# If x out of y choices are more likely to happen, how likely are the x choices?

by Chuck   Last Updated October 19, 2019 20:19 PM - source

If there are y choices that survey takers can pick, and generally all are equally likely, but for a certain subgroup, x of those choices are a certain percent more likely, what is the probability of someone in that subgroup choosing an option with a higher likelihood.

As a concrete example, in case I'm not clear, there are five choices (a, b, c, d, e), and generally they're equally likely, so for most people, each has a 20% chance of being chosen. But there's a demographic that is 25% more likely to pick a or b than they are the other three.

By brute force, I think this means that if I give a and b five chances to be selected and give c, d, and e four chances each (so that there are 25% more a's and b's than the others), that's a total of 22 choices, 5 of which are a, making the chance of that demographic selecting a 22.7% and the chance of choosing c 18.2%. and this seems right, because 18.2% * 1.25 = 22.7%

It's not too difficult with 25% increase in likelihood because it's an easy fraction of 100%. I can see how I could do similarly for other percentages that aren't as easy, but I'm wondering if there's some algorithm that can pop out the percentages with just the information that there are five options, and selecting a or be is 25% more likely than selecting c, d, or e.

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