Socks

[factor-fiction]

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?

[jim360]

Doesn't matter what sock you pull out at first, so long as the second matches, which will occur 1/3 of the time. In general the answer is 1/(n-1) where n is the total number of socks.

At least, I think it's that.

[bhg481]

I would argue 1 in 3.

The first sock can be either colour. There are then 3 socks remaining in the drawer and only 1 of them will do to make the pair.

[Prudie]

I'm nervous to give an answer but make it 1/3

[factor-fiction]

Most people I asked thought it would obviously be an even chance, arguing you can either get 'white white', 'white black', 'black black' or 'black white'

[jim360]

Yes I can see why people might come up with that -- the point here is that the combinations are "right right", "right wrong", "right wrong" -- the first sock doesn't matter.

Probability is only confusing if you make it so. Drawing a probability tree makes it rather a lot easier, always. That and paying attention to the actual question, which everyone has a problem with pretty much all the time in maths.

[Retrochic]

You need a pair and it does'nt stipulate if the pair is a black pair or white pair so the choices are a pair (W/W, or B/B) OR no pair ( W/B). I would say the chances are 1 in 2

[jim360]

Definitely 1/3 -- draw the full probability tree and you'll see that you can get "no pair" twice as often as you get a pair.

[jim360]

If we label the socks as W, w, B, b, then the full event space is:

Ww WB Wb wW wB wb Bb BW Bw bB bW bw

ie 12 possible draws of two socks, all of equal probability 1/12=(1/4)*(1/3). Only four of these combinations lead to matching pairs, for a total chance of 4/12 or 1/3.

[woofgang]

I agree, once you have picked a sock, its one in three.

[factor-fiction]

And it's one in three even before you pick a sock isn't it?

[sunny-dave]

Yes - I just did the same analysis as Jim (using B1 B2 and W1 W2 as it happens) - the confusion about 50/50 is because that ignores the fact that there are twice as many ways of getting WB or BW than WW or BB.

W1 W2
W1 B1
W1 B2

etc

[joggerjayne]

Yes, it is, ff.

If socks 1 and 2 are black, and socks 3 and 4 are white ...

Before you dip into the drawer, there are only six "two sock" combinations that you can pull out ...

1,2 ... 3,4 ... 1,3 ... 2,4 ... 1,4 ... 2,3

Of those six combinations, two of them (1,2 and 3,4) will make a pair.

[Retrochic]

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.

[jim360]

Do the analysis yourself retrochic, you'll see it's 1/3. It has to be: more simply, if you think about what you are left with after pulling out the first sock, there are three left, of which only one matches what you already have.

[sunny-dave]

There are twice as many 'wrong' pairs as 'right pairs' in the drawer, retrochic.

[factor-fiction]

Thanks everyone. I'm glad I went for the slimmed down version of the question. (The one I saw in The Guardian was the same except it involved 4 black and 4 white socks so would have been a 3 in 7 chance I think)

[jim360]

If you're insisting that the socks are drawn out at the same time, then the event space would only reduce to:

Ww, WB, Wb, wb, wB, Bb

ie six combinations of which only two match -- still 1/3, still twice as many wrong pairs, etc.

[sunny-dave]

But what if they were striped socks, factor?

[jim360]

Black with white stripes or white with black stripes?

[Retrochic]

ok I must be reading this question incorrectly. 'what are the chances you will pull out a pair of matching socks' Does this mean you get more than one go at getting a pair? There are only two variables -pair or non pair, if you have a non pair and are then allowed to go back in and pick another sock then that changes the equation. Doesn't it?