Maximizing Probability of Drawing a Yellow Ball from Two Urns

[akoska]

Urgent probability questions!

Please help!

How can 5 black and 5 yellow balls be put into two urns to maximize the probability that a yellow ball is drawn from a randomly chosen urn?

I got:
P(draw yellow)=P(draw yellow intersect urn1)+P(draw yellow intersect urn2)
=P(draw yellow|urn1)P(urn1) + P(draw yellow|urn2)P(urn2)
=1/2((y/y+k)+(5-y)/(10-y-k))

where y=number of yelow balls in urn 1 and k=number of black balls in urn 2
and y, k<=5

Then, I think I should plug in k=1, 2, 3..., differentiate the above, and set it to zero. Find critical value--but I never get a solution! I get something like -9=0...


Also, if there are b black balls and y yellow balls in an urn, and a person takes one out, replaces it, and adds in another of the same color, prove that at the nth trial, the probability of picking a black ball is b/b+y

 

[EnumaElish]

Let F = P(draw yellow|urn1)P(urn1) + P(draw yellow|urn2)P(urn2) = (y₁/n₁ + y₂/n₂)/2.
You also have
y₁ + y₂ = 5
n₁ + n₂ = 10.
Therefore you can write y₂ = 5 - y₁ and n₂ = 10 - n₁ and substitute into F.
Now, how do you simultaneously set y1 and n1 to maximize F?

 

[D H]

Let F = P(draw yellow|urn1)P(urn1) + P(draw yellow|urn2)P(urn2) = (y₁/n₁ + y₂/n₂)/2.
You also have
y₁ + y₂ = 5
n₁ + n₂ = 10.
Therefore you can write y₂ = 5 - y₁ and n₂ = 10 - n₁ and substitute into F.
Now, how do you simultaneously set y1 and n1 to maximize F?

This doesn't work for a couple of reasons:

• The solution (y1=2.5, n1=5) is not an integral solution.
• More importantly, the solution is a saddle point. Its neither a minimum nor a maximum. If you had six of each color rather than five, the derivative-based solution is acheivable but is not maximal.

The resultant resultant probability for the derivative-based approach is 50%. There are many realizable arrangements that yield a 50% probability, for example

• One of each color in one jug, the rest in the other jug
• Two of each color in one jug, the rest in the other jug
• One yellow and four black balls in one jug, four yellow and one black in the other jug.

There are several arrangements that do better than 50%. For example, two yellow and four black balls in one jug, three yellow balls and one black ball in the other yields a probability of 1/3 for jug A, 3/4 for jug B, or 13/24 for a random jug. There are ways to do even better. If you think about the problem for a bit one arrangement should pop out as special. Of course, you could brute-force it; there are only 21 combinations to worry about.

 

[EnumaElish]

I was wrong; y1=3, n1=5 is a saddle point. Thanks for catching it.