probability

Apologies if the answer to this is readily found elsewhere. My math and stats are weak and thus I don't even know the search terms for what I'm trying to do . . . I have b anonymous indistinguishable buckets into which I'm putting i identical items. I want to know all possible distributions and their probabilities. For example, if I have 3 buckets and 3 items, the answer I want is: [3,0,0] -> 1/9 [2,1,0] -> 6/9 [1,1,1] -> 2/9 Notice that the buckets are anonymous and thus I want identical distributions to be combined as I have done above. For example, the [2,1,0] case is actually the sum of

When implementing a hash table using a good hash function (one where the probability of any two elements colliding is 1 / m, where m is the number of buckets), it is well-known that the average-case running time for looking up an element is Θ(1 + α), where α is the load factor. The worst-case running time is O(n), though, if all the elements end up put into the same bucket. I was recently doing some reading on hash tables and found this article which claims (on page 3) that if α = 1, the expected worst-case complexity is Θ(log n / log log n). By "expected worst-case complexity," I mean, on

For a list of 10 ints, there are 10! possible orders or permutations. Why does random.shuffle give duplicates after only 5000 tries? >>> L = range(10) >>> rL = list() >>> for i in range(5000): ... random.shuffle(L) ... rL.append(L[:]) ... >>> rL = [tuple(e) for e in rL] >>> len(set(rL)) 4997 >>> for i,t in enumerate(rL): ... if rL.count(t) > 1: ... print i,t ... 102 (7, 5, 2, 4, 0, 6, 9, 3, 1, 8) 258 (1, 4, 0, 2, 7, 3, 5, 9, 6, 8) 892 (1, 4, 0, 2, 7, 3, 5, 9, 6, 8) 2878 (7, 5, 2, 4, 0, 6, 9, 3, 1, 8) 4123 (5, 8, 0, 1, 7, 3, 2, 4, 6, 9) 4633 (5, 8, 0, 1, 7, 3, 2, 4, 6, 9) >>> 10*9*8*7*6*5*4*3*2

This question already has answers here : Closed 8 years ago . Possible Duplicates: Random weighted choice Generate random numbers with a given (numerical) distribution I have a list of list which contains a series on numbers and there associated probabilities. prob_list = [[1, 0.5], [2, 0.25], [3, 0.05], [4, 0.01], [5, 0.09], [6, 0.1]] for example in prob_list[0] the number 1 has a probability of 0.5 associated with it. So you would expect 1 to show up 50% of the time. How do I add weight to the numbers when I select them? NOTE: the amount of numbers in the list can vary from 6 - 100 EDIT In

I have an integral to evaluate "x^(-0.5)" ; x in [0.01,1] for which I am using Importance Sampling MC : The theory says that an approximate PDF has to be used to compute the expected value (which will almost surely converge to the mean - value of the integral) After plotting the given integral, and exponential PDF, based only on the plots, I chose the rexp and dexp to generate the PDF - and my code looks like this - #Without Importance Sampling set.seed(1909) X <- runif(1000,0.01,1) Y <- X^(-0.5) c( mean(Y), var(Y) ) #Importance sampling Monte Carlo w <- function(x) dunif(x, 0.01, 1)/dexp(x

So i'm implementing a heuristic algorithm, and i've come across this function. I have an array of 1 to n (0 to n-1 on C, w/e). I want to choose a number of elements i'll copy to another array. Given a parameter y, (0 < y <= 1), i want to have a distribution of numbers whose average is (y * n). That means that whenever i call this function, it gives me a number, between 0 and n, and the average of these numbers is y*n. According to the author, "l" is a random number: 0 < l < n . On my test code its currently generating 0 <= l <= n. And i had the right code, but i'm messing with this for hours

My brother turns 21 in a couple of weeks and my parents and I are taking him to Las Vegas. For my 21st, I brought $200 to gamble in Vegas and came home with around $450, mostly from playing craps. I plan on bringing $200 again for this trip and before I go I thought I'd run some craps simulations to see if I can double my money again. I've read from several sources that the house has the smallest advantage in craps when placing a passline bet with maximum odds. From my memory, and as surveyed by Wizard of Odds , most casinos on the Strip are 3-4-5 odds with a $5 minimum. Taking this into

The classic Fisher Yates looks something like this: void shuffle1(std::vector<int>& vec) { int n = vec.size(); for (int i = n - 1; i > 0; --i) { std::swap(vec[i], vec[rand() % (i + 1)]); } } Yesterday, I implemented the iteration "backwards" by mistake: void shuffle2(std::vector<int>& vec) { int n = vec.size(); for (int i = 1; i < n; ++i) { std::swap(vec[i], vec[rand() % (i + 1)]); } } Is this version in any way worse (or better) than the first? Does it skew the resulting probabilities? Yes it's even distribution assuming rand() is. We will prove this by showing that each input can generate