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(2): my solution....QUESTION by Ady TZIDON)

The problem is that the second person does indeed have s-1 ways to
choose a different number than the first person, but the third person
is left with two cases. If the second person successfully chose a
different number, then the third person has s-2 choices left to be
different. If the seconde person failed to choose a different
number, then the third person has only s-1 choices left.
Therefore your formula for the probability that all N choices are
distinct is correct, but since you are ordering your selections, there
are multiple ways to reach, for instance N-1 distinct (i.e. 2 could
have been a repeat, 3 could have been a repeat, etc. N-1
possiblities, each with the probability you mentioned, so they'd need
to be added up to give the probability of exactly 1 indistinct choice).

Hope this helps!