I have a urn full of M balls with an unknown mix of white and black balls. I draw N balls out (without replacement) and get a particular proportion of white vs black balls. From the looks of it there should be a way for my to calculate the probability the balls I have draw have the same proportion (white vs black) as the balls in the urn.

Am I also able to calculate the probabilities of the various proportions based on what what I've draw?

For example if I have a urn full of 10 balls and I draw out 5 white balls then it would be very unlikely the initial composition of the urn would be 5 white/5 black.

  • $\begingroup$ With replacement or without? Usually in this sort of problem, we have a prior distribution, that is, a distribution for the number of each kind. Uniform prior can be used, but it is not necessarily the most appropriate prior. Then, using our experimental result, we can find the posterior distribution. $\endgroup$ – André Nicolas Jul 18 '16 at 6:25
  • $\begingroup$ Clarified that there is no replacement. However the prior is unknown. $\endgroup$ – Q the Platypus Jul 18 '16 at 6:28
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    $\begingroup$ Without a prior, classical conditional probability has nothing to offer. $\endgroup$ – André Nicolas Jul 18 '16 at 6:32
  • $\begingroup$ Can you explain why? I can enumerate all possible urns and all possible draws from that urn with a perticular proportion and then count the number of situations where the two are the same. Am I missing something? $\endgroup$ – Q the Platypus Jul 18 '16 at 7:51
  • $\begingroup$ Yes, you are. You're missing the fact that counting the number of favourable situations only allows you to calculate a probability if all situations are equiprobable. Since you don't know the prior, you don't know that they are. $\endgroup$ – joriki Jul 18 '16 at 8:05

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