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Hopefully I can describe this problem correctly.

Let's say you have 2 objects, and you want to randomly choose between the two objects. You can do this by devising a function that can give a 50% chance to choose one of the objects.

Now, let's say you have chosen this object, and you now introduce a new object. You want to make a new choice, between this already-chosen object, and this newly introduced object. You want to devise a function such that overall, there was still an equal likelihood of choosing each of the three object. However, at this point, you only have the result of the first choice, and the newly-introduced third object.

I was told the answer was to give a 2/3 weighting to the already chosen object, and give a 1/3 weighting to the third object. However, my gut is telling me this is wrong, but I can't explain why. Is this actually correct?

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  • $\begingroup$ You have not made it clear exactly what you want to accomplish at the second stage. If you only want each of the three object to have an equal chance of being drawn at the second stage you just give all three probability 1/3. But maybe you have some other condition to try to balance the two selections. $\endgroup$ Sep 18 '12 at 22:01
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It is correct. Call the original two A and B, of which you picked A, and the new one C. One way to look at it is that you should have $\frac 13$ chance of picking C, so have to keep A with chance $\frac 23$. Another way is that in the original selection you had $\frac 12$ chance of picking A. Now you want the chance (knowing the fact that you will make thetwo picks) to be $\frac 13$. Since $\dfrac {(\frac 13)}{(\frac 12)}=\frac23$ that is the chance that you should keep A.

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  • $\begingroup$ Thanks. I went thru the thought process, and I figure that if you added a new object D, then D would have a 1/4 chance to get picked. But C would have 1/3 * 3/4 = 1/4 chance of getting picked as well, so it looks like it really is correct. $\endgroup$
    – steve8918
    Sep 18 '12 at 22:58

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