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I have 500 tries to draw 1 from 250 balls, all labelled (1 to 250), from an urn (with replacement). Each of the 250 balls are equally likely to be selected.

What is the probability that I will, with the 500 tries, draw all 250 different balls at least once?

bryan.blackbee
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  • The win condition is fairly hard to calculate. Can you think about how to calculate the probably that you won't draw all the balls at least once? What's the likelihood that at least one ball doesn't get drawn at all? – Ian Coley Mar 29 '13 at 19:59

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Your question is the coupon collector's problem with $n=250$ and $k=500$.

For the coupon collector's problem with $n$ objects, let $D$ be the number of draws needed to get a complete set. Then we have the formula $$\mathbb{P}(D\leq k)=n^{-k}\ n!\ \left\lbrace {k\atop n}\right\rbrace.$$ Here the braces indicate Stirling numbers of the second kind. Plugging in $n=250$ and $k=500$ gives $$\mathbb{P}(500 \mbox{ draws suffice})=250^{-500}\, 250!\,\left\lbrace{500 \atop 250}\right\rbrace =4.49220\times 10^{-18}\approx 0.$$ This means that the chance is practically zero.

The average number of draws to get a full collection with $n=250$ is $$250\left(1+{1\over 2}+{1\over 3}+\cdots+{1\over 250}\right)=1525.168,$$ so you need over 1500 draws to get a halfways decent chance to get all 250 balls at least once.

  • Wow, I actually rephrased this question as part of a database assignment, and learnt something new in probability theory. I initially tried to model it with the "birthday problem" but it seemed futile...right now, I have a new model to learn. Cheers man! – bryan.blackbee Mar 30 '13 at 04:32
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    @bryansis2010 Glad to help. There is a lot of material on the coupon collector's problem available. You may start here: http://en.wikipedia.org/wiki/Coupon_collector%27s_problem –  Mar 30 '13 at 04:45