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Richard Walker

Sock Problem Solution

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Edited by Richard Walker, Thursday, 8 May 2025, 20:15
The question didn't say how many pairs of sock are involved, so suppose we have 2 socks of each colour and n colours. so there are 2n socks altogether.

Then as the socks are drawn one by one from the draw it is possible that socks 1 and 2 match, socks 2 and 3 match, and so on, up to socks 2n - 1 and 2n. In all there are 2n - 1 possible pairs of consecutive socks whose colour could match.

In each consecutive pair once the colour of the first sock is known there is then a 1/(2n - 1) probability that the next sock with be the same colour, or in other words the mean expectation of the colours matching is 1/(2n - 1).

But this applies to all 2n - 1 pairs and by the principle known as linearity of expectation we can find the overall expectation simply by adding the 2n - 1 individual expectations. But (2n - 1) x 1/(2n - 1) = 1 and so we conclude the expected number of matches is 1, whatever the number of pairs of socks in the draw!

I wrote a Python program and ran it, and sure enough as predicted the result is always close to 1.

n = 20
trials = 10000
sox = list(range(n)) * 2
matches = 0
import random
for i in range(trials):
    random.shuffle(sox)
        for j in range(2*n - 1):
        if sox[j] == sox[j + 1]: # do socks match?
            matches = matches + 
print('Mean =', matches/trials)
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Richard Walker

A Problem With Socks

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In my sock drawer I have some pairs of socks. Each pair is a different colour from every other pair but otherwise the socks are all identical. Originally they were in neat matching pairs but now they are all jumbled up.

Suppose I pull socks out one at a time, with my eyes closed, until the draw is empty. How may times can I expect two consecutive socks to be the same colour?


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Richard Walker

An Unusual Word - Gongoozler

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I came across in the Times "Word watch" (Times 2, 01/05/25). According to the OED it originally meant "an idler who stares at length at activity on a canal", and by extension someone who stares at anything for a long time. 

The OED classifies it as a dialect word and cites the Glossary to Bradshaw's Canals & Navigable Rivers Eng. & Wales (1904), which suggests a Lake District origin. Here is the entry, which I located with the help of Google Books


I thought I'd see if I could find out more about the origin, so I tried asking DeepSeek AI and it suggested the first print appearance as being in "The Folk-Speech of South Cheshire" by Egerton Leigh. This seemed promising! After some digging I located such a book and was getting quite excited - but alas! gongoozler was not there, DeepSeek was wrong.

The OED's summary is that the word of unknown origin, but does offer a tantalising third possibility – 

"... compare Lincolnshire dialect gawn ‘stare vacantly or curiously’, gooze (also goozen) ‘stare aimlessly, gape’."

–  and perhaps that's the best guess we have.

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Richard Walker

The Battle of Fennel

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A few days ago I was surprised, when reading a recipe, to discover the Greek name for the herb (and vegetable) fennel is marathon.

This sounds very similar to the name of the famous Battle of Marathon (490 BCE), at which an Athenian army, with a smaller contingent from Plataea, resoundingly defeated a much stronger Persian force, in an encounter often seen as the epitome of a brave defence overcoming a powerful and aggressive adversary.

And indeed the place is named after the plant. I once visited Marathon and there wasn't a whole lot to see really, and I can't remember any fennel, but the place was noted for it in the 5th century BCE. The word marathon itself has a long history, because it is attested in a tablet or tablets from Mycenaean times, so around a millennium earlier than our battle. The tablet was written in Linear B script, not Greek characters, although the language was early Greek, and the word is transcribed like this


 
To return to the story: the Athenians had sent a runner, Pheidippides, to the other pre-eminent Greek city at the time, Sparta, to ask for assistance, but Sparta was in the middle of a solemn religious festival whoch prevented them for responding straightaway.

There was not enough time to wait for the delayed Spartan help to arrive, so the Athenian army and 1000 soldiers from Plataea fought alone. The Greeks attacked the Persians and eventually won, against the odds.

Now although Pheidippides' epic run seems to have been to Sparta, as described above, in time a second version of the story emerged: that Pheidippides ran instead to Athens, bearing news of the victory. Marathon to Athens is - well, about the length of a modern Marathon. When the Olympic Games were resurrected in 1890, a race over this distance was included, although it had not been part of the ancient Games.

