2
$\begingroup$

My problem is:

If an urn contains balls of $10^7$ different colors, namely $K_1, K_2, \ldots K_{10^7}$, and there are 1000 balls of each color, so that the total number of balls in the urn is $10^{10}$, then if I draw $10^8$ balls, how do I calculate the probability of having extracted at least one ball of each color? Many thanks in advance

| cite | improve this question | | | | |
$\endgroup$
  • $\begingroup$ sorry for my ignorance, could you please write me down the equation? $\endgroup$ – Danny Bennett Jul 18 '13 at 15:30
  • $\begingroup$ I read but it is not easily applicable to my case (at least for me), as in there there are only 1 type of coupons while I have 1000 of each type of colored ball. I am posting this questions because I do not have a Statistical background. I would really appreciate if you could show me where to position my number into the equation to obtain the probability. $\endgroup$ – Danny Bennett Jul 18 '13 at 15:37
  • $\begingroup$ Why not? If I draw a fixed number of x balls, shouldnt be there only a limited numbers of ways to obtain at least one balls of each color (calculated using combinations) to add up to a probability? Again, I based my comments on some answers seen on similar but not applicable problems. $\endgroup$ – Danny Bennett Jul 18 '13 at 15:47
1
$\begingroup$

To a good approximation, you can pretend the population in the urn doesn't change as you draw, which is the same as drawing with replacement. The chance that color $1$ is represented is $1-\left(\frac {10^7-1}{10^7}\right)^{10^8}\approx 1-4.54\cdot 10^{-5}$. To a less good approximation, you can consider each color independently. This ignores the fact that if you draw several of one color, the chance of getting another color is reduced. The chance that every color is represented is $(1-4.54\cdot 10^{-5})^{10^7}\approx 6 \cdot 10^{-198}$

| cite | improve this answer | | | | |
$\endgroup$
  • $\begingroup$ The approximation from the coupon-collector problem (your first approximation)‌ says that the expected number of drawings after which you've drawn all colours is $E[T] = 10^7 H_{10^7} \approx 1.67 \times 10^8$. It's surprising that the probability of having drawn all colours after $10^8$ drawings instead of $1.6 \times 10^8$ drawings would be as low as $10^{-198}$, though that's possibly just my poor intuition for these things. $\endgroup$ – ShreevatsaR Jul 18 '13 at 16:23
  • $\begingroup$ @ShreevatsaR: It is surprising, but changing the exponent to $1.7\cdot 10^8$ changes the chance for one color to $1-4.14\cdot 10^{-8}$ Big exponents can move things quickly. $\endgroup$ – Ross Millikan Jul 18 '13 at 16:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.