ChatterBank1 min ago
Socks
54 Answers
Wildwood's query about dice and probabilities reminded me of a problem I came across recently which showed that even fairly able mathematicians can get confused by probability.
Suppose you have a sock drawer which contains just 1 pair of white socks and 1 pair of black socks. These black socks and white socks are all the same size and feel exactly the same. Now imagine that the 4 separate socks had been thrown in straight from the tumble drier and so are not paired up. Also imagine that you get dressed in the dark and pull out two socks at random. What are the chances you will pull out a pair of matching socks?
Suppose you have a sock drawer which contains just 1 pair of white socks and 1 pair of black socks. These black socks and white socks are all the same size and feel exactly the same. Now imagine that the 4 separate socks had been thrown in straight from the tumble drier and so are not paired up. Also imagine that you get dressed in the dark and pull out two socks at random. What are the chances you will pull out a pair of matching socks?
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I had a bloke in to do a few - do you think this will sell?
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It's a common error, retrochic -- basically you are applying a correct point, that colour doesn't matter, but then applying it too early, with the effect that, somewhere along the line of working, you've set 2=4.
What's gone wrong? I am now going to launch into a lecture, but hopefully an instructive one.
In all problems of discrete probability, the most sensible approach to finding a solution goes something like as follows:
- write down all possible distinct outcomes in the most specific way possible;
- computer the probabilities of each of these outcomes individually;
- Only once this has been done can you turn to considering the specific question, and grouping together various outcomes according to how they fit in with the question, and computing the probability of the groups you are interested in.
Applying this procedure to the problem in question would lead us to the following worked solution:
1) I have four socks, of which I will end up selecting two.
2) I can do this first by drawing one sock, then by drawing another [Note -- this will NOT change the probabilities compared to drawing two at once, as I'll demonstrate later.].
3) In this problem we can identify two white and two black socks and I will start by treating them as distinct. Label this W, w, B and b and we obtain twelve possible specific pairs: Ww, WB, Wb, wW, wB, wb, BW, Bw, Bb, bW, bw, bB.
4) The probabilities for each of these outcomes can be determined as follows: Assume that the experiment is fair, so that the probability of drawing a sock is the same as the probability of drawing any other sock. Then the probability of drawing the first sock is always 1/4 (one sock/total number of socks); and the probability of drawing the second sock is 1/3, so that the probability of any of these pairs is exactly 1/12 -- and all 12 pairs have this probability (total probability = 1, so we have indeed exhausted the possibilities).
5) Now we should check that we haven't messed the problem up by stipulating that we draw one sock then the other. We can do this by noting that in the end, the pairs wB and Bw are the same. Thus we could instead group the possible final pairs as P(W and w) = P(w then W) +P(W then w) = 1/12 + 1/12 = 2/12 = 1/6, and so on. The relative probabilities of any distinct pairing are the same, as in Ww and Wb, etc are all still equally likely pairings. The conclusion we can reach is that from the point of view of what we end up with, whether the socks are drawn together or separately doesn't affect things.
6) NOW we apply the conditions of the specific problem: "...pull out a pair of matching socks." This means that the specific colour doesn't matter, and we can relabel the combinations as follows:
Ww = S
WB = D
Wb = D
wB = D
wb = D
Bb = S
Where S stands for same and D for different. This relabelling, crucially, is just a relabelling of the outcome and has not affected the probability of each outcome, which is still 1/6 for each.
7) Since we are interested only in matching pairs, we identify two outcomes that work, out of a total of 6 equally likely outcomes. Thus P(matching pair) = P(S =Ww) + P(S=Bb) =1/6 + 1/6 = 2/6 = 1/3.
By setting up the approach to the problem as generically as possible, we have ensured that we've not missed anything about the problem, and are led to the solution of 1/3.
What's gone wrong? I am now going to launch into a lecture, but hopefully an instructive one.
In all problems of discrete probability, the most sensible approach to finding a solution goes something like as follows:
- write down all possible distinct outcomes in the most specific way possible;
- computer the probabilities of each of these outcomes individually;
- Only once this has been done can you turn to considering the specific question, and grouping together various outcomes according to how they fit in with the question, and computing the probability of the groups you are interested in.
Applying this procedure to the problem in question would lead us to the following worked solution:
1) I have four socks, of which I will end up selecting two.
