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Jan 13, 2015 at 6:05 history edited random_guy CC BY-SA 3.0
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Jan 12, 2015 at 15:11 comment added whuber (+1) Using the formula in the duplicate for $P(X_{10}(A)=50)$ gives the answer as $$\frac{104586258561652151699128857}{3881770614586446846243219560109206166400000000}\approx 2.6942926\times 10^{-20}.$$ It also shows that one must conduct $41$ draws in order to have a better than $1/2$ chance to collect all $50$ numbers (the chance then is approximately $0.501362$). It will require $81$ draws to have better than a $99\%$ chance of collecting them all, etc.
Jan 12, 2015 at 15:09 comment added whuber The duplicate thread answers that question, too.
Jan 12, 2015 at 15:05 comment added user1478335 Is it possible to estimate how many trials would be necessary to draw all fifty numbers, as a matter of interest?
Jan 12, 2015 at 15:05 vote accept user1478335
Jan 12, 2015 at 13:19 comment added user1478335 Thank you. I thought that I was being naive and that the probability would be low - didn't 'expect' it to be so vanishingly small though!
Jan 12, 2015 at 12:34 history edited random_guy CC BY-SA 3.0
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Jan 12, 2015 at 12:26 history edited random_guy CC BY-SA 3.0
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Jan 12, 2015 at 12:18 history answered random_guy CC BY-SA 3.0