I'm trying to get better at prob&stats, so this was a good puzzle for me. I'm not certain I have the right answer, but it seems straightforward enough after thi

I'm trying to get better at prob&stats, so this was a good puzzle for me. I'm not certain I have the right answer, but it seems straightforward enough after thinking it through. Spoilers follow!! For question one: A and B will each have, on average, 100 drivers. Suppose we also know the std. deviation from that average. Then the more popular road can be expected to have (100 + stddev) drivers, and the less popular road (100 - stddev) drivers. (This doesn't change the expected # of drivers for each road, since either road has an equal chance of being the more popular one). The binomial distribution (https://en.wikipedia.org/wiki/Binomial_distribution https://en.wikipedia.org/wiki/Binomial_distribution) ("given n coin flips each with chance p of being heads, what's the probability that k of them were heads?") perfectly models the road choice, and its variance (stddev^2) is conveniently available in closed form: np(1-p), which given n=200 cars and p=0.5 is 50. So the expected number of drivers on more popular road is 100+sqrt(50), and the less popular road 100-sqrt(50). For question two: the article gives away the trick, for the most part -- we count the number of cars each driver observes, not the number of cars each road observes. There are 200 drivers, and (100+sqrt(50)) of them observe the other (100+sqrt(50)) drivers on the more popular road, and similarly on the less popular road. So the expected number of cars a driver will see is 1/200 * (100+sqrt(50))^2 + (100-sqrt(50))^2, which works out to 114.6. TLDR: If there are two equivalent roads and you and 199 other drivers make an independent coin flip on which road to take, you should expect to observe 115 cars on the road you choose.