A Hairy, Ballsy Problem

dividavi also solved it and got the right answer, with a unique approach, I might add. Since we now have three people who've solved it, one here on the forum, I'll go ahead and post my full solution (I rounded in the original, but since everyone's posting the full probabilities I'll do that here too):

My solution:

Bag A is most probable. It's 55.56% likely to be the bag you pulled the blue ball from.

Reasoning: with no additional information, each bag has a 25% probability of being the correct bag. Once you pull the blue ball, that probability changes:

Bag A predicts you'll pull a blue ball 100% of the time, so 1*.25 = .25
Bag B predicts you'll pull a blue ball 50% of the time, so .5*.25 = .125
Bag C predicts you'll pull a blue ball 20% of the time, so .2*.25 = .05
Bag D predicts you'll pull a blue ball 10% of the time, so .1*.25 = .025

To find the probability of each bag, divide the new number by the total (.25+.125+.05+.025 = .45).

Bag A: .25 / .45 = .5556
Bag B: .125 / .45 = .2778
Bag C: .05 / .45 = .1111
Bag D: .025 / .45 = .0555

In Bayes form (for Bag A): P(A|B) = P(B|A)P(A) / (P(B|A)P(A)+P(B|~A)P(~A))

P(A|B) = Probability Bag A given you pulled blue ball
P(B|A) = Probability you pulled blue ball given Bag A (1)
P(A) = Probability Bag A (.25)
P(B|~A) = Probability you pulled blue ball given NOT Bag A (.2667)
P(~A) = Probability Not Bag A (.75)

P(A|B) = 1*.25 / (1*.25+.2667*.75)
P(A|B) = .25 / .45
P(A|B) = .5556