# Posterior Odds of 99:13.7 Stated As A Probability

The material I am working with:

The example that I am working on can be found on page 2, see picture at bottom of post. It is stated that "Prior odds of 99:1 ...change by the result to posterior odds 99:13.7, a 12% probability." Later it says the posterior odds are 13.7:1, so 6.7% still rather different."

What am I missing here?

Could this not be described as decimal too:

.073 * .99 = .07227

In other words 7.2%, a number different than stated in the paper

EDIT: still having trouble with final part of the problem

now solving (1-x)/x = 0.072993, we arrive at x = 93.2%


and

so (1-x)/x = 7.226 or x = 1/8.226 = 12.16% probability


are throwing me off

• note that an odds ratio less than one isn't commonly notated as a percent ... just like an odds ratio greater than one isn't likely to be expressed as a percent. if anything, an odds ratio might be followed by a multiply symbol, e.g. 0.07227× ... but that might get confusing when there's a variable $x$ lurking about. the $aa$:$bb$ notation is probably best for odds ratios – phdmba7of12 Sep 11 '18 at 22:17

if the prior odds ratio is 99:1 (with a corresponding pre-felt probability of (1-x)/x = 99 ... x = 1% chance that the aunt has the ability)... multiplication by the likelihood results in posterior odds ratio of 99:13.7 = 7.226

odds are ratios of probabilities that add to unity

so (1-x)/x = 7.226 or x = 1/8.226 = 12.16% probability that p = 0.75 vs the alternative hypothesis of p=0.5 with a probability of 88.8% of the aunt not having the ability

if, instead the prior odds ratio was 1:1 (a priorly believed probability of 50%), after multiplying by the likelihood, you'd end up with a posterior odds ratio of 1/13.7 or 0.072993

now solving (1-x)/x = 0.072993, we arrive at x = 93.2% probability that p = 0.75 vs the alternative hypothesis that p = 0.5

note that 100%-93.2% = 6.7%

so in the first case a prior probability of 1% that the aunt has the ability is increased to 12.2% by the data likelihood

in the second case, the prior probability of 50% that the aunt has the ability is increased to 93.2%

• the pdf author should be comparing 12.2% (case 1 ... up from a prior 1%) chance to 93.2% (case 2 ... up from a prior 50%) chance that the aunt has the ability – phdmba7of12 Sep 11 '18 at 21:59
• thank you so much for the answer ! I am almost there. see edit, please walk me through the now solving (1-x)/x = 0.072993, we arrive at x = 93.2% & so (1-x)/x = 7.226 or x = 1/8.226 = 12.16% probability. need a hand hold step by step! LOL see my edits for the chart i am building to do this / where i am. TY – learnAsWeGo Sep 12 '18 at 5:17
• if we called the odds ratio $r$, then $$(1-x)/x = r$$ solved for $x$ is step1. $$xr=1-x$$ step2. $$x(r+1)=1$$ step3. $$x=1/(r+1)$$ – phdmba7of12 Sep 12 '18 at 5:31
• TY I will be throwing up questions as I work through this paper. you will probably be able to gain +500 points from this trail of questions LOL – learnAsWeGo Sep 12 '18 at 15:15