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How can $20$ balls, $10$ white and $10$ black, be put into two urns so as to maximize the probability of drawing a white ball if an urn is selected at random and a ball is drawn at random from it?

The correct answer is to put $1$ white ball in an urn and the remaining $9$ white and $10$ black balls in the other urn. How does one prove this?

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    $\begingroup$ One calculates the probability of drawing a white ball in all different configurations, and compares the results. I'm sure there's a clever way to go about it, but no matter how clever your argument is, that's at the bottom. $\endgroup$ – Arthur Dec 7 '16 at 17:50
  • $\begingroup$ I am looking for a more elegant method. $\endgroup$ – David Dec 7 '16 at 17:52
  • $\begingroup$ Then perhaps try something recursive: Show that in any other configuration, moving a ball from the urn with fewest to the one with most balls increases the probability, as long as there is at least one white in each urn. Or just straight-up find a formula for the probability given the number of white and black balls in the first urn. $\endgroup$ – Arthur Dec 7 '16 at 17:53
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The probability of choosing a white ball is

$$P(\mbox{w}) = \frac12\frac{\# \mbox{white in urn 1}}{\mbox{total in urn 1}} + \frac12\frac{\# \mbox{white in urn 2}}{\mbox{total in urn 2}}.$$

If each urn has an equal number of white and black balls, then $P(\mbox{w})$ is always $50\%$.

Otherwise, one of the urns has more black balls than white, let's say urn 2. Therefore, $$\frac{\# \mbox{white in urn 2}}{\mbox{total in urn 2}} < \frac12.$$ Now let's also assume that there is atleast one black ball in each urn. In this case,
$$\frac{\# \mbox{white in urn 1}}{\mbox{total in urn 1}} \leq \frac{10}{11} \hspace{.3in} \mbox{(in case 10w,1b)},$$ therefore $$P(\mbox{w} \ \big| \ \mbox{atleast one black in each}) < \frac12 \frac{10}{11} + \frac12 \frac12 \approx .705.$$

Finally, we check the case where there is not atleast one black ball in each urn. So let urn 1 have 1w,0b and urn 2 have 9w,10b, and check that the probability of white in this case is $>.73$. Also note that any more white in the first urn are redundant: it won't change the probability of getting white if the first urn is chosen (it's already 100%), and it will always lower the probability of getting white in the second urn. Therefore the best option is 1w,0b in one urn, and 9w,10b in the other.

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  • $\begingroup$ From that I understood that one urn ought to contain $0$ black balls, but to my understanding it does not prove that the urn with $0$ black balls should contain exactly $1$ white ball. I suppose I could calculate the probabilities of the cases where the urn with only white balls has $1$ to $10$ balls and that way complete the proof but it seems like a hassle. $\endgroup$ – David Dec 9 '16 at 19:26
  • $\begingroup$ Well, any more white in the first urn are redundant: it won't change the probability of getting white if the first urn is chosen (it's already 100%), and it will always lower the probability of getting white in the second. I think it is enough just to state this observation, rather than compute all the individual cases. $\endgroup$ – Michael Harrison Dec 9 '16 at 19:30
  • $\begingroup$ That makes sense, thank you very much. $\endgroup$ – David Dec 9 '16 at 19:32
  • $\begingroup$ No problem, I updated the answer just for completeness. $\endgroup$ – Michael Harrison Dec 9 '16 at 19:32
  • $\begingroup$ I cannot award the bounty for 22 hours so you'll get it tomorrow. $\endgroup$ – David Dec 9 '16 at 19:33

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