--> The probability of you being a winner of a season and also winning the most legs is 2/5

This point intrigued me somewhat as a mathematician to work out (not quite) true probabilities for the likelihood of winning and winning the most legs, so let's actually break down the maths of this (for simplicity purpose, let us assume NELs on leg's 3, 6, and 9, and single eliminations otherwise. In addition, we shall assume every team is equally likely to win a given leg (in practice this is not the case, since there are naturally stronger/weaker teams, the presence of Fast Forwards, U-Turns etc.), and each leg is independent of the next (you can think of this as each leg travelled to a new country with an equaliser at the airport i.e. all teams on the same flight, or an Hours of Operation at the next destination)).

First, the probabilities that arise from this vary so let's take the two extreme examples; winning every leg (I) and winning 2 legs whilst every other team only wins 1 leg (II) (for the calculation we'll assume we win leg 1 as our other leg since this gives us the lowest probability).

For (I), the probability of you winning

*EVERY* leg is Pmin = (1/11) * (1/10) * (1/9) * (1/9) * .... * (1/4) * (1/3) = 1/6286896000 (so, an extremely tiny chance, but possible!).

For (II), assuming that P(does not win a leg) = 1 - P(winning a leg) (and as before stating that we win Leg 1 and Leg 12, and NEL's are the same as above), our calculation becomes (1/11) * (9/10) * (8/9) * (8/9) * ... * (3/4) * (1/3) = 34836480/628689000, approximately equal to 0.005541126... (so again, very improbable, but can happen!). Note that I chose Leg 1 deliberately as this is the lowest probability for this scenario, the highest probability comes from winning the last two legs (why you might ask? Because we are effectively replacing a 3 for a 10, so the numerator becomes bigger hence a bigger probability). In this scenario, we actually end up with a probability of Pmax = 116121600/628689000, approximately equal to 0.018470418..., so the range of probabilities for our overall event is Pmin <= P(winning final leg and winning most legs) <= Pmax, given the set up of our scenario.

To further prove that Pmax is indeed the maximum we can get for our given set up, to get the numerator we multiplied 10 * 9 * 8 * 8 * 7 * 6 * 6 * 5 * 4 * 4 * 1 * 1 = 116121600. If we were to win another leg, we would have to change one of these numbers to a 1 (i.e. P(not winning leg 1) = 10/11 so P(winning leg 1) = 1/11, so the 10 in the calculation would be changed to a 1), giving us a value of 11612160 < 116121600. Any set up involving winning more than 2 legs will yield a strictly smaller numerator hence a lower probability.

If we wanted to be a little more sneaky with how we calculate "true" Pmin and Pmax, we would need to manipulate the scenarios slightly (and use separate scenarios for each one to manipulate the probability to be as small/big as possible). For Pmin, assuming NELs Legs 1-3 and we win every leg, the probability of this occurring is a staggeringly small 1/26564630400 (realistically it is 0 since when has there every been a race with 3 NELs in the first 3 legs of the race??).

Pmax however presents a new set of issues as the numerator will need to be big relative to the denominator (fortunately for Pmin the numerator was going to be 1 so all we had to do was make the denominator as big as possible). Initially my first thought was to have NELs Leg 1, 2, 3 and we win Leg 11 and Leg 12, but this actually was shown to have a smaller value than that calculated earlier! Unfortunately since it is now past 2am and my brain is knackered from doing more mathematics than I care to admit I may have another look at this problem another time.