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Paradox Probability Puzzle - 21 October

This is a famous paradox which has caused a great deal of argument and disbelief from many who cannot accept the correct answer. Four balls are placed in a hat. One is white, one is blue and the other two are red. The bag is shaken and someone draws two balls from the hat. He looks at the two balls and announces that at least one of them is red. What are the chances that the other ball he has drawn out is also red?

1. 1/6

probabilities are WB,WR,WR,BR,BR,RR

2. 1/5 ... of the 6 possibilities one (White & Blue) has been eliminated leaving one red & red combo from the remaining five.

3. You can come up with a different answer based on how you look at it. 1/6, 1/5, 1/3 are all possible answers (I think). The reason is because the person who made this was evil enough to make it vague, so that multiple answers can be derived and people can argue with eachother, marriages will break up, wars will start, etc. etc. I had an argument with a friend for hours on a similar question. (something about daughters)

If you assume that you start with nothing and try to calculate the probability of getting both reds, one at a time, the probability is 1/2 and 1/3, making it 1/6. If you look at it with the idea that drawing one blue and one white is impossible, it would still be 1/6 (I think). Because if your first drawing is not red, then it is guaranteed that your second drawing is red, but it doesn't matter since that isn't what you wanted. So, the first drawing has to be red (1/2) and second drawing has to also be red (1/3), making it 1/6.

If you look at it from the perspective of the "end results", the possible endings are: BW BR BR2 RW RR2 RB R2W R2R R2B WB WR WR2. These are permutations. The reason I listed out reds as separate is because if I merged them, then some results will have a higher probability but will be reduced as an equal probability as the rest. By doing this, they all have 1/12 chance of the result. If you removed BW and WB due to the statement, then you'll have 1/10 chance on all. So the chance of RR2 or R2R is 1/5.

Now, if you look at it from the perspective of one was already drawn and it was red, and you are looking for the probability of whether the second one will be red, the probability is 1/3. Because one is for sure red, leaving the second drawing (order does not really matter) with 1/3 chance. Making it 1/3.
Another way to rewrite the situation is that if the person said "at least one was red" and revealed one as truly red, and hid the second ball and then asked the probability, then it would be 1/3. The probability of the first ball (red) should not factor in since we have already been given an answer making a probability of 1 or 100% multiplied by the second ball 33% or 1/3.

4. The chances indeed depend on wether the balls are taken one at a time, or both at the same time.

5. No it doesn't.

6. AnonymousOctober 28, 2011 at 11:09 PM

Your 3rd paragraph is the correct explanation.

In the second paragraph, you incorrectly assumed that throwing out the possibility of getting the WHITE/BLUE combination does not change the probabilities. It in fact does change the probabilities.

In your final paragraph, you correctly found the probability of getting red FIRST, and then getting a second red ball. However, I could also draw out white first, and then red. Or I could draw out blue first, and then red. By including these other ways of having at least one red, the probabilities again are changed. The problem did not specify that the first ball was red -- just that at least one of them was red.

7. The correct answer is 1/3.

Of the 6 possible equally likely outcomes: RR.RW,WR,RB,BR,WB the condition is not "at least 1 is red" but "you are told at least 1 is red" which is not the same thing.

If RR Prob("you are told at least 1 is red") = 1
If RW Prob("you are told at least 1 is red") = 1/2
If WR Prob("you are told at least 1 is red") = 1/2
If RB Prob("you are told at least 1 is red") = 1/2
If BR Prob("you are told at least 1 is red") = 1/2
If BW Prob("you are told at least 1 is red") = 0

1. It is common practice in mathematics to assume the information you have been given is correct. Hence, if they say a red ball has been drawn then a red ball has been drawn. If you don't trust the question then you put into doubt alot of other things, such as, were there really only 4 balls? maybe there was an orange one as well.

Point is, you must trust the information in the question is correct.

2. I am assuming the information that I've been given is correct, that a red ball has indeed been drawn.

The answer is still 1/3 for the reason I explained in my previous post: just because I've been told "at least 1 is red" doesn't change the prior probability that I could have been told something else if the balls drawn were RW or RB

3. You're overthinking this. In setting up problems like this, one not only wants to account for all relevant info, but to disregard an infinitude of irrelevancies. Read the problem again.

"[S]omeone draws two balls from the hat. He looks at the two balls and announces that at least one of them is red."

You've been told at least one is red. What you could have been told under other scenarios only matters in other puzzles. It doesn't matter here, anyway. Two balls were drawn, and then the statement is made as the speaker is looking at the pair. I don't know how this could have been written more clearly. As he's speaking, he's looking upon one of five pairs of balls, with WB (in either order) effectively noted as impossible at this point. He's either holding W and one of the R, W and the other R, B and one of the R, B and the other R, or both R.

I suspect you probably follow us "one-fifthers," but I have such difficulty understanding the factors you consider, I wonder if maybe you've done some problems in the past tricky enough to induce in you a kind of story problem paranoia. I've been tricked into tabulating probabilities of balls of a certain color being drawn from urns when I'd been led to assume a particular distribution of the balls chosen to go into those urns to begin with, and it wasn't at all safe to make that assumption, which put me on edge for a little while. Nothing weird like that is going on with this problem.

4. I'm not overthinking it, the answer is 1/3 for the reasons already stated.
(Applying your reasoning to the Monty Hall Problem -"I pick Door1, Monty opens Door2, what's the probability the car is behind Door3?" - you'd get an answer of 1/2, since you'd disregard the possibility of Monty opening Door3 if the car was behind Door1)

Although it's a necessary condition for a red ball to be drawn in order to "announce that at least one of them is red" it's not a sufficient condition which is what.the "one-fifthers," are assuming

8. If we assume he had one ball RED then the other chances are either a RED or BLUE or WHITE as such the possibility to get another RED ball will be 1/3 = 33.33%