Facey McFaceFace Facey McFaceFace - 1 year ago 84
C++ Question

Generate a random number less than 4 digits but the probability of it having 1, 2, or 3 digits is equal

I'm currently using

1 + (int)(rand() * 999.0 / RAND_MAX)
to generate a random number between 1 and 999 inclusive but the two and one digit numbers don't occur as often as the three digit numbers.

How can I fix this?

Note that although the original code gives a range of 0 to 999 inclusive, I actually want a range of 1 to 999 inclusive.

Answer Source

Your observation that one digit numbers don't occur as often as two and three digit numbers is not surprising.

There are only 9 one digit numbers (not including zero), but there are 90 two-digit ones, and 900 three digit ones. So a uniform random number generator will draw numbers in that frequency.

To generate random numbers in the range [1, 999] such that the probability of their having 1, 2, and 3 digits is equal, use your favourite generator to generate a random number p, say, in the range [0, 1) (see the new random library functions in C++ to do that), and transform it using std::pow(1000, p);.

You should note that the resulting distribution will not be piecewise-uniform: that is to say that the probability of drawing a number with a certain number of digits is not the same as the probability of drawing any other number with that number of digits. But it does have a continuous and differentiable cumulative density function which can be important mathematically.

(For the mathematically-inclined, the transformation I'm applying is the quantile function of the distribution that the OP needs).

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