probability

Suppose that I have an n-sided loaded die where each side k has some probability p k of coming up when I roll it. I'm curious if there is good algorithm for storing this information statically (i.e. for a fixed set of probabilities) so that I can efficiently simulate a random roll of the die. Currently, I have an O(lg n) solution for this problem. The idea is to store a table of the cumulative probability of the first k sides for all k, them to generate a random real number in the range [0, 1) and perform a binary search over the table to get the largest index whose cumulative value is no

In some code I want to choose n random numbers in [0,1) which sum to 1 . I do so by choosing the numbers independently in [0,1) and normalizing them by dividing each one by the total sum: numbers = [random() for i in range(n)] numbers = [n/sum(numbers) for n in numbers] My "problem" is, that the distribution I get out is quite skew. Choosing a million numbers not a single one gets over 1/2 . By some effort I've calculated the pdf, and it's not nice. Here is the weird looking pdf I get for 5 variables: Do you have an idea for a nice algorithm to choose the numbers, that result in a more uniform

I want to generate a random number with a given probability but I'm not sure how to: I need a number between 1 and 3 num = ceil(rand*3); but I need different values to have different probabilities of generating eg. 0.5 chance of 1 0.1 chance of 2 0.4 chance of 3 I'm sure this is straightforward but I can't think of how to do it. The simple solution is to generate a number with a uniform distribution (using rand ), and manipulate it a bit: r = rand; prob = [0.5, 0.1, 0.4]; x = sum(r >= cumsum([0, prob])); or in a one-liner: x = sum(rand >= cumsum([0, 0.5, 0.1, 0.4])); Explanation Here r is a

Say you have two hashes H(A) and H(B) and you want to combine them. I've read that a good way to combine two hashes is to XOR them, e.g. XOR( H(A), H(B) ) . The best explanation I've found is touched briefly here on these hash function guidelines : XORing two numbers with roughly random distribution results in another number still with roughly random distribution*, but which now depends on the two values. ... * At each bit of the two numbers to combine, a 0 is output if the two bits are equal, else a 1. In other words, in 50% of the combinations, a 1 will be output. So if the two input bits