Past exam of the year 1 of the computer science course of the University of Oxford / 2014 / Discrete mathematics, continuous mathematics and probability / 7 / b

Each position is equally likely to receive any coin, so the law of total probability gives $\mathbb{P}(A_i)=(1/3+1/2+2/3)/3=1/2$. For two specified positions, the ordered pair of distinct coin biases is uniform over six possibilities. Assuming independent tosses conditional on the order,Thus the events are not independent, even pairwise. The covariance of draws without replacement explains the difference: the head indicator random variables have covariance $13/54-1/4=-1/108$. Independent physical tosses given the assigned coins do not imply independence after that hidden assignment is averaged out.

Let $C_j$ identify the selected coin and let $H$ be the event of two heads. The prior coin probabilities are equal, and the conditional probabilities of $H$ are $1/9,1/4,4/9$. By Bayes' theorem,The full posterior distribution over the three biases is $(4,9,16)/29$.

Given the coin, the next toss is independent of its earlier tosses and has that coin's head probability. Averaging using the posterior distribution gives the posterior predictive probabilityObserving two heads raises the probability of having chosen a high-bias coin.

If the selected coin has bias $p$, the other two biases sum to $3/2-p$. A uniform choice between those two coins therefore has conditional head probability $3/4-p/2$. Average over the posterior distribution, using the previous posterior predictive probability:The observation makes a high-bias coin more likely to have been removed from the remaining pair.