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.)

only big N justifies the statemeht:

...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)...........

My solution:

Denote the number of all different 6-tuples out of 49 by s, s=C(49,6).

The first person chosing an arbitrary 6tuple leaves for the second person s-1 choices out of the s possible to avoid repetition, the 3rd person has a probability of (s-2)/s to pick a new 6tuple etc

Thus for N PEOPLE THE PROBABILITY OF HAVING ALL DISTINCT CHOICES IS

p=(s-1)*(s-2)*(s-3)*...(s-k )...*(s-N+1)/(S^(N-1))

and expected number of DIFFERENT combinations is

N*p

If anyone would like to provide numerical results-

analytical or simulation-based- BMG (=be my guest).

ady

*Edited on ***January 15, 2005, 7:42 am**