Ducklings in a Lake

We'll look first at the general case of $N$ ducklings resting in the lake.

For each of the $N$ ducklings, extend a radial line from the center of the lake through the duckling's position to a point on the lake's perimeter. (If a ducking happens to be precisely at the center of the lake, (an event that occurs with zero probability, anyway), then it would be on the same half of the lake as any other duckling by default.)

With this mapping of $N$ duckling positions in the lake to $N$ points on the lake's perimeter, the desired probability is then the same as the probability that $N$ points chosen randomly on (the perimeter of) a circle lie on the same semicircle.

First choose any of the $N$ points. Now for each of the other $(N - 1)$ points the probability that they are within $\pi$ radians going clockwise of the chosen point is $\frac{1}{2}$ , and so the probability that all the remaining $(N - 1)$ points lie on the (clockwise) semicircle that has the initially chosen point as the 'leftmost' point is $(\frac{1}{2})^{N - 1}$ .

Since the choice of the initial point was arbitrary, this result holds for each of the $N$ points on the circle, and since these events are disjoint, (i.e., we can't have two different points serving as the leftmost point of the clockwise semicircle on which all $N$ points lie), we can simply sum these $N$ probabilities to find that the probability that all the points lie on the same semicircle is $P(N) = N * (\frac{1}{2})^{N-1}$ .

With $N = 2016$ we then have that $P = 2016 * (\frac{1}{2})^{2015} \Longrightarrow \frac{63}{P} = \dfrac{2^{2015}}{\frac{2016}{63}} = \dfrac{2^{2015}}{32} = 2^{2010}.$

Finally, we then find that $\log_{2}(\frac{63}{P}) = \log_{2} 2^{2010} = \boxed{2010}$ .

Here is a more elegant way of setting it up, which is equivalent to Brian's approach.

First, the probability that there is a duck in the center, or two ducks in the same spot, or two ducks on a diameter, is 0, so we can ignore those cases.

Now, for any configuration of 2016 ducks in positions $d_i$ , let's also consider the antipodal point (reflection through the center) denoted as $d_i ^*$ . Let's look at the probability in these $2 ^ {2016}$ cases where duck $i$ is in either $d_i$ or $d_i ^*$ .

There are $2 \times 2016$ cases in which the ducks are in a semicircle, which correspond to picking 2016 consecutive points taken about the circle. Thus, the probability is $\frac{ 2 \times 2016 } { 2 ^ {2016} }$ .

Since this probability is constant for any starting configuration, if follows that the probability is always going to be $\frac{ 2 \times 2016 } { 2 ^ {2016} }$ .