Have You Seen This Urn Problem Before?

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SUMMARY

The discussion centers on a probability problem involving an urn containing red, green, blue, and orange balls, where a sampling machine draws pairs of different colored balls with replacement. The participants clarify that the problem does not have a specific name but relates to various urn problems in probability theory. To estimate the proportion of each colored ball, one can divide the number of pairs containing a specific color by twice the total number of drawn pairs. This method allows for the estimation of proportions despite only drawing two colors at a time.

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I'm interested if anyone has seen something like this urn problem before and if it has a name?

The premise is:

An urn contains red, green, blue and orange balls. I have a sampling machine that can draw samples from the urn, but will only draw balls of any two colors at a time. For example, the machine can draw red and green balls and I will get a sample that tells me the ratio of red to green but not the proportion of green balls in the urn as a whole.

Thanks!
 
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I'm assuming your problem is about drawing a pair of two different color balls each time, right? Do you replace the balls back to the urn before the next drawing?
 
Hi Viralux, yes, different colored balls, and with replacement. Thanks
 
I don't think this problem has an specific name since there are many different urn like problems in the probability theory.

In this case you want to estimate the proportion of each colored ball; if you draw one at a time you simply divide the number of of each colored ball n by the total number of drawn balls N, that is n/N.

Since you are drawing two at a time and they are different color each time you have only 6 possible outcomes {r,g} {r,b} {r,o} {g,b} {g,o} {b,o} and you can estimate then the proportion for each pair as if they were just one color.

But if you want to estimate the proportion of just one of the colors you simply need to divide the number of pairs having that color by twice the total number of drawn pairs, for example, the estimation for the red ball would be: (#{r,g}+#{r,b}+#{r,p})/(2*N)
 

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