I have a black box that contains 4 balls. 3 are red and 1 is blue.
I remove two from the box without seeing them.
I look at one of them, it is red.
What is the probability that the other one is red?
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4$\begingroup$ Computer simulation strongly supports a probability of 2/3. $\endgroup$– user2357112Commented Nov 11, 2014 at 19:04
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$\begingroup$ I don't understand why you keep flipping between accepted answers :P $\endgroup$– DarkgammaCommented Nov 11, 2014 at 19:14
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$\begingroup$ @Darkgamma Fair point, I don't like it either.. but the explanation and answers keep improving. lesson: wait longer before accepting $\endgroup$– d'alar'copCommented Nov 11, 2014 at 19:20
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$\begingroup$ It seems to me that the two solutions of enumerating the answers or thinking about just looking at the remaining three balls are logically / mathematically equivalent. Sort of like the two approaches to the Two Trains puzzle. $\endgroup$– bannceeCommented Nov 12, 2014 at 15:29
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$\begingroup$ Yes, I agree that they are logically equivalent - because when 3 of the 12 possibilities were eliminated it is now the same as discounting the 1st ball. But there's something satisfying about seeing it all enumerated out (which is OK for such a small problem) imo $\endgroup$– d'alar'copCommented Nov 12, 2014 at 15:39
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6 Answers
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The procedure of drawing 2 balls and looking at 1 can produce 12 equally likely outcomes. We can draw balls in 6 different ways, and we can choose either of the 2 balls we drew to look at. Labeling the red balls R1, R2, and R3, and using an arrow to mark the one we look at, the possibilities are
R1,R2 R1,R3 R2,R3 R1,B R2,B R3,B
^ ^ ^ ^ ^ ^
R1,R2 R1,R3 R2,R3 R1,B R2,B R3,B
^ ^ ^ ^ ^ ^
When we see a red ball, this evidence eliminates 3 possibilities:
R1,B R2,B R3,B
^ ^ ^
Of the 9 possibilities consistent with the evidence, 6 of them have the other ball be red. Thus, the probability that the other ball is red
is 2/3.
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2$\begingroup$ Interesting extra: If I look at both and say "at least one of these is red" instead, the probabilities change to 50% (since I don't eliminate any combinations) $\endgroup$– FalcoCommented Nov 12, 2014 at 9:43
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My solution was to disregard the first ball, which we know is red.
The second ball we extracted came out of a pool of 3 balls (2 red and 1 blue).
Therefore, it has a chance of exactly
2/3
to be red.
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$\begingroup$ This is originally how I thought of it +1. But I found it a little bit unconvincing for some reason. $\endgroup$ Commented Nov 11, 2014 at 19:31
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2$\begingroup$ @d'alar'cop I find it very convincing. The first extraction has no effect on the second, other than the fact that it leaves one less red ball in the heap. $\endgroup$ Commented Nov 11, 2014 at 19:36
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$\begingroup$ I understand the reasoning and I think it's true as well, but I find the accepted version more "complete" (I guess that's why I feel a little more convinced by it). $\endgroup$ Commented Nov 11, 2014 at 19:47
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1$\begingroup$ @d'alar'cop You're right, that looks like the proper way of computing the odds. I consider this to be a shortcut towards the answer and found it interesting enough to post. $\endgroup$ Commented Nov 11, 2014 at 19:51
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It's equivalent to taking one of the balls, then looking at it, then taking out another and looking at it. 66% is my guess.
Edit: I am going to risk ending up stupid and complain: why 50%? Maybe the combinatorics data ends up showing that but to me it doesn't make any sense. I'm going to stick with 66% due to the fact that, if we assume we have balls labelled R1 R2 R3 B, giving six possible matches, when we pick out two balls and look at one, we can determine that one of the balls is, for example, R1. This means that there are three valid matches: R1R2 R1R3 R1B and three invalid ones: R2R3 R2B R3B. R1R2 and R1R3 thus make up two thirds of the possible matches.
