# Waterfall diagrams and relative odds Imagine the following situation: on the left is a red waterfall and on the right a blue waterfall, whose waters are flowing down at 20 litres/second and 80 litres/second respectively. Then, some water is diverted and 90% of the red water mixes with 30% of the blue water in a single waterfall.  The question of interest is the following: if you take water from the middle part, what is the proportion of it that comes from the red waterfall? You could solve it using [[Bayes' rule|Bayes’ rule]] and conditional probabilities. But I stumbled upon a much more elegant solution on the Arbital wiki[^1] which relies on likelihood ratios. Indeed, by multiplying prior proportions of $1:4$ for red vs. blue by relative likelihoods of $3:1$, we end up with: $ (1 \cdot 3) : (4 \cdot 1) = 3 : 4 $ Therefore, the proportion of red water is $3/7$, or about 43%. Why does it work? Let $A$ be the event "The water is from the red waterfall" and $B$ the event "The water is in the middle part". The above equality can be rewritten as: $ (P(A) \cdot P(B\mid A)) : (P(\bar{A}) \cdot P(B\mid \bar{A})) = P(A\mid B) : P(\bar{A}\mid B) $ If you divide the left member by $P(B)$, you end up with two instances of Bayes' rule: $ \frac{P(A) \cdot P(B\mid A)}{P(B)} : \frac{P(\bar{A}) \cdot P(B\mid \bar{A})}{P(B)} = P(A\mid B) : P(\bar{A}\mid B) $ This proves the correctness of the method. --- ## 📚 References [^1]: [Waterfall Diagram](https://arbital.com/p/bayes_waterfall_diagram/?l=1x1&pathId=77121).