Quote:Original post by PenanceQuote:Original post by Sneftel
Waitasec.Quote:Original post by PenanceYou'd love to use a map to do addition for you?
Say I want to simulate 2d6. That means there are 6^2 = 36 possibilities that map to 11 buckets (2-12). I'd love to be able to call rand() once and get 0-35 as a result, and then map that back to the 2-12 range while preserving the proper probability distribution.
Not sure what you mean. =(
Quote:Original post by alvaro
A very low-tech solution that saves you lots of calls to rand() is to extract more than one die roll from each call to rand().
*** Source Snippet Removed ***
Quote:Original post by SneftelQuote:Original post by PenanceQuote:Original post by Sneftel
Waitasec.Quote:Original post by PenanceYou'd love to use a map to do addition for you?
Say I want to simulate 2d6. That means there are 6^2 = 36 possibilities that map to 11 buckets (2-12). I'd love to be able to call rand() once and get 0-35 as a result, and then map that back to the 2-12 range while preserving the proper probability distribution.
Not sure what you mean. =(
output = input%6 + input/6 + 2 . It's just addition.
Quote:Original post by PenanceWhat's your point? Iterate through the inputs 0-35, and you will get the correct number of 2s output, the correct number of 3s ouptut, and so on up to the correct number of 12s output.
16%6 + 16/6 + 2 = 8, not 7
#include <cmath>#include <iostream>#include <map>int dice_roll(int number_of_dice, int number_of_sides, int seed) { int accumulator = 0; for (int i = 0; i < number_of_dice; i++) { accumulator += (seed % number_of_sides); seed /= number_of_sides; } return accumulator + number_of_dice;}int main(int, char **) { std::map<int, int> results; for (int i = 0; i < 6*6*6*6; i++) { ++(results[dice_roll(4, 6, i)]); } for (std::map<int, int>::iterator itr = results.begin(); itr != results.end(); ++itr) { std::cout << itr->first << " - " << itr->second << std::endl; } return 0;}Quote:Original post by SneftelQuote:Original post by PenanceWhat's your point? Iterate through the inputs 0-35, and you will get the correct number of 2s output, the correct number of 3s ouptut, and so on up to the correct number of 12s output.
16%6 + 16/6 + 2 = 8, not 7
Quote:Original post by Penance
If I generalized this, it would be input%d + input/d + n for n d-sided dice with input [0, d^n - 1] right?
Quote:Original post by alvaro
Also, if you are rolling 20D6, is it that important that you get the exact distribution? You start running pretty quickly into the law of large numbers, which tells you that a normal distribution is a very good approximation to your roll. Then just use an algorithm that samples a normal distribution.
In the case of rolling 20D6, you can use N(70,7.63762615825973334425) as a very good approximation. Then round the result, or something (I am not sure what these numbers are being used for).
Quote:Original post by no such userQuote:Original post by Penance
If I generalized this, it would be input%d + input/d + n for n d-sided dice with input [0, d^n - 1] right?
No, see the code sample directly above your post.
