TITLE: What's 4 times more likely than 80%?
QUESTION [143 upvotes]: There's an 80% probability of a certain outcome, we get some new information that means that outcome is 4 times more likely to occur.
What's the new probability as a percentage and how do you work it out?
As I remember it the question was posed like so:

Suppose there's a student, Tom W, if you were asked to estimate the
  probability that Tom is a student of computer science. Without any
  other information you would only have the base rate to go by
  (percentage of total students enrolled on computer science) suppose
  this base rate is 80%.
Then you are given a description of Tom W's personality, suppose from
  this description you estimate that Tom W is 4 times more likely to be
  enrolled on computer science.
What is the new probability that Tom W is enrolled on computer
  science.

The answer given in the book is 94.1% but I couldn't work out how to calculate it!
Another example in the book is with a base rate of 3%, 4 times more likely than this is stated as 11%.

REPLY [7 votes]: To solve the exercise you should be familiar with the concept of odds. Unfortunately the term "odds" is used without consistency by different groups of peoples and therefore it causes often confusion. 
Quoting Wikipedia: 

In statistics, odds are an expression of relative probabilities,
  generally quoted as the odds in favor. The odds (in favor) of an event
  or a proposition is the ratio of the probability that the event will
  happen to the probability that the event will not happen.

The initial odds that Tom is a computer science student are $$8:2$$ where the symbol $:$ should be read as to. To see that these odds correspond to the probabilities $80\%$ (for) and $20\%$ (against), divide both numbers with their sum: $$(8/10):(2/10) = 0.8:0.2 = 80\%:20\%$$
If it is $4$ times more likely that Tom is enrolled in computer science then the above ratio becomes $$(4\cdot8):2 = 32:2$$ which corresponds to the probabilities $$\frac{32}{32+2}:\frac{2}{32+2} = 0.9411:0.0589 = 94.11\%:5.89\%$$ 

Similarly if an event is $3%$ likely to occur, then it's odds can be expressed as $$3:97$$ Now if this event becomes $4$ times more likely, the odds become $$(4\cdot3):97= 12:97$$ which corresponds to a probability of $$\frac{12}{12+97}=0.11=11\%$$ for and a probability of $\frac{97}{12+97}=89\%$ against.