Let’s say I want to find the expected number of sixes on six rolls of a fair die. I can do a probability distribution using the binomial theorem for each possibility ( no sixes, one six, two sixes etc…), then multiply each probability by it’s associated number. (i.e. one times the prob. of one six, two times the prob of two sixes etc.) added up, I get the expected number.

But what if I have sixty rolls…or one thousand? In principle the technique should be the same, but it seems pretty tedious. Is there any shortcut that involves only pencil and paper? (I assume there are programs for this)

If there is a solution and it’s over my head, as happened with my last math question, that’s fine, I’d just like to know.

Once again, thanks in advance to the board mathematicians for answering my stupid question!

Well, the quickest shortcut is to notice that since the expectation is a linear operator and the rolls are all independent identical Bernoulli trials, the expected number of sixes in N rolls is N times the probability of a six in a single roll, just like you’d expect. (Formally, if S[sub]n[/sub] is the Bernoulli random variable, 1 if a six is rolled on trial n and 0 otherwise, then you want to know E{S[sub]1[/sub]+…+S[sub]N[/sub]} = E{S[sub]1[/sub]} + … + E{S[sub]N[/sub]}.)

If you just want to know more generally how to compute sums of the form
(sum)[sub]n[/sub] ([sub]N[/sub]C[sub]n[/sub])np[sup]n/sup[sup]N-n[/sup],
this can be done pretty easily as well. Consider the sum
(p+q)[sup]N[/sup] = (sum)[sub]n[/sub] ([sub]N[/sub]C[sub]n[/sub])p[sup]n[/sup]q[sup]N-n[/sup].
Differentiating this equation with respect to p, then multiplying by p and evaluating at p+q=1, gives a simple formula for the sum above.

I’m not sure what you’re asking, but is it something like this? You want to know what’s the probability of rolling between 150 and 200 sixes with 1000 die rolls? You can calculate each number with the formula for the binomial distribution, but that’s a lot of figuring to cover that range. It would be nice if there was an easy way to do that.

Now this could possibly be the question that Omphaloskeptic’s answering, but his answer is a little dense for me, 22 years out of college.

CurtC, something like what you’re asking (the probability of rolling between 150 and 200 sixes in 1000 rolls) is often handled by using the normal distribution to approximate the binomial distribution. With a large enough number of trials (like 1000), the number of “successes” out of n trials pretty closely approximates the normal (Gaussian, bell-shaped) distribution.

But actually, it looks like the OP asked a simpler question. When you have n independent trials (like n die rolls), and each one has a probability p of “success,” the expected number of successes is np.

Either I’m misunderstanding the OP’s question, or some of you are making it a lot more complicated than it has to be.

Sorry that was a really bad example. After messing around with binomial distributions and the formula for mathematical expectation I discovered that the expected number of sixes is in fact, one. (for six throws, that is.) just what you get in two seconds using Thudlow Boink’s common sense answer! :smack:

I feel like a frackin’ idiot. I was looking at some problems in basic probability which led me to way overthink this question. Sorry, guys.

Nonetheless, I think there is a valid question in there somewhere, when dealing with less obvious probability distributions. I’m not sure what it is yet. :o

Actually you do not require independence between the trials (rolls) as both you and Omphaloskeptic suggest. For any random variables E[x + y] = E + E[y] regardless of whether x and y are independent or not.