Conditional probability question about choosing 2 balls at random

Question

The question is as follow:

There is a bag with 20 black balls and 20 white balls. You close your eyes and draw two at random. You see one of the balls you chose is a black one, so what is the probability the second one is also black?

I was thinking this problem over with a few people and we are all getting different answers.

I thought this was a conditional probability question where we compute

$$P(\text{second ball is black}| \text{first ball is black})$$

But it was pointed out this assumes the first ball is black only, which may not be the case. It could be the case we see the second ball we picked is black.

Another answer we considered was 20/40*19/39, but I am not quite sure this makes use of the information given? What would be the proper method for discerning the answer?

Answer 1

You are getting different answers probably because the problem statement is ambiguous: it can be interpreted in at least three different ways. At the end of your answer you were expressing how it was not clear to you what the given information is and how different answers depend on different assumptions ... and all of that is indeed just not clear! Consider the three following scenarios:

1. You closed your eyes when drawing the balls, but then you opened them, saw both balls, and see that exactly one of them is black. (this is analogous to you keeping your eyes closed, and someone telling you: "Hey, exactly one of the balls you drew is black")

2. You closed your eyes when drawing the balls, but then you opened them, saw both balls, and see that at least one of them is black. (this is analogous to you keeping your eyes closed, and someone telling you: "Hey, at least one of the balls you drew is black")

3. You closed your eyes when drawing the balls, then you put one behind your back, and then you opened your eyes and see that the one in front of you is black. (this is analogous to you keeping your eyes closed, but you giving one of the balls to another person and that person telling you: "Hey, this ball is black")

Note that all 3 scenarios are compatible with the statement that "You see one of the balls you chose is a black one" (analogously: you know that one of the balls you chose is a black one), but you will arrive at a different probability for the second ball being black in each case:

Scenario 1:

$$P = 0$$

Scenario 2:

$$P = \frac{P(BB)}{P(BB) + P(BW) + P(WB)} = \frac{\frac{20}{40}\cdot \frac{19}{39}}{\frac{20}{40}\cdot \frac{19}{39}+\frac{20}{40}\cdot \frac{20}{39}+ \frac{20}{40}\cdot \frac{20}{39}}$$

Scenario 3:

$$P = \frac{19}{39}$$

Notes:

This problem is analogous to the "Boy or Girl Paradox": "Mr.Jones has 2 children, one of which is a girl. What is the chance the other child is a girl?" Math teachers love to give that problem to students, and when the student answers the intuitive $\frac{1}{2}$, the teacher will say 'Got ya, it is $\frac{1}{3}$'.

(so, if you got the socks problem from your instructor, it is very likely that the instructor is looking for the answer compatible with scenario 2 above (hey, it contains the most math symbols!) ... but the best answer is to point out the ambiguity and answer according to each of the possible intepretations, as I did)

However, the problem is that the problem is often not unambiguously stated, so that in fact there are interpretations where $\frac{1}{2}$ makes perfect sense, and other situations where it should be $\frac{1}{3}$. (see Boy or Girl Paradox for a discussion)

Did you get to the Monty Hall problem yet? That one is likewise very often stated in an ambiguous way, so that the right answer cannot be determined. (see Monty Hall Problem for a discussion on that one)