 Solution

Let $N$ denote the mattress count and $U$ the event of undisturbed sleep. Use the intended model in which the presence of the pea is independent of $N$, with probability $1/2$, and without a pea sleep is undisturbed. Conditioning on $N$ givesThe law of total probability and the equal prior probabilities giveBy Bayes' theorem, the posterior probabilities of $N=6,7,8$ given $U$ are respectively $5/18,6/18,7/18$. Therefore the conditional expectation isUndisturbed sleep shifts the posterior toward larger mattress counts because those counts make the observation more likely.