# Bayesian Odds

Bayes' theorem

$p(D|C) = \frac{p(C|D)\cdot p(D)}{p(C)}.$

joins in one formula the a priori probabilities of events $C$ and $D$ and the posterior probabilities $p(C|D)$ of event $C$ provided $D$ took place - and $p(D|C)$, the probability of $D$ provided that $C$ is known to take place.

On occasion when there are two events, say $A$ and $B$, whose comparative posterior probabilities are of interest, it may be more advantageous to consider the ratios, i.e. the odds:

$\frac{p(A|C)}{p(B|C)} = \frac{p(C|A)}{p(C|B)}\cdot \frac{p(A)}{p(B)}.$

Ward Edwards gives a simple example where the latter formula comes in handy:

There are two bags, one containing $700$ red and $300$ blue chips, the other containing $300$ red and $700$ blue chips. Flip a fair coin to determine which one of the bags to use. Chips are drawn with replacement. In $12$ samples, $8$ red and $4$ blue chips showed up. What is the probability that it was the predominantly red bag?

Author Edwards writes

Clearly the sought probability is higher than $0.5$. Please don't continue reading till you have written down your esimate.

Better yet, try to derive your estimate formally. Make use of Bayes' theorem, of course.

Solution

### References

1. W. Edwards, Conservatism in Human Information Processing, in Judegment under uncertainty: Heuristics and biases, D. Kahneman, P. Slovic, A. Tversky (editors), Cambridge University Press, 1982

There are two bags, one containing $700$ red and $300$ blue chips, the other containing $300$ red and $700$ blue chips. Flip a fair coin to determine which one of the bags to use. Chips are drawn with replacement. In $12$ samples, $8$ red and $4$ blue chips showed up. What is the probability that it was the predominantly red bag?

### Solution

Let $A$ be the event of selecting the first bag. Let $B$ be the event of selecting the second bag. Finally, let C be the result of the experiment, i.e., drawing $8$ red and $4$ blue chips from the selected bag. Clearly,

\begin{align} p(C|A) & = \big(\frac{7}{10}\big)^{8}\big(\frac{3}{10}\big)^{4}{12 \choose 8} \\ p(C|B) & = \big(\frac{7}{10}\big)^{4}\big(\frac{3}{10}\big)^{8}{12 \choose 4} \end{align}

so that $\displaystyle\frac{p(C|A)}{p(C|B)} = \big(\frac{7}{3}\big)^{4} \approx 29.642$. Now, $p(A) = p(B) = 0.5$, implying that

$\frac{p(A|C)}{p(B|C)} \approx {29.642}.$

This is the odds of selecting one of the bags relative to the other. From $p(A|C) + p(B|C) = 1$, it follows that $p(A|C) \approx \displaystyle\frac{29.642}{30.642} \approx 0.967$.

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