Bag and balls revisited

[bonanova]

A bag contains two blue balls and an undetermined number of red balls.

A ball of unknown (red or blue) color is added to the bag.

Finally, a ball is drawn at random from the bag.

If the drawn ball is blue, what is the probability that the added ball was red?

[phil1882]

i would be very curious to see the logical answer to this... my guess would be...

near 0. with only that information, however, it seems impossible to tell.

[bushindo]

A bag contains two blue balls and an undetermined number of red balls.

A ball of unknown (red or blue) color is added to the bag.

Finally, a ball is drawn at random from the bag.

If the drawn ball is blue, what is the probability that the added ball was red?

I think the crux of the puzzle lies in the part that is highlighted red above. I assume that by 'undetermined number of red balls', you mean that the precise number of red balls is unknown. However, the solution to the puzzle requires that we assume something (a priori information) about the generating distribution of the red balls (e.g., poisson distribution, exponential distribution, gamma distribution, etc.). We could also start going into improper priors, but I'd rather not =). This puzzle seems dangerously close to a subjective argument about what type of prior distribution does 'undetermined number of red balls' mean, maybe the OP can clarify that a bit?

[benjer3]

At first, I thought it would be impossible to determine anything from just one draw, but then I realized that there is a chance that you draw the added ball, which can give us a probability for the added ball.

I first imagined that there are 2 blue balls, 1 red ball, and 1 unknown, and that we drew a blue ball. The chances that the added ball is red depend on the ball drawn as follows:

blue1: 50%

blue2: 50%

red1: 50%

unknown: 0%

Each possible scenario can then be averaged together to give a probability of 1/3rd that the added ball was red.

From there, it was easy to realize that the drawn blue ball will always be either one of two first blue balls, or the added ball. That would make it so the probability that the added ball was red is 1/3rd no matter how many red balls we started out with.

I do have my doubts about my solution, but it's the best I got. And after all, I thought it was impossible at first.

Edited July 29, 2012 by benjer3

[kbrdsk]

Taking a hint from benjer. There are five possible scenarios where we draw a blue ball: A red ball is added and you draw blue1 or blue2, or a blue ball is added and you draw blue1, blue2, or the newly added ball. So a red ball is added in 2/5 scenarios where a blue ball is drawn, therefore the chance is 2/5.

[bushindo]

I was incorrect in the post above. Turns out we don't really need the prior distribution of the red balls. In the words of CaptainEd, "Durn, hit by Bayes again" :-)

[jim]

EXTENSION PROBLEM--only slightly more difficult. Instead of assuming the added ball had an original probabilty of 1/2 red and 1/2 blue assume a probability of r red and b blue, with r+b = 1 of course.

[plasmid]

Suppose the bag originally had N red balls and 2 blue balls.

If a blue ball were added, then the probability of drawing a blue ball would be

P(drawn ball is blue, given that the added ball was blue) = 3 / (N+3)

If a red ball were added, then the probability of drawing a blue ball would be

P(drawn ball is blue, given that the added ball was red) = 2 / (N+3).

I'll say that there's a probability R that the added ball was red and (1-R) that the added ball was blue (in the absence of any knowledge about what ball is drawn from the bag at the end of this scenario). And I'll restate the problem of "If the drawn ball is blue, what is the probability that the added ball was red?"... it's equivalent to the probability that (the added ball is red AND the drawn ball is blue) divided by the probability that (the drawn ball is blue).

The probability that the added ball was red AND the drawn ball is blue is

P(added ball is red) x P(drawn ball is blue, given that added ball is red)

= R x (2 / [N+3])

= 2R / (N+3)

The probability that the drawn ball is blue is

[P(added ball is red) x P(drawn ball is blue, given that added ball is red)] + [P(added ball is blue) x P(drawn ball is blue, given that the added ball is blue)]

= R x (2 / [N+3]) + (1-R) x (3 / [N+3])

= 2R / (N+3) + (3-3R) / (N+3)

= (3-R) / (N+3)

Now to solve the problem. The probability that (the added ball is red AND the drawn ball is blue) divided by the probability that (the drawn ball is blue)

= The first equation solved above / The second equation solved above

= [ 2R / (N+3) ] / [ (3-R) / (N+3) ]

= 2R / (3-R)

Again, this doesn't depend on how many red balls were in the bag. In the special case that there was initially a 50/50 chance that the added ball was either blue or red, that would come out to be (2 x 0.5) / (3 - 0.5) = 1 / (2.5) = 2/5 = 40%.

[kbrdsk]

Probability of the added ball being red would be 2r/(3b+2r) where r and b are the initial probabilities of the ball being either red or blue.

Forgot exactly how I got there but here goes.

Let n be the number of balls in the bag, then the possibility of selecting any one ball after adding a red ball is r*1/(n+1), similarly the possibility of selecting any one ball after adding a blue ball is b*1/(n+1). If you add a blue ball there are 3 possibile scenarios where you draw a blue ball, with red there are two so 3b/(n+1) + 2r/(n+1) is the probability you'll draw a blue ball at all.

Therefore the ratio of the probability of drawing blue after red to drawing blue at all is 2r/(n+1) : 3b/(n+1)+2r/(n+1), assuming a blue ball is drawn by setting the latter half of the ratio to equal one and simplifying we get 2r/(3b+2r) : 1.

Edited July 29, 2012 by kbrdsk

[jim]

EXTENSION 2--Same problem but now assume we add two balls red with probability r and blue with probability b. What about when we add three balls? This time it will we convienent to assume we start with 10 blue balls--but it is easy to generalize to k blue balls, just a little messy to write up the answer.

[CaptainEd]

I was incorrect in the post above. Turns out we don't really need the prior distribution of the red balls. In the words of CaptainEd, "Durn, hit by Bayes again" :-)

BIT by Bayes, Bushindo! Thanks for the citation

[bonanova]

After missing the boat on my own puzzle, I am pleased to be able to host this conversation.