The urns

Problem statement

There are 11 urns labeled by $u \in {0, 1, 2, . . . 10}$ , each containing ten balls. Urn $u$ contains $u$ black balls and $10 - u$ white balls. Fred selects an urn $u$ uniformly at random and draws $N$ times with replacement from that urn, obtaining $n_{B}$ blacks and $N - n_{B}$ whites. Fred's friend, Bill, looks on. If after $N = 10$ draws $n_{B} = 3$ blacks have been drawn, what is the probability that the urn Fred is using is urn $u$ from Bill's point of view?

Solution

Firstly, we know how to express the probability distribution of the number of blacks given the urn and the total ball count:

P (n_{B} | u, N) = (\begin{matrix} N \\ n_{B} \end{matrix}) f_{u}^{n_{B}} (1 - f_{u})^{N - n_{B}} (1)

Where we define [ $f_{u} := ?$ ].

Adding the $u$ dimension creates the joint probability distribution of the random variables $u$ and $n_{b}$ , $P (u, n_{B} | N)$ . It can be written as:

[

P (u, n_{B} | N) = P (?) P (u)

]

Notices how $P (u)$ appears, the probability that Fred selects a given urn, which is uniform at $\frac{1}{11}$ as per the question.

From here, we need Bayes Theorem, which is:

[

P (A | B) = \frac{P (A, B)}{P (B)} = \frac{?}{P (B)}

]

From the joint probability of $u$ and $n_{B}$ we can obtain the conditional distribution of $u$ given $n_{B}$ using Bayes Theorem:

\begin{aligned} P (u | n_{B}, N) & = \frac{P (u, n_{B} | N)}{P (n_{B} | N)} \\ = \frac{P (n_{B} | u, N) P (u)}{P (n_{B} | N)} \end{aligned}

We so far know 2 of the 3 expressions in that quotient. The third, the denominator, $P (n_{B} | N)$ , is the marginal probability of $n_{B}$ which we can obtain using the sum rule:

P (n_{B} | N) = \sum_{u} P (u, n_{B} | N) = \sum_{u} P (u) P (n_{B} | u, N)

So the conditional probability of $u$ given $n_{B}$ is:

\begin{aligned} P (u | n_{B}, N) & = \frac{P (n_{B} | u, N) P (u)}{\sum_{u^{'}} P (n_{B} | u^{'}, N) P (u^{'})} \\ = \frac{P (n_{B} | u, N) P (u)}{P (n_{B} | N)} \\ = \frac{1}{P (n_{B} | N)} \frac{1}{11} (\begin{array}{c} N \\ n_{B} \end{array}) f_{u}^{n_{B}} (1 - f_{u})^{N - n_{B}} \end{aligned}

We must calculate and sum all probabilities of the form (1) above, then p the sum by the probabilities from (1) for the given urn.

By the way, the sum for when $n_{B}$ is 3 is: $P (n_{B} = 3 | N = 10) = 0.083$ .

Todo: compare this to the bent coin.

24.08.2019