Label one disc as “1”, two discs as “2”, three discs as “3”,…., sixty three discs as “63”.
These 1+2+3+...+63 = 2016 labeled discs are put in a box.
Discs are then drawn from the box at random without replacement.
(i) What is the minimum number of discs that must be drawn in order to guarantee drawing at least ten discs with the same label?
(ii) Will the answer change if discs were drawn from the box with replacement?
(In reply to
re: Solution by Ady TZIDON)
This is not a probability or expected value problem. The puzzle looks for a minimum number of drawings of a disk to guarantee that at least one number will have been drawn at least 10 times. The replacement then allows any number whatsoever, even 1 or 2, to be drawn unlimited numbers of times. So, regardless of how unlikely it is, 1 can be drawn 9 times, 2 nine times, 3 nine times, ..., 63 nine times, before the next draw guarantees a tenth draw of one of those numbers. As Jer said, there could be 9*63=567 draws without any number being drawn more than 9 times, but the 568th draw has to match one of those already drawn 9 times, making a tenth time.

Posted by Charlie
on 20160825 14:08:39 