Bayes by counting

Idea: A belief is a count in disguise

Replace every probability with a crowd of equally likely cases, and cross out what the evidence rules out.

• Read the answer from what is left
• Pick a crowd that divides by every denominator

Example: The price alert

$1$ day in $5$ is volatile. The alert fires on $\frac{3}{4}$ of volatile days and $\frac{1}{4}$ of calm ones. It fired. Chance the day is volatile?

1. $100$ days: $20$ volatile, $80$ calm
2. Alerts: $20\times \frac{3}{4}=15$ and $80\times \frac{1}{4}=20$
3. Volatile: $\frac{15}{35}=\frac{3}{7}$

Formula: Bayes' rule

$P(H\mid E)=P(E\mid H)\times \frac{P(H)}{P(E)}$

The group you want over everyone who shows the evidence. P(E) is the sum of the groups that survive.

Common mistake: Swapping the conditionals

The alert fires on $\frac{3}{4}$ of volatile days, yet a firing alert means volatile only $\frac{3}{7}$ of the time.

Instead Say each one out loud separately: "given volatile" versus "given an alert".

Idea: More causes, same method

Give each cause a row: prior times likelihood, then add the rows and divide.

• $3$ urns: split the crowd in $3$
• Causes need not be urns: dice, coins, days

Example: Which die was it?

A bag has a $4$-sided and a $6$-sided die. Pick one, roll it, see at most $4$. Chance it is the small die?

1. $24$ picks: $12$ per die
2. Small always qualifies: $12$; large: $12\times \frac{4}{6}=8$
3. Chance: $\frac{12}{12+8}=\frac{3}{5}$

Quick check: A double-headed coin

A bag holds $1$ double-headed and $3$ fair coins. Pick one and flip it twice: $2$ heads. Chance it is double-headed?

Idea: How to say it

Name the crowd out loud, give both group sizes, and only then divide.

• The interviewer can follow every step
• A slipped number shows up as a group that looks too large