If I am thoroughly wrong in my methodology, might someone explain why? The other answers I've seen don't really cut it for me but that might just end up being chalked up to pickiness or stubbornness.
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2$\begingroup$ Nice and simple. I find enumerating the possibilities to be ugly, and this answer abstracts as much as it can. In both cases, the only thing we can say about the second ball is that it's not the first... Physically picking up the second ball after looking at the first is no different than picking it up along with the first. $\endgroup$ Commented Nov 11, 2014 at 21:57
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$\begingroup$ I agree, this answer cuts directly to the heart of the problem and solves it in plain words, which is why I like it. I would prefer if the stated answer were 2/3 for the sake of exactness, rather than 66%. $\endgroup$ Commented Nov 11, 2014 at 21:59
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$\begingroup$ The big difference is, by looking only at one of the balls, you eliminate some possibilities: By combinatorics you would have 50% chance of both being red. But the cases where you picked a blue ball are not as likely because the ball I looked at was not blue, so some cases are excluded and lower the chance of these combinations to 1/3 $\endgroup$– FalcoCommented Nov 12, 2014 at 9:47
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This is how I see it Basically you have 4 balls. What is the probability that you will pick 2 red balls out of the 4?
There are only 6 ways to choose 2 balls out of the 4. And only 3 of those will give you both red balls.
So my answer is:
3/6 = 1/2
EDIT:
Since @w0lf posted the clarification I am going to try to do this one with the proper math notation.
Let A denote that first ball is red.
Let B denote that the second ball is red.
Then P(B|A) = P(A and B)/P(A)
Read P(B|A) as probability of B happening after A has happened.
P(A and B) i have solved above. which is equal to 1/2. And P(A) is 3/4. Therefore P(B|A) = (1/2)/(3/4) = 2/3
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3$\begingroup$ This answers the question: "What is the probability of extracting two red balls?". OP's question is "Knowing that one ball is red, what is the probability of the other one being red too?" $\endgroup$ Commented Nov 11, 2014 at 19:24
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1$\begingroup$ So which is your answer, 1/2 or 2/3? If you discarded the first answer it might be good to edit it out of the answer altogether or strikethrough it or something. $\endgroup$– MoyliCommented Nov 12, 2014 at 8:00
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$\begingroup$ @Moyli The second answer of 2/3 is correct. I kept the first answer since I refer to it in my second answer. $\endgroup$– stackErrCommented Nov 12, 2014 at 13:45
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Classical conditional probability (or "Bayesian" probability) problem. Bayes formula is well known, but just for simplicity purposes,one may simply think: I have one ball FIRMLY in hand and it IS red. So the problem has transformed to: Draw one ball out of 3 (2 red, 1 blue) probability of drawing a red, clearly= 2/3
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To expand on DarkGammas answer
Label the balls R1, R2, R3 and B1 if you grab two of them the possible combinations are
R1/R2 R1/R3 R1/B1
R2/R1 R2/R3 R2/B1
R3/R1 R3/R2 R3/B1
B1/R1 B1/R2 B1/R3
For a total of 12 combinations.
You take two out and look at one which happens to be red. We will call this one R1.
Now if you know the one you looked at was the first ball grabbed you can narrow it down to:
R1/R2 R1/R3 R1/B1 (3 combinations; 2 combinations have the second ball being red; 2/3)
Now if you know the one you looked at was the second ball grabbed you can narrow it down to:
R2/R1 R3/R1 B1/R1 (3 combinations; 2 combinations have the first ball being red; 2/3)
If you do no know which one you are looking at your possible combinations are:
R1/R2 R1/R3 R1/B1 B1/R1 R2/R1 R3/R1. (6 combinations; 4 combinations have both red;2/3)
Therefore I believe 66.6% to be the answer.
(I was initially working under the premise that the last example would yield a result of 50%. once I found that to be wrong, I didn't want to waste the time I put into this answer and posted it anyways.)
answered Nov 11, 2014 at 19:49