19 Mar '10 00:14

If your sock drawer has 6 black socks, 4 brown socks, 8 white socks, and 2 tan socks, how many socks would you have to pull out in the dark to be sure you had a matching pair?

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20 Mar '10 09:20

Not everyone here at RHP has started their first year (if I understand 'first year' correctly). But the pidgeon hole principle is a problem everyone can grasp. I use it myself to entertain guests in a party. Questions like stockings in a dark room where right and left doesn't matter, and gloves where it does matter, and such. Very appreciated.*Originally posted by neil67d***Shouldnt all this bollocks be covered in first year?**- Joined
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e420 Mar '10 23:58Thread 107865

Feb 09. Then it was 10 socks.

Becuause of that thread I now buy all black socks and no other colour

so if I need socks in a power cut I can just take out two.- Joined
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podunk, PA22 Mar '10 01:092 edits

Woops, I went right for the combinatorics ( which apparently I haven't the skill to use either).*Originally posted by AThousandYoung***There are only four colors. Once you have one of each...then what?**

Just out of curiosity can anyone show me how to come up with the correct answer using a combinatorial approach? I just think that I was way off logically, and managed to arrive in the realm of the correct solution by luck.- Joined
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22 Mar '10 08:44first sock = colour a.

chance of second sock being colour a = 1/4

if not, second sock is colour b

chance of third sock being colour a or b = 1/2

if not, third sock is colour c

chance of fourth sock being a or b or c = 3/4

if not, fourth sock is colour d

chance of fifth sock being a or b or c or d = 1

So chance of getting a match by sock number n is:

N : Chance

1 : 0

2 : 1/4

3 : 1/4 + (1-1/4) * 1/2 = 5/8

4 : 5/8 + (1-5/8) * 3/4 = 29/32

5 : 29/32 + (1 - 29/32) * 1 = 1- Joined
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127.0.0.122 Mar '10 15:59

Your math makes it seem like there are infinitely many socks in the drawer and they are evenly distributed between the 4 different color.*Originally posted by iamatiger***first sock = colour a.**

chance of second sock being colour a = 1/4

if not, second sock is colour b

chance of third sock being colour a or b = 1/2

if not, third sock is colour c

chance of fourth sock being a or b or c = 3/4

if not, fourth sock is colour d

chance of fifth sock being a or b or c or d = 1

So chance of getting a match by sock num ...[text shortened]...

3 : 1/4 + (1-1/4) * 1/2 = 5/8

4 : 5/8 + (1-5/8) * 3/4 = 29/32

5 : 29/32 + (1 - 29/32) * 1 = 1

The problem does not reflect that.

There could be a million black socks, 19 red socks, 2 blue socks, and 1 green sock and the answer would still be 5.- Joined
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tinyurl.com/y2c6j2t322 Mar '10 21:26*Originally posted by joe shmo***Woops, I went right for the combinatorics ( which apparently I haven't the skill to use either).**

Just out of curiosity can anyone show me how to come up with the correct answer using a combinatorial approach? I just think that I was way off logically, and managed to arrive in the realm of the correct solution by luck.**If your sock drawer has 6 black socks, 4 brown socks, 8 white socks, and 2 tan socks, how many socks would you have to pull out in the dark to be sure you had a matching pair?**

There are 20 socks. Suppose you pull a white sock first. You cannot pull a white sock with certainty. We can model certainty as having atrocious luck. This means that you will now fail to pull a white sock 2nd.

Now we have 20 - 8 = 12 possible socks to pull without getting a pair. Suppose we pull a black sock. By the same reasoning, the third sock comes from a pool of only 6 possible "losing" socks. Say it's brown. The fourth must be the tan sock (we're talking worst possible luck here).

There are no colors left.

I don't know if that's just a long winded version of my previous post or not but I tried ðŸ™‚