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.)
Each of the 13,983,816 combinations has 13,983,815/13,983,816 probability of not being chosen by a particular person, and (13,983,815/13,983,816)^N of not being chosen by anyone.
While the choices are not independent, that is, if one combination is known not to have occurred, the others are all the more likely to have occurred, the expected value of the total of all these 1's (present) and 0's (absent) is still the sum of the expected values of each event (i.e., the event that the first combination, second combination, etc. is present).
So the expected number of nonchosen combinations is 13,983,816*(13,983,815/13,983,816)^N, and the expected number of different combinations chosen is then 13,983,816*(1(13,983,815/13,983,816)^N) **
** Edited, as inadvertently I had typed N*(...., rather than 13,983,816*(... in these two formulas.
Edited on January 15, 2005, 4:56 am

Posted by Charlie
on 20050114 19:35:20 