Variance - How to calculate?

I think I understand the concept of house edge. The thing that I keep seeing mentioned on these forums is variance. So... starting with betting red on a roulette table with no zero (basically a coin toss, for simplicity's sake):

How much of a deviation from 50/50 can I expect after -

10 spins?
100 spins?
1000 spins?
10000 spins? etc...

Is there a formula to calculate this?

What about with the zero - 18/37 (48.65%) to win, 19/37 (51.35%) to lose.

I guess this is to do with the "standard deviation" term that I keep seeing. Any detailed explanation of this would be very appreciated.

You might need to wake up the stuff that's been dormant in your brain since your last math class, but for a good, detailed explanation, Wikipedia does a great job:
http://en.wikipedia.org/wiki/Standard_deviation
http://en.wikipedia.org/wiki/Variance

Thanks DJ!

I'll check it out - I've been meaning to wake up the dormant areas in my brain anyways :P

Quote: DJTeddyBear

You might need to wake up the stuff that's been dormant in your brain ...

Oh I do, I do!! Unfortunately, the stuff that's been dormant insists on remaining dormant and each day more and more of the remaining brain cells insist upon joining them!

Here's the basics:

For a single coin flip, the average number of heads which will show up is 1/2. The standard deviation computes out to be 1/2, also. Which kinda makes sense, since the only possible outcomes are 0 (mean - standard deviation) and 1 (mean + standard deviation).

For multiple trials, the mean increases proportionally to the number of trials. For 100 flips, the average number of heads is 1/2 x 100 = 50. But the standard deviation grows proportionally to the square root of the number of trials. So the standard deviation is 1/2 x sqrt(100) = 5.

For a large number of trials, outcomes are typically within 1 standard deviation, usually within 2 standard deviations, and almost always within 3 standard deviations of the mean. So for 100 flips, you'll usually have between 45 and 55 heads, although results between 35 and 65 are possible.

For your roulette question, the mean win drops from 50% to 48.65%, as you specify. The standard deviation is still going to be very close to 1/2. Actually, its

sqrt((.4865^2 + .5135^2)/2) = 0.5002

So for many trials, the house edge will increase with the number of trials. But the variance is almost identical to a zero-edge game.

Thanks for the help, this is much simpler than the wikipedia page - I guess my brain cells have been dormant for longer than I thought because I'm having trouble making much use of it.

With the formula you supplied though:

Quote: PapaChubby

sqrt((.4865^2 + .5135^2)/2) = 0.5002

Am I right in assuming this is the standard deviation for 1 trial? I'm not sure I understand where to plug in the value to figure it out for any larger number of trials.

Thanks again!

Quote: DJGenius

Thanks for the help, this is much simpler than the wikipedia page - I guess my brain cells have been dormant for longer than I thought because I'm having trouble making much use of it.

With the formula you supplied though:



Am I right in assuming this is the standard deviation for 1 trial? I'm not sure I understand where to plug in the value to figure it out for any larger number of trials.

Thanks again!

Yes, this is the standard deviation for one trial. The standard deviation for multiple trials is this times the square root of the number of trials. For 100 trials, the standard deviation is 5.002. For 10,000 trials, the standard deviation is 50.02.

So for 10,000 spins, the average number of red results is 4865. The standard deviation is 50. Since the payout is 1:1, you'd need 5000 red results to break even. The break even point is 2.7 standard deviations from the mean, indicating that you will lose money about 99% of the time. All because of that one pesky green spot!