Puzzle page


Unfortunately I am lazy and disorganised. One symptom of this is that I can never be bothered to fold up my socks in pairs when they come out of the wash. I just chuck them in the drawer. Another symptom is that I always wake up late for work and end up having to rush. Given that I only have white and black socks, how many socks do I have to grab out of my drawer at random to make sure the collection I've grabbed contains a matching pair?

This puzzle is part of the Hands-on risk and probability show, an interactive event culminating in Who Wants to be a Mathionaire? workshop sessions, which you can book to perform at your school. The puzzle appeared in the book How any socks make a pair? by Rob Eastaway.

For some challenging mathematical puzzles, see the NRICH puzzles from this month or last month.



first one is lets say black
second is (worst case scenario) white
third one will make a pair with either the first or second sock

This is a simple application of the pigeonhole principle. If you have k pigeonholes and k+1 pigeons, then one pigeonhole will contain more than one pigeon.




Likewise, an extension of the problem is 'how many socks do I have to pull out' if I have three colors? Four? n colors? When you see the solution for all of these, the answer, elegant (to me), is always 'colors + 1'. --That last one will always duplicate one of the previous colors for what is in this case defined as "pair". (Not that I would like to pull out six socks [for five colors] each morning, leaving the remains all over the sink, simply because I was too lazy to pair them in the first place.)

A then more challenging question is 'how many pulls' for two colors, to make THREE identical socks? Or three colors, three socks, etc.

The answer to above problem is c*(n-1)+1. Because the answer is strictly greater than c*(n-1) as the combination c11,c12,...,c1n;c21,c22,...,c2n;...;c(n-1)1,c(n-1)2,...,c(n-1)n (where cij is color j) is a possible outcome and contains no n-group of any color. Now, let us assume that we took out c*(n-1) socks and still there is no n-group than I claim that the outcome is a permutation of above mentioned outcome.
Proof of claim:
let ni, for i=1,2,3,...c be number of socks there are which have color j. And assume that there is no n-group.
for all i, ni<=n-1
summing up for all i:
Now let us suppose that color I does not have n-1 number, then the sum(ni)

first lets say white
second black
third black or white either one it makes a pair

Haha. Even after reading the question several times, I spent ages trying to figure out how to guarantee the guy gets one of each colour, instead of just an individual pair. Anyway my answer for that (wrong) question was 1/2 all socks + 1. ^0^

Ho ho.

2 and a spare in cause you get a black and a white so you then will either have a pair of white or a pair of black socks

if sock a is white,
and sock b is black,
then sock c can be either black or white and would make a pair

at first I was like 2 but then I thought this is supposed to be hard so then this puzzle got me thinking and I said it depends on the colour so for example you can have 1 black and 1 white you need another black or white sock to make a full pair so therefore the answer is ..... 3 ( in this case ) but logically you need 2 cause you only have 2 feet and to make a PAIR you need 2 like to apples make a pair 2 carrots make a pair etc. And that is my answer and I think most of you will agree !!!!

By ... Zainab !!!