A Lotto bet is picking 6 numbers out of 49 -- if you pick the correct combination, you get the jackpot!
If N persons play, there will be many repeats, since it's highly probable that some combinations will be chosen by two persons or more. (This is known as the "birthday paradox".)
What's the expected number of DIFFERENT combinations that will be chosen, if N persons play? (Assume these persons pick their combinations totally randomly.)
(In reply to
re: A simpler way of getting to solution by Jay Schamel)
Sure! Thanks for asking.
If k combinations are covered, then the probability of somebody
winning is k/C, where C is the total number of possible combinations.
Let k be a random variable, and p(k) be the probability that that number of distinct combinations are covered.
Then the probability that somebody wins = Σ(p(k)*k/C) =
(1/c)*Σp(k)*k = (1/c)*(the expected number of distinct combinations).
You are correct that in general, the expected value of f(k) is not the
same as f(expected value of k). It is true in this case, because
the probability that somebody wins is directly (linearly) proportional
to the number of distinct combinations.
I like this approach, because it avoids all the questions about
summing up the expected values of variables that are not independent of
each other.
re(2): A simpler way of getting to solution
