How can I make a random selection from an inversely-weighted list?

Given a list of integers, e.g. 1, 2, 3, 4, I know开发者_开发百科 how to select items based on their weight. The example items would have probabilities of 10%, 20%, 30%, and 40%, respectively.

Is there an equally simple method of selecting items based on the inverse of their weight? With this method, the example list would be equal to a weighted list of 1, 1/2, 1/3, 1/4 (48%, 24%, 16%, 12%), but I want to avoid the conversion and use of floating-point arithmetic. (Assume all of the integers are positive and non-zero.)

You could divide the numbers' least common multiple by each number and get integral proportions.

For [1, 2, 3, 4], this is 12. Your weights are 12/1=12, 12/2=6, 12/3=4, 12/4=3.

You could also multiply them all together and not bother with the LCM as well. The numbers will be higher but the proportions will be the same: 24/1=24, 24/2=12, 24/3=8, 24/4=6.

First get the sum of the weights, call it S (e.g. 1 + 1/2 + 1/3 + 1/4 = 2.083). Then to find the probability of weight w_i, you divide w_i by S (e.g. 1/2.083 = 48%.

I don't think there's a nice, closed-form formula for this expression for general sequences of numbers.

The sum of the weights are harmonic numbers. For large n, the sum converges to ln(n)+gamma where gamma is the Euler–Mascheroni constant (~0.577). So for large n, you could use this formula to approximate the sum.

EDIT: There are ways to reduce floating point errors. One such way is to calculate the sum from the smallest term up to the largest term (e.g. 1/n + 1/(n-1) + ... + 1). This allows the intermediate calculations to maximize the number of bits of precision. By doing this, rounding issues should not be a problem.