VB 2005 RANDOM WITH PERCENTAGE

[suguimoto]

Hello there, I would like to know if it´s possible to randomly pick a number based on a percentage, for example: from 1 to 5. nº % 1 10% 2 20% 3 10% 4 20% 5 40% I have 40% to pick 5 instead of the others. Is it possible to be implemented in VB2005 ?

[Luc Pattyn]

Hi, Of course it is possible. Here is the simplest approach: - allocate an array of 100 elements; - now fill it according to your specs (10 ones, 20 twos, 10 threes,...); order does not matter. - now get a random number from Random in the range [0,99]; - use it as an index into the array, and get the element. Obviously, you do not really need the array, you could replace it with a couple of conditionals, in pseudo-code:

// note: the thresholds are cumulative
if num<10 then return 1
if num<30 then return 2
if num<40 then return 3
if num<60 then return 4
return 5

:)

Luc Pattyn [Forum Guidelines] [Why QA sucks] [My Articles]

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I only read code that is properly indented, and rendered in a non-proportional font; hint: use PRE tags in forum messages

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[T2102]

The easiest general solution for a very small # of discrete states is to invert the CDF. Generate a uniform on [0,1]. When it's between [0, .1) map to 1, [.1, .3) map to 2, [.3, .4) map to 3, [.4, .6) map to 4, [.6, 1) map to 5.

[Luc Pattyn]

And how is that any different from what I posted? :)

Luc Pattyn [Forum Guidelines] [Why QA sucks] [My Articles]

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I only read code that is properly indented, and rendered in a non-proportional font; hint: use PRE tags in forum messages

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[T2102]

It's very similar, but most likely is less efficient given the implementation of Random. (Yes I have designed several simulation algorithms before). Random generators are typically designed to generator random unsigned longs of uniforms efficiently. The generators that give you a random integer have an unnecessary multiplication thrown unless of course they are specialized for powers of two where they use bitwise operators.