If x out of y choices are more likely to happen, how likely are the x choices? 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.
 A: Using R as a calculator:
prop = c(1.25, 1.25, 1, 1, 1)
p = prop/sum(prop);  p
[1] 0.2272727 0.2272727 0.1818182 0.1818182 0.1818182

So you need to divide each of the proportions by the sum of all five proportions. That gives the probabilities you want.
Application: I hope that answers your question. In case you're curious, here is an application
of the procedure you're asking about.
Simulating a survey with specified proportions. The sample  function in R allows the use of a vector of
proportions that are to be used in taking random samples.
In sample(1:5, 1000, rep=T, p = prop), the first argument is
The population, which has five categories (with integer labels 1 through 5); second, we seek a random sample of size 1000; the third
argument specifies sampling with replacement; and the fourth argument specifies the proportions to be used in sampling.
In effect, R turns proportions into probabilities, as in your question.
So one simulated random sample to your specifications can
be taken as follows:
set.seed(1019)
prop = c(1.25, 1.25, 1, 1, 1)
x = sample(1:5, 1000, rep=T, p=prop)
table(x)
x
  1   2   3   4   5 
237 221 191 194 157 

A chi-squared test of the tabled results. With a sample as small as 1000, we can't expect exact
proportions, but we can use a chi-squared test to check
if results of the simulation are a plausible match for the intended probabilities. The null hypothesis is that the sample was taken according to your proportions.
The observed and expected counts used in the test are:
obs = tabulate(x); obs
[1] 237 221 191 194 157
exp = p*1000; exp
[1] 227.2727 227.2727 181.8182 181.8182 181.8182

The chi-squared statistic $Q = \sum_{i=1}^5\frac{(O_i-E_i)^2}{E_i} = 5.257.$ For a sample as large as ours, sampled using the specified proportions, the distribution of $Q$ is well approximated by a chi-squared distribution with
$5-1=4$ degrees of freedom. If $Q \ge 9.488,$ we would reject the sample is inconsistent with your proportions. In this test at the 5% level, we conclude that the 
simulated data are consistent with those proportions.
q = sum((obs-exp)^2/exp); q    
[1] 5.257
qchisq(.95, 4)
[1] 9.487729

The density function of $\mathsf{Chisq}(4)$ is shown in the plot below.
The heavy vertical line shows the observed value of $Q.$ The dotted vertical line shows the 5% critical value $9.488$ of the test.

