# drawing at least 90% of colors from urn with large populations

My problem is: suppose I have an urn containing balls of $n = 10000000$ (i.e., $10^7$) different colors, with $1000$ balls of each color (so the total number of balls is $1000n = 10^{10}$). Suppose I draw $100000000$ (i.e. $10^8 = 10n$) balls. My question is:

how do I calculate the probability that I have drawn at least 90% (in this case $1000000$ or $10^6$) of the different colors?

• I think you should merge this with your other question about the urn with $10^{10}$ balls of $10^7$ colors. – MJD Jul 18 '13 at 15:05
• So you mean you have $1000\cdot 10000000$ balls, colored equally amongst $10000000$ different colors? Because the first sentence strongly implies that there are only $10^7$ balls. – Thomas Andrews Jul 18 '13 at 15:05
This is not a simple question to answer. The naive approach would be to say we can pick the colors not to see in ${10^7 \choose 10^6}$ ways, then each pull has a $0.9$ chance of avoiding them, so we get a probability of ${10^7 \choose 10^6}0.9^{10^8}$. Unfortunately, this double counts the events were we miss $10^6+1$ of the colors and overcounts even worse when we miss more colors. But it is an upper bound, and so small we don't need to worry about the overcount. Alpha gives $9.61097262395 \times 10^{-3163936}$ so you have a great chance of seeing at least $90\%$. In your other question you see there is a poor (but much better chance than missing more than $10\%$) chance of seeing them all.
• Yes, that is a calculation of the chance of seeing less than $90\%$, but is too high. – Ross Millikan Jul 18 '13 at 17:14