Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2009 / Paper 1 / 12

Neither biscuit type can run out in this case. Treat each freely made choice as an independent Bernoulli trial, with success meaning chocolate. The number taken after $n$ choices has a binomial distribution, soThe binomial coefficient counts the possible success positions, and every such arrangement has the same probability. Independence of successive choices is the probabilistic model implicit in the question; identical marginal preferences alone would not imply it.

Write the chocolate count as $K_n={\sum }_{j=1}^n{I}_j$ with independent Bernoulli random variables. Each has expected value $p$ and variance $p(1-p)$. By linearity of expected value and additivity of variance for independent variables,

For the last chocolate to be taken on choice $n$, the first $n-1$ choices must include exactly $C-1$ chocolate choices, and choice $n$ must take chocolate. Before that final success there is still chocolate available, so no forced plain choice has yet occurred. The negative binomial stopping argument givesThis is the all-trials convention of the negative binomial distribution. The finite examination scenario also has a remaining probability that the tray still contains chocolate after all $F$ choices; the displayed probabilities need not sum to one over $C\le n\le F$. At $p=0$ the last chocolate is never taken, and at $p=1$ it is taken on choice $C$.

A forced plain choice occurs exactly when the last chocolate is taken before the final Fellow is served: if it is taken by choice $f\le F-1$, choice $f+1$ has no chocolate available, whatever that person's preference would have been. Exhaustion precisely on choice $F$ forces nobody. The distinct possible exhaustion times are disjoint events, henceEquivalently, imagine independent preferred choices even after stock is exhausted. The event is that at least $C$ preferred chocolate choices occur among the first $F-1$ people, so the binomial distribution gives $P_{\mathrm{Disaster}}={\sum }_{k=C}^{F-1}\left(\genfrac{}{}{0ex}{}{F-1}{k}\right)p^k(1-p{)}^{F-1-k}$. The success-count and waiting-time duality explains equality of these two sums. Plain-biscuit depletion cannot interfere because $D>F$.