Probability Question

I have what is likely a fairly simple question that I can't figure out how to answer. It has to do with the chances of rolling a 2 on 2 six sided dice over an arbitrary number of rolls. Given n rolls, I know the chances of rolling a 2 at least once are 1-(35/36)^n, correct? My thinking is to calculate the chance of an event happening at least once, you calculate the chance it never occurs and take its complement.

But how do I calculate the chances of rolling a 2 at least twice? Or 3 times? Etc. I can pretty easily just simulate it to find out but I'm more interested in understanding the math. Is there a simple formula for figuring this kind of question out?

Stephen.

The odds of rolling a 2 is 1 / 36.

Then we look at the binomial probability, which states

P(k out of n) = (n! / ((k!)(n-k)!)) * p^k * q^(n-k), where

P = probability
k = number of instances of desired result
n = number of trials
p = probability of trial being true
q = probability of trial being false = (1-p)

So lets say you have 72 rolls and you want the odds of rolling 3 twos in those 72 rolls.

P(3 of 72) = 72!/3!69! * (1/36)^3 x (35/36)^69
P(3 of 72) = (72*71*70)/(3*2*1) * .0000214335 * .143160
P(3 of 72) = .1830

You can chart this nicely in Excel. n! is a formula FACT(n). Using the fixed references, you can put p in a cell, n on the left and k across the top, and create a nice table that lists P based on k successes over n trials based on the probability of success p.

The problem is that you can only go up to about n=170 in excel. After that you need to deal with the binomial approximation.

Thanks boymimbo. However, after building the Excel chart and reading up on the binomial probability, it seems like that gives me probability of rolling my number exactly k times in n tries. So if k = 3, then the binomial probability excludes results where I rolled a 2 four times in n rolls.

How can I figure out the chances of rolling a number *at least* k times (but possibly more) in n iterations?

Just take 1 - p(k=3, 2, 1) to get the probability of more then 4 times.

I was looking at the Pai Gow Poker odds and was trying to figure out how you came up with the percentages. The Wizard of Odds states that there are 154,143,080 total combinations. It also says that there are 64,221,960 combinations of a pair equaling 0.41663862. I entered this information into a conversion calculator and got this:

64221960 is what percent of 154143080? Answer: 41.6638619132023

I was curious why you guys show 0.4166... and the calculator shows 41.66...

Thanks,
justagambler

boymimbo, thanks again for the help but I'm challenged here. Could you break it down a little more for me? I don't entirely follow you.

Let's go back to original example of n = 72. In that case, p(k=1) is .27063, p(k=2) is .2745, and p(k=3) is .183. Using that data, how can I derive the probability that in 72 rolls, I roll a 2 at least 3 times?

Oh, and justagambler: 0.4166 is the same as 41.66%. Probabilities are usually expressed as decimals rather than percentages, the idea being that the probabilities of all of the possible outcomes for an event add up to 1 (which is the same as 100%). Translating between them is easy: just move the decimal over two places.

P(k>=3) = 1 - P(k=2) - P(k=1) - P(k=0)

boymimbo: Ah hah! I get it, then. I wasn't thinking about the P(k=0) case.

Quote: boymimbo

The problem is that you can only go up to about n=170 in excel. After that you need to deal with the binomial approximation.

Or you could use BINOMDIST(k,n,p,0) :^)

(Or BINOMDIST(k,n,p,1) for up to k times instead of exactly k times.)

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