# Random number weighted probability - where highest probability is least likely

I want to do a standard picking a random number by weighted probability problem.

Example Data Set:

Person A    70
Person B    30
Person C    40


So normally, these are the probabilities of being picked randomly:

Person A    50%
Person B    21%
Person C    29%


Now I want to make it so it's inverted, where the higher your probability, the less likely you are to be picked. How do I invert it?

The way I tried:

Invert the probabilities

Person A    50%
Person B    79%
Person C    71%


Add those up .5 + .79 + .71 = 2

Divide them by the new total:

Person A    .5/2  = 25%
Person B    .79/2 = 40%
Person C    .71/2 = 35%


I'm not confident in that answer though - is that the correct way of donig it? Or am I thinking about it wrong?

• You might want to change your terminology: it's confusing to say "the higher your probability, the less likely you are to be picked". Instead, how about "the higher your weight" or "the higher your score"? – Théophile Aug 22 '18 at 16:43

There's no right way to do this - any method of inversion will have its pros and cons. However, a way to do this is to invert the weights before computing probability, i.e., $$\textrm{New Weight} = \frac{100}{\textrm{Old Weight}}$$