20 Mar 10
Originally posted by neil67dNot 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.
Shouldnt all this bollocks be covered in first year?
20 Mar 10
Thread 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.
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22 Mar 10
2 edits
Originally posted by AThousandYoungWoops, I went right for the combinatorics ( which apparently I haven't the skill to use either).
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.
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22 Mar 10
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 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
22 Mar 10
Originally posted by iamatigerYour math makes it seem like there are infinitely many socks in the drawer and they are evenly distributed between the 4 different color.
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.
22 Mar 10
Originally posted by joe shmoIf 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?
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.
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 🙂