2) I can do this first by drawing one sock, then by drawing another [Note -- this will NOT change the probabilities compared to drawing two at once, as I'll demonstrate later.].
3) In this problem we can identify two white and two black socks and I will start by treating them as distinct. Label this W, w, B and b and we obtain twelve possible specific pairs: Ww, WB, Wb, wW, wB, wb, BW, Bw, Bb, bW, bw, bB.
4) The probabilities for each of these outcomes can be determined as follows: Assume that the experiment is fair, so that the probability of drawing a sock is the same as the probability of drawing any other sock. Then the probability of drawing the first sock is always 1/4 (one sock/total number of socks); and the probability of drawing the second sock is 1/3, so that the probability of any of these pairs is exactly 1/12 -- and all 12 pairs have this probability (total probability = 1, so we have indeed exhausted the possibilities).
5) Now we should check that we haven't messed the problem up by stipulating that we draw one sock then the other. We can do this by noting that in the end, the pairs wB and Bw are the same. Thus we could instead group the possible final pairs as P(W and w) = P(w then W) +P(W then w) = 1/12 + 1/12 = 2/12 = 1/6, and so on. The relative probabilities of any distinct pairing are the same, as in Ww and Wb, etc are all still equally likely pairings. The conclusion we can reach is that from the point of view of what we end up with, whether the socks are drawn together or separately doesn't affect things.
6) NOW we apply the conditions of the specific problem: "...pull out a pair of matching socks." This means that the specific colour doesn't matter, and we can relabel the combinations as follows:
Ww = S
WB = D
Wb = D
wB = D
wb = D
Bb = S
Where S stands for same and D for different. This relabelling, crucially, is just a relabelling of the outcome and has not affected the probability of each outcome, which is still 1/6 for each.
7) Since we are interested only in matching pairs, we identify two outcomes that work, out of a total of 6 equally likely outcomes. Thus P(matching pair) = P(S =Ww) + P(S=Bb) =1/6 + 1/6 = 2/6 = 1/3.
By setting up the approach to the problem as generically as possible, we have ensured that we've not missed anything about the problem, and are led to the solution of 1/3.
Retrochic at 9:30 - "You only have one go to pull out a pair, you can't ' double dip'. You have two choices -to pull out a matching pair or an unmatched pair -forget what colour they are its irrelevant, you have 1 in two chances of getting a pair".
You have a white sock in your hand and a drawer containing 99 black socks and 1 white sock. Do you still think you have a 1 in 2 chance of picking a pair?
Similarly, if the drawer contains 99 white socks and 1 black sock, your original argument states that you have a 1 in 2 chance of pairing the white one in your hand. I don't think so!
It is clear that the chance of pairing the sock you already have depends on the ratio of black/white socks in the drawer.
If you have a white sock in your hand and there is 1 white sock and 2 black socks in the drawer you have 1 chance to be right and 2 chances to be wrong.
You have a white sock in your hand and a drawer containing 99 black socks and 1 white sock. Do you still think you have a 1 in 2 chance of picking a pair?
Similarly, if the drawer contains 99 white socks and 1 black sock, your original argument states that you have a 1 in 2 chance of pairing the white one in your hand. I don't think so!
It is clear that the chance of pairing the sock you already have depends on the ratio of black/white socks in the drawer.
If you have a white sock in your hand and there is 1 white sock and 2 black socks in the drawer you have 1 chance to be right and 2 chances to be wrong.
I'm resurrecting this to pass on a theory put to me yesterday. I went riding yesterday with a friend who is a retired professor of mathematics at Bristol Uni . I put the problem to him as best I could remember it. Firstly he asked if the colour of the pairor socks was stipulated -I replied 'no, the probability is calculated on a pair only, colour is not a factor. He asked if sequence was a factor in probability ie I've picked out a white sock what is the probability I'll pick out a match -No this was not stipulated in the question. After a whole narration which frankly went a bit over my head he came to the same conclusion as a few on here that were lambasted for 'intuitive incorrectness' Your chances of pulling out a pair (any colour) is 50/50. You either get a pair or you don't. (W/W B/B) OR (W/B) . He also said that one of the commonest reasons for failing to solve a probability problem correctly is not reading the question fully.
