Hi
@vaisman
Sure, frankly I try to respond quickly especially if I think we have an error (as above). Plus, we consider it a contribution (to help us fix mistakes or otherwise improve). In regard to the follow-up: We are given the assumption that "outperformers beat the market 75% of the time" which statistically can be written as a
conditional probability, Pr[ Beat | Out-performer]. The above is retrieving the probability that the outperforming manager beats the market in exactly 6 out of 10 years. If the manager has a 50% probability of beating each year, you could think of this as the probability of flipping a coin and getting exactly 6 heads out of 10 flips (except this manager has a 75% probability). Because each year's outcome is a Bernoulli (i.e., either beat or miss, only two outcomes) and this is a series of 10 Bernouillis, this a binomial distribution. So, the formula is applying the binomial distribution (
https://en.wikipedia.org/wiki/Binomial_distribution) to find the probability of exactly 6 successes out of 10 trials given p = 0.75. The (10 6) is the binomial coefficient (
https://en.wikipedia.org/wiki/Binomial_coefficient) that is "inside" the binomial. Here it equals 10!/(6!*4!) = (10*9*8*7)/(4*3*2) = 210; it means there are 210 combinations of six items that can be drawn from a set of 10 items. The 0.75^6*0.25^4 = 0.000695229 is the probability of an exactly (single) sequence of 6 beats + 4 misses; e.g., "BMBMBMBMBB". But we want to count any sequence with six beats, and there are 210 different sequences to the 210* 0.000695229 = 14.60% is the probability of a combination (as opposed to a permutation) of 6 beats out of 10 years. I hope that's helpful!
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