Now -athon has become a productive suffix denoting some activity taking place over an unusually long distance or time, so in the OED you can find danceathon, skate-a-thon, cut-a-thon (hairdressers) and schmoozathon, and Collins gives walkathon, drinkathon and sale-a-thon as examples.
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Richard Walker

More Beautiful Geometry

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Here's another nice geometrical theorem. I hadn't seen it before today but it seems it's usually called Thébault's problem II and is credited to Victor Thébault (1882–1960), a mathematician who published over 1000 problems

Here's how it goes (this is a slight extension of the original problem, which had a square where the following requires only a parallelogram). Given a parallelogram ABCD as shown, erect equilateral triangles on two adjacent sides. Then the two free vertices of those triangles together with the vertex of the parallelogram opposite form an equilateral triangle.

The following is my proof. I haven't looked at any published ones, but I'm sure this is one of the standard ones and I imagine we could use complex numbers quite nicely as well, but this uses only ideas that would have been familiar to Euclid.

In this diagram there are two different lengths, marked with | and ||. Call these L1 and L2 and suppose the angle BCD is Îą. Then using the properties of equilateral triangles (all angles are 60) and parallelograms (adjacent angles add up to 180) we can mark in a series of angles.


The angle EDF is 60 + Îą by subtracting the other three angles at D from 360.

Now if we consider triangles EFD, FCD and BAE, we see they all have an angle of 60 + ι between sides of length L1 and L2, which means these are all congruent and it follows immediately that the three dotted distances are equal and BEF is equilateral as claimed.

ThÊbault's theorem has attracted some deal of interest, because it is so neat and quite simple, but seemingly was not spotted for over 2,000 years. It has ben generalised at least twice, most recently in 2019, see The Mathematical Gazette, Vol. 103, No. 557 (July 2019), pp. 343-346.

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Richard Walker

A Lovely Geometry Theorem

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Suppose equilateral triangles ABE, ACF and ADG share a common vertex A as shown. Then the midpointss of the segments EC, FD and GB form the vertices of an equilateral triangle.

The special case where points BECFDG lie on a circe centred at A was proved by Alon Amit (April 5 2020) in response to a question posed on Quora:

A hexagon is inscribed in a unit circle such that three alternate sides are of unit length with the other three arbitrary. Connect the midpoints of the three arbitrary length sides. Can you prove those points form an equilateral triangle?

Amit's clever proof, using complex numbers, does not in fact depend on the points lying on the unit circle and holds as long as we have the configuration in the figure above, and this more general case is even more surprising than the original version.


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Richard Walker

Pet Lovers 🐈 🐕

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I've just helped towards a Kitty's expensive dental bill. I was happy to do it, because I'm an Ailurophile. The word is relatively modern (1914) and comes from Ancient Greek ἀίλουρος "ailouros" = cat, which is thought to originally mean something like "wavy tail", rather poetic to my mind. A somewhat earlier term for a cat lover was Philofelist, dating from 1843.

A dog lover is a Cynophilist, or Cynophile, from Greek κυνος "kynos" = dog, from Proto-Indo-European kwon, which is also the root of words such as hound, kennel, and canine.

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Richard Walker

Daffynitions

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Edited by Richard Walker, Monday, 7 Apr 2025, 21:46

Condescending: prison inmate using rope to escape.

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Richard Walker

Apple Blossom Time

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Richard Walker

My Latest Startup

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I’ve stated a business that checks Christmas cribs are up to standard. It’s called “QA in a Manger”.

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Richard Walker

Too Little Information

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Edited by Richard Walker, Tuesday, 1 Apr 2025, 23:53
I love puzzles where there seems to be too little information. I found this one on the Mind Your Decisions YouTube channel.


If the radius of the circle is 8 units, find x.

How can we solve this when we don't know where on the diameter point P lies? Well, the mere fact that we aren't told where but are still expected to solve the problem, supplies the extra information we need. The length x must be the same wherever P is.

Very well then, we can choose anywhere we like. So let's pick the centre of the circle.

Now the angle marked in the triangle is 60 degrees, by subtraction from 180, and the two sides adjacent are radii of the circle, so their length is 8 units. Because of the 60 angle the triangle must be equilateral, and so the length of its third side is also 8.

So the answer is x = 8 units.
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Richard Walker

Household Chores

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Richard Walker

Solution to "Five Points of a Sphere"

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See https://learn1.open.ac.uk/mod/oublog/viewpost.php?post=295536

If there are five points on the surface of a sphere, then no matter how they are arranged at least four of them lie in the same hemisphere.

Proof: Pick any two of the points. These two points, taken with the centre of the sphere, define a plane that cuts the sphere into two hemispheres, both containing the points we picked.

Three points remain, and of these at least two must lie in the same hemisphere, which will then contain the required four points.

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Richard Walker

Tom Swifty

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"Thank you so much for the lovely sofa and matching armchairs", said Tom sweetly.
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Richard Walker

Five Points of a Sphere Puzzle

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Suppose we mark five points of the surface of a sphere.

