? on Standard Deviation in a Double 0 or American Roulette

How do I calculate the std dev for the frequency of a 5-number section occurring (i.e. neighbors or 5-consecutive numbers on the wheel) or even 6-number section, etc. up to 9-number section?

I've been watching an Interblock Roulette wheel and I've seen a 5-number section that didn't hit for 67 consecutive spins. I know the average hit frequency is simply 38/5 or 7.6. I also know that the probability of these 5-number section not hitting for 67 consecutive is (33/38)^67 or about 1 in 12,737.

Then I recently witness some really rare events that makes me think maybe I really don't know gambling statistics.
-- a 8-number section went zero for 49 consecutive spins -- (30/38)^49 -- or about 1 in 107,264 (please note the average hit frequency is 38/8 or 4.75)
-- a 6-number section went zero for 74 consecutive spins -- (32/38)^74 -- or about 1 in 333,340 (please note the average hit frequency is 38/6 or 6.33)

Assuming my math is correct, my next thought was to calculate the std dev for the frequency these bets but I am not sure how to do. I was curious to see how many sigma's std dev these rare events were. To put things into perspective, the average RF in video poker occurs once in about 40,000 hands. I find it hard to believe a wheel with 38 numbers could have such reach such rare frequencies that often.

If someone could showed me how to calculate the std dev for a 5-number section, I think I can translate it to 6-number, 7-number, 8-number, and 9-number sections.

Thank you.

binomial standard deviation
Square root of (N * P * Q)
N = number of trials
P = probability of success
Q = 1-P

EV = P * N

Let us know what you find out
Good Luck!

Sorry, but that answer can't be correct since it gives a population-specific standard deviation answer.

I am interested in the hit frequency std deviation since I already know the mean (i.e. the mu is 38/5) and there must be a formula that solves the std dev of the hit frequency without regards to "N" trials. For example, if I am interested in knowing the std dev of a 6-side die and I know a "5" occurs once (on average) in six spins, I would like to know the std dev of the frequency of "5" occurring and this is the same logic in my roulette question. That is to say the correct answer is independent of the number of trials.

I am really interested to know how many std dev it was when a 6-number sector bet took so many spins before it hit again.

Thanks again.

The reason it depends on the number of trials is that what the standard deviation is is the square root of the variance, which adds up with each trial. When you talk about the mean, it sounds like you're talking about the mean per bet (and I'm not sure exactly what 38/5 is...), and that you can only have as the variance.

A 5-number section (i.e. neighbors) occurs on average once in 7.6 spins or 38/5. This mean occurrence DOES NOT change with the number of trials. If the mean does not change with the number of trials, then the dispersion or std dev also does not change with the number of trials. The ABSOLUTE DISPERSION will change based on the number of trials -- however, I am not asking about the absolute population-specific std dev.

Thanks.

Ah, I see what you mean. In some large sample of trials, the mean number of spins before hitting would be 7.6. Okay, that's right.

The standard deviation, then, would be... hmm. I believe the mean of the square of the number until you hit would be equal to 2698/25, making the variance 1254/25 = 50.16, making the standard deviation roughly 7.082. To work it out for yourself, and for any equation, you'll want to use the formula that the mean of the square of the number of tries until you hit will be equal to (2-p)/p^2, (where p, of course, is the chance of hitting on a given try), so subtract from that the square of the average number of tries before you hit, and the square root of that difference will be the standard deviation.

I think there's only so much use you can get out of that number, though, due to the shape of the distribution, unless you were actually doing a number of random trials. If you're trying to work out probabilities, I recommend you just do it directly.

I may need a reference or weblink on your approach.

If I understand what you did to arrive at a std dev number of 7.082 for a 5-number section (and mean of 7.6), then
1 S.D. would be 7.6 + 7.082 or 14.682 spins -- basically the mean plus the std dev
2 S.D. would be 7.6 + 2*(7.082) or 21.764 spins
3 S.D. would be 7.6 + 3*(7.082) or 28.846 spins
4 S.D. would be 7.6 + 4*(7.082) or 35.928 spins
5 S.D. would be 7.6 + 5*(7.082) or 43.010 spins
6 S.D. would be 7.6 + 6*(7.082) or 50.092 spins

Or stated another way, 67 consecutive misses betting a 5-number section or neighbors would be about a 8.4 S.D. event??? [(67 minus the mean of 7.6) / 7.082] = 8.4

You look at Wikipedia (source: http://en.wikipedia.org/wiki/Standard_deviation) and a 7 S.D. event is extremely rare.

Clearly I am missing something. I am going to try posting this question on the Math forum to see if someone can find out where I went wrong.

Thanks.

Quote: Ardent1

Clearly I am missing something. I am going to try posting this question on the Math forum to see if someone can find out where I went wrong.

Thanks.

OK.
The approach is the negative binomial distribution for one success.

You are assuming that the distribution for the WAIT TIME of a binomial random variable is a "normal distribution".
FAR from it. I can post a few pics when I get back to my own computer.

Most normal distributions, the mean, mode and median are about the same. Centered in the distribution and very easy to see.
Not with a wait time dist.
The mode is the first trial, the median is about 70% of the mean and the mean is not even close to the peak of the curve.

For just one success in N trials you already know the probability distribution.
After more successes,more than one, the distribution starts to look more like a normal distribution.

More after the holidays if you have not figured it out.

7craps: Thank you, you are the first person to answer my question. I am partly to blame for wording the question poorly.

I am asking about the std dev on a priori basis while everyone answered it on a posteriori basis, i.e. on a population-specific answer or an answer that depended on the number of trials. My question is really, before these trials begin, we must already have a known std dev (hence, it is a constant) and after the first hit or first successful trial, we look to see what it is based on the apriori std dev.

> For just one success in N trials you already know the probability distribution

Exactly, but I don't know how to convert that "probability distribution" that into a std dev number, hence the std dev on a priori basis for the first hit or first successful trial.

I am going to explain the problem in more detail on the Math forum since that is where this question ought to belong.

Thanks again.