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 Playing Lotto (Posted on 2005-01-14)
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.)

 See The Solution Submitted by Federico Kereki Rating: 3.7500 (4 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 re(2): A simpler way of getting to solution Comment 13 of 13 |
(In reply to re: A simpler way of getting to solution by Jay Schamel)

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.

 Posted by Steve Herman on 2005-01-18 17:17:10

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