What's 4 times more likely than 80%?

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

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%.

Solutions Collecting From Web of "What's 4 times more likely than 80%?"

The most reasonable way to match the answer in the book would be to define the likelihood to be the ratio of success over failure (aka odds):
then the probability as a function of the odds is
In your case the odds are $4:1$ so $4$ times as likely would be $16:1$ odds which has a probability of
This matches the $3\%$ to $11.0091743119266\%$ transformation, as well.

Bayes’ Rule

Bayes’ Rule for a single event says that
O(A\mid B)=\frac{P(B\mid A)}{P(B\mid\neg A)}\,O(A)
where the odds of $X$ is defined as earlier
O(X)=\frac{P(X)}{P(\neg X)}=\frac{P(X)}{1-P(X)}
This is exactly what is being talked about in the later addition to the question, where it is given that
\frac{P(B\mid A)}{P(B\mid\neg A)}=4

Daniel Kahneman’s book mentions Bayesian reasoning. An answer using Bayesian reasoning is as follows:

Let $C$ be the event that Tom is compsci, $N$ be the event that he has a “nerdy” personality.

We are given $P(N|C)/P(N|\neg C)= 4$, which implies that $P(N|\neg C) = P(N|C)/4$.

By Bayes Theorem (and using the theorem of total probability to expand the denominator)

&=& \frac{P(N|C) P(C)}{ P(N)} \\
&=& \frac{P(N|C) P(C)}{P(N|C)P(C) + P(N|\neg C) P(\neg C)} \\
&=& \frac{P(N|C) P(C)}{P(N|C)P(C) + 0.25 P(N|C)P(\neg C)} \\
&=& \frac{P(C)}{P(C) + 0.25 P(\neg C)} \\
&=& \frac{0.8}{0.8 + 0.25 \times 0.2} \\
&\approx& 0.9411765

Similar reasoning in the 3% case leads to $P(C|N) = 0.03 / (0.03 + .25*.97) \approx 0.1100917$.

Well, I’d say 80% chance of success means failure 1 out of 5 times. 4 times more likely means failure only 1 out of 20 times, so the new probability would be 95%.

The statement of the context (in my words) is as follows:

If you believe that $80\%$ of graduate students are enrolled in computer science (base rate), and you also believe that the description of Tom W is four times more likely for a graduate student in computer science than for a graduate student in other fields, then Bayes’s rule says you must believe that the probability
that Tom W is a computer scientist is now $\approx94.1\%$.

Here is how to perform the Bayesian reasoning. Let $\rm CS$ be the event that a student is enrolled in computer science, and $\rm desc$ the event that [description] holds true of a graduate student. Then

  • The base rate says that $P(\rm CS)=80\%$.
  • The relative statement says that $P({\rm desc|CS})=4P({\rm desc|\neg CS})$

Thus $P(\neg{\rm CS})=0.2$ and $P({\rm desc|\neg CS})=0.25P({\rm desc|CS})$. Bayesian reasoning says that

$$\begin{array}{ll} P({\rm desc}) & = P({\rm desc~\&~CS})+P({\rm desc~\&~\neg CS}) \\
& =P({\rm desc|CS})P({\rm CS})+P({\rm desc|\neg CS})P(\neg{\rm CS}) \\
& =(0.8+0.25\cdot0.2)P({\rm desc|CS}) \\
& =0.85P({\rm desc|CS}) \end{array}$$

Bayes rule says that

$$\begin{cases} P({\rm desc|CS})= \frac{P({\rm desc~\&~CS})}{P({\rm CS})} \\ \phantom{blah} \\ P({\rm CS|desc})=\frac{P({\rm CS~\&~desc})}{P({\rm desc})} \end{cases}$$


$$P({\rm CS|desc})=\frac{P({\rm desc|CS})P({\rm CS})}{P({\rm desc})}=\frac{0.8}{0.85}=0.9411764705882352\dots\approx94.1\% $$

Similarly, if the base rate was $3\%$ instead of $80\%$, the calculation would go as follows:

$$\begin{array}{ll} P({\rm desc}) & = P({\rm desc~\&~CS})+P({\rm desc~\&~\neg CS}) \\ & =P({\rm desc|CS})P({\rm CS})+P({\rm desc|\neg CS})P(\neg{\rm CS}) \\ & =(0.03+0.25\cdot0.97)P({\rm desc|CS}) \\ & =0.2725P({\rm desc|CS}) \end{array}$$

$$P({\rm CS|desc})=\frac{P({\rm desc|CS})P({\rm CS})}{P({\rm desc})}=\frac{0.03}{0.2725}=0.1100917431192660\dots\approx11\% $$

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.

The only way I see to make sense of this is to divide by $4$ the probability it does not happen. Here we obtain $20/4=5$, so the new probability is $95\%$.