# Conditional Probabilities - The Mad Kings' Draft

[I have also posted this question on mathoverflow]

The Problem of the Mad King's Draft:

Suppose there is country which is ruled by a king who can be either 'mad' or 'normal.' The king rules a a large country with a continuum of citizens who have names j from [0,1]. The citizens do not know whether the king is mad or not but believe both characters are equally likely.

The king drafts citizens. The mad king drafts 2 citizens while the nice king drafts 4 citizens at random, that is, each citizen is 'equally likely' to be drafted. Drafted citizens do not know how many other citizens are drafted.

What is the posterior of a citizen j who is drafted by the king? This is an important question, since drafted citizens want to escape the draft if the king is mad.

It seems the obvious answer is that the correct posterior should be 1/3. But why?

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Ideally, I would like to find a general definition of conditional probabilities that applies to the Mad King's Draft and that is analogous to the standard definition of conditional probabilities, found, for example, on Wikipedia(conditional expectation).

None of these standard definitions apply, which is not satisfying.

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I'm going to go out on a limb and presume you meant "What is citizen j's posterior probability that the king is mad?". "what is the posterior of a citizen j?" could be construed as having a different meaning. – Michael Hardy Mar 30 '12 at 2:36
What is the prior probability of the king being mad? Is it 1/2, or is the citizen expected to extract it from his posterior? – deinst Mar 30 '12 at 2:48
- Hardy: Yes, what is citizen j's posterior probability that the king is mad if citizen j is drafted. – Nahpetz Mar 30 '12 at 3:12
-deinst The prior probability that the king is mad is 1/2. – Nahpetz Mar 30 '12 at 3:12

If the population is $N$ then the likelihood of being drafted is $\frac{2}{N}$ or $\frac{4}{N}$ depending on the state of the king. The prior porbabilities are $\frac12$ and $\frac12$ so the posterior probability of the king being mad is

$$\dfrac{ \frac12 \times \frac{2}{N}}{\frac12 \times \frac{2}{N} +\frac12 \times \frac{4}{N}} = \dfrac13$$

and I would agree that this is the obvious answer.

For somebody not drafted the posterior probability of the king being mad is

$$\dfrac{ \frac12 \times \frac{N-2}{N}}{\frac12 \times \frac{N-2}{N} +\frac12 \times \frac{N-4}{N}} = \frac{N-2}{2N - 6}$$ which is higher than $\frac12$ for $N \ge 4$.

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Thank you. This is a "limiting procedure" of approximating the zero probability event by a positive probability event in a finite population. I am looking for a general ("measure-theoretic") definition that applies to this problem and that is analogous to the standard definitions of conditional probabilities, for example, those on wikipedia. The Mad Kings's Draft might be simplistic but still one cannot apply those definitions. Thank you! – Nahpetz Mar 30 '12 at 11:53
Thanks again, Henry. – Nahpetz Apr 4 '12 at 13:37