The Student Puzzle Corner contains problems in statistics or probability. Solving them may require a literature search. Student IMS members are invited to submit solutions (to bulletin@imstat.org with subject “Student Puzzle Corner”). The deadline is November 15, 2018. The names of student members who submit correct solutions, and the answer, will be published in the following issue. The Puzzle Editor’s decision is final.

## Here’s Anirban DasGupta’s latest puzzle, on coin tossing.

After our well deserved summer intermission, we need to get resolute again, and this time on a statistics problem. The inferential premise of our problem is completely standard, but the nature of the data is not quite standard. You will deal simultaneously with probability and inference.

It has been noted in the probability literature that students of probability grossly underestimate how many consecutive heads one is likely to obtain in a given number of tosses of a (fair) coin. Very few would be adventurous enough to write more than three or four consecutive heads in, say, 100 imaginary tosses of a coin. Actually, the expected value here is almost 6. The distribution of the length of the longest head run is, in principle, known.

Here is our exact problem this time.

Suppose a coin with an unknown probability p for heads, in a single trial, is tossed 10 times, and the length of the longest head run is found to be 1.

(a) Find the MLE of p based on just this data.

(b) Find a moment estimate of p based on just this data.

Anirban’s solution to the previous puzzle follows.

Joyce Cahoon

Student member Joyce Cahoon at North Carolina State University sent serious answers to last month’s puzzle, and we commend her efforts.

If we denote $S_n = \sum_{i=1}^n x_i$, then by straightforward calculations, the probability that the $n$ residents
have no common friends is
$\frac{(N-x_1)!\,(N-x_2)!\,\cdots (N-x_n)!}{(N!)^{n-1}\,(N-S_n)!}.$
For part b), denote by $B_i$ the event that resident $i$ of the town is a friend of each of the $n$ residents in consideration. In part a), we already have the probability that none of the events $B_1, B_2, \cdots , B_N$ occurs.
Apply de Moivre’s formula to find the probability that exactly one of the events $B_1, B_2, \cdots , B_N$ occurs.
For $N = 10^6, n = 15, x_i \equiv 4\times 10^5$, the numerical value is approximately $0.37$.
Add it to the value from part a) to get a probability of $.71$ that at most one resident of the town
is a friend of the $n$ residents in question.

For part c), denote by $W = W_{N,n}$ the cardinality of the intersection of the n acquaintance sets. Then, $E(W) = \frac{x_1\,x_2\,\cdots x_n}{N^{n-1}}$ (just use indicators). When $n = N$ and each $x_i = N-\log N$, this is seen without trouble to be $1+o(1)$ as $N \to \infty$. Thus, an approximate value of the probability that $W$ equals $1$ is $e^{-1} = .37$.