General Probability Theory Question (a bit OT...)

I thought I'd post this here rather than in "Other Topics" which is currently filled with discussion of world politics...


Say there is an event with a possible outcome O. The probability is x of O occurring and (1-x) of it not occurring. The events are independent. Now say this event happens n times and outcome O occurs m times, where m/n does not equal x. How can I figure out the probability of either this or a worse deviation from the norm occurring?


That was a little confused, so here's a practical demonstration: I roll a die 120 times. The probability of rolling a six is 1/6, same as every other number, so in theory I will observe 20 sixes. Let's say I actually roll 35 sixes. How can I work out what the probability was that I would roll anywhere from 35 to 120 sixes? (If I roll less than 20 sixes, I can just work out the same thing for non-sixes).


Note: I know basic probability, but I know virtually nothing about statistical theory, so if the answer involves standard deviations, or combinations, or anything of that kind, you'll have to remind me how to calculate them too.


Thanks


Chris

I'll redefine your variables for a minute...

Let s = how many successes you want

Let f = how many failures you want

Let Ps = probability of success

Let Pf = probability of failure

THEN

the probability of exactly s successes and f failures is

(s+f)C(s). Ps^s. Pf^f

Note that (s+f)C(s) also equals (s+f)C(f), so you can swap your idea of success and failure without affecting the result.


I'm not explaining combinations here, maybe get a book, or someone else can explain it. Microsoft excel has a function, combina, which will evaluate it for you though.


So in your example, you want 35 6s and 85 non-6s...

P = 120C35 * (1/6)^35 * (5/6)^85


I haven't figured out how to do 35 6s or better yet, other than to just spread them out on a spreadsheet and sum the individual probabilities.


I'll let you know if I come up with anything.

For these problems you use the normal approximation to the binomial distribution. Using your dice problem, the mean is 20 and the SD is SQR of NPQ. Here that is SQR of 120 x 1/6 x 5/6 which is a little above 4. 35 or more successes is 15 more than expected, about 3.7 SDs more. I think that is close to 500-1. Look it up in a normal distribution table.

Here is a link to a good paper on the subject.


http://www.math.tamu.edu/~marvit/finite/notes/ch8sec456.pdf


All that is left is for someone smart to give us the proof of why a binomial distribution can be approximated by a normal distribution, and I'll be dancing in the streets.

Look up the central limit theorem. It says that the sum of N random variables approaches the normal distribution under certain conditions (I believe only finite variance is needed).