that matter, Wikipedia's "Standard Deviation" page: see link below) will tell you, the SD is the square root of the sum of the winning variance times its probability and the losing variance times its probability. For our case, we get:
SD = SQRT(W*(9-mu)^2+L*(-1-mu)^2)
SD = SQRT(0.09676*(9-(-0.0324))^2 + 0.90324*(-1-(-0.0324))^2)
SD = 2.956...
Thus, the SB has an EV of -0.0324 with a SD of 2.956...
Now, for your second question, you wanted to know the percentage of time you'd go 20, 30, or 45 rounds without winning the SB.
I assume here, you're counting ONLY those rounds on which you PLAY the SB, since for me, the probability is exactly 1 because I never play this SB ;-)
At any rate, the probability that you will lose the SB the next n times in a row is simply L^n, so for 20, 30, and 45 we get probabilities of 0.13064, 0.04722, and 0.01026, respectively.
Hope this helps!
Dog Hand