Can you prove that however the points are distributed it is always possible to draw a hemisphere that include at least four of the five points in its interior or on its boundary?

This puzzle appears in many places and was included in a maths competition as recently as the early 2000s. but I think it must go back further and may have first been published by Martin Gardner, although I don't have the reference.
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Richard Walker

The Royal Order of Adjectives

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Do you like my grammar hardback red new fat big lovely book?



Most native speakers of English will instinctively agree the order of the adjectives in my sentence above is wrong—weird even—without necessarily being able to describe exactly why; it flouts some rules that we all know but are not normally conscious of. These rules are difficult to write down precisely but roughly speaking follow a sort of semantic spectrum. The Cambridge Dictionary gives this order

opinion - size - physical nature - shape - age - colour - origin - material - type - purpose

If we rewrite my initial sentence using this order we get

my lovely big fat new red hardback grammar book

which sounds perfectly normal, albeit a bit wordy perhaps.

Mu new dictionary—the one pictured—calls the rules "Adjectival order", although "Royal Order of Adjectives" is a more colourful name. 
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Richard Walker

Etymology of "coleslaw"

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Eating my coleslaw on Monday, I wondered about the etymology. The first element is recognisable as meaning "cabbage", as in kale, Old English cole, German kohl, and Italian Cavalo, etc. all stemming (pun intended) from Latin Caulus, stalk or cabbage.

But "slaw" seemed a bit mysterious. 

It turns out to be from Dutch, sla, slaw, a contraction of "salad", and we borrowed "coleslaw" directly from Dutch koolsla,"cabbage salad".
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Richard Walker

Daffynition

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punctual:  I’ll fight everyone 
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Richard Walker

Reedbeds at Fowlmere

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Richard Walker

Answer to "Another Classic Problem"

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See here for the problem I posed.

The answer is the expected proportion of Heads and Tails is 50:50.

A plausible intuition is that the strategy described in the problem with result in more Heads being observed than Tails, but this intuition is misleading

Each coin flip is independent and over a large number of trials Heads and Tails will each turn up with about the same frequency, and no strategy we might decide to follow can change this ratio.

I wrote a Python program to simulate the problem and sure enough if we run it Heads and Tails each appear in with about the same frequency.

import random
H = 0
for i in range(1000000):
throw = random.choice(['H','T'])
    if throw != 'T':
        H = H + 1
    while throw != 'T':
        throw = random.choice(['H','T'])
        if throw  != 'T':
            H = H + 1
print('Heads per 1000000 Tails', H)

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Richard Walker

One Liner

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I was going to buy a cul-de-sac, but I decided it was a dead end.
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Richard Walker

Another Classic Problem

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Suppose 1,000 people take part in the following experiment.

Each participant is given a fair coin, which they proceed to flip one or more times, recording the outcome. As long as their coin turns up heads they keep flipping it. As soon as a tail comes up (but not before) they immediately stop flipping their coin.

What is the expected proportion of heads recorded versus the number of tails?

(This problem is usually framed a bit differently but I think the version given above is preferable to the traditional one.)
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Richard Walker

"Turkey Tail"

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We found this striking Turkey Tail fungus Trametes versicolor growing on a dead log in our garden.


From Wikipedia, it is found world-wide and lives by digesting lignin, a key structural component in plants= tissues. The fungus has been used in traditional Chinese medicine and a range of beneficial properties have also been claimed for it in the setting of conventional medicine. For a summary see this article. I haven't read enough to assess how strong the evidence is, and I think opinion is probably divided.

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Richard Walker

Tom Swifty

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Edited by Richard Walker, Sunday, 2 Mar 2025, 13:43

"I'm counting how many pots of tea you make", said Tom brutally.

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Richard Walker

A classic puzzle from Martin Gardner

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Edited by Richard Walker, Sunday, 2 Mar 2025, 22:44

This puzzle can be found in many places in the internet but was first published by Martin Gardner in his 'Mathematical Recreations' column in Scientific American. It goes like this (this wording is mine, not the original which I don't have to hand.)

Two mathematicians are sitting outside a café, and one say to the other, “I have three children, and the product of their ages is 36. Can you work out their ages?”

“No”, say the second mathematician, “I don’t have enough information”.

“Then what if I tell you the sum of their ages is equal to the number of that house directly opposite?” the first mathematician replies.

“Still not enough information”, answers the second.

“Well what if I tell you the oldest is called Bill?”, say the first.

“Aha, now I know the ages!” says the second mathematician.

1. Can YOU work out what the ages are? (Solution appears further down.)

2. Can you find a number different for 36 but which would still let the second mathematician deduce the ages if they knew the sum of the ages and the fact that the oldest child is called Bill?



PS The second triple is wrong see comments.

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