JPryde

01.27.2014
, 04:01 AM
 #2


Part 1 is rather easy:
The chance to win at least one is determined as 1(chance to win none).
Chance to win none is
995/1000*994/999*993/998*992/997*991/996*
990/995*989/994*988/993*987/992*986/991*
985/990*984/989*983/988*982/987*981/986
= 985*984*983*982*981 / (1000*999*998*997*996)
~ 0.9270
Chance to win at least one ticket is therefore 10.9270 = 0.072925 if each ticket is allowed to win (and thus every ticket you do not own can be a winner).
Chance to win at least one price is 7.29% with 5 tickets and 15 draws.
Working on part 2, which is a lot more tricky.

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