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.)

Let C = number of distinct combinations = 13,983,816
Let D = expected number of distinct combinations

The probability that nobody wins = ((C-1)/C)^n

The probabilty that nobody wins is also 1 - (D/C).

Therefore, setting the two equal and solving,
D =  C(1-((C-1)/C)^n)
   =  13,983,816(1-((13,983,815)/13,983,816)^n)

Posted by Steve Herman on 2005-01-16 12:02:45