So, how many single socks might we expect to have at various times in our pairing exercise?
We need first to ask this question in a more precise manner. There are two related but separate versions of this question.
But before we start, some ground rules…
Just don’t get me started on odd socks – finding the planet where they go is another much more complicated problem, which I’ll leave to astrobiologists and ufologists. So, here we’re thinking only about sock piles with no odd socks.
We will also ignore our natural ability to easily see the two bright orange socks in the pile and pick them out as a pair straight away. That is, we assume each sock is chosen from the pile completely at random, then checked to see if it matches any other previously selected single sock.
Suppose we have N pairs of socks. After choosing s socks, let the number of unpaired socks at any point be u. We want to know the average value of u for any given value of s.
As the value of u obviously depends on the value of s, we say that u is a function of s, and so (formally speaking) we want a formula for the function u(s).
A function u(s) should at best provide an exact value, but at least give us a good approximation.
We can then ask…
After pulling s socks out of the pile, on average how many of these would be unpaired?
We note that
There are many mathematical functions which follow this pattern, starting small, increasing in the middle to some maximum value, then decreasing again. Let’s try the simplest function following this pattern – a quadratic.
A general quadratic may be written as
u(s) = as^{2} + bs + c
with a, b and c being (as yet undetermined) constants.
Substituting the combinations of s and u as above – (s, u) = (0, 0), (1, 1) and (2N, 0) respectively – into the above equation, we get
0 = c
1 = a + b + 0
0 = a(2N)^{2} + b(2N) + 0
=> b = 2aN
from which
a = 1/(2N – 1)
b = 2N/(2N – 1)
So we then have
u(s) = s(2Ns)/(2N1)
This function has a maximum average^{*} value for u when
s=N => u = N^{2}/(2N1)
This latter u is the average (also called the expected) number of unpaired socks after exactly half the socks have been sorted. (Note that this is not the expected maximum number of unpaired socks for the entire sock sorting exercise – see Question 2 below.)
In fact, these formulas give the exact values for the average number of unpaired items and the maximum value of this average. This may come as a surprise, but formulas are often no more complicated than they need to be.
In any case, try it with a pack of cards, treating as pairs those cards with the same value and colour e.g. 7 spades and 7 clubs or Q diamonds and Q hearts. Your count of unpaired cards should, on average, follow reasonably close to the following graph:
It is amazing that we can develop an exact formula for what might seem a very complicated scenario, simply by using three almost trivial data points, together with basic assumptions about simplicity and symmetry – even if the formula is for an average value rather than an exact value.
* Note that we said “maximum average“. We will deal with “average maximum” in an update to this page.
Any 10 digit number I tried seemed to be a phone number – are there any which are not?
Even some 11 digit numbers suffer (in this context) the same fate. (This has changed quite a bit since 2008 – see here.) But, after 20 minutes searching, I did find… wait! If I write it down here as is, then, when Google indexes this page, it will cease to be not on Google.
I could write it as 19×97×11469109 (as found by the excellent Wolfram Alpha site), but I’d like to record your even smaller numbers in an easily comparable way. In any case, if your smaller number is prime, then that wouldn’t work.
I’ll just say my number is 567887 preceded by 21137.
To keep it simple, the rules are:
I know Google responses often depend on the user location and the particular country search page used. An unmatched number in one place may be a postal order number in another.
So I’ll add the minimum number found for each country version of the Google search page.
Tell me whether you want your name or nickname acknowledged or not.
For starters, I have
Google search page  Date  Second part  preceded by  First part  Found by 
www.google.com  10 Apr 2016 
567887 
preceded by 
21137 
Mark Michell 
www.google.com.au  10 Apr 2016 
567887 
preceded by 
21137 
Mark Michell 
www.google.com  11 Jun 2017 
563286 
preceded by 
20874 
Daniel Philpott 
Off you go. Tell me if you followed any method e.g. odd numbers, lots of one digit etc.
4 x 4 = 16
That was also the case on 8 February 2016, and will be again on 2 August 2016. How often do this or similar events happen?
Let’s give the name product dates to these and other combinations of month, day, hours, minutes and/or seconds to match a year number.
Each century has 213 of these dates (see this and other detailed counts in the table below).
We might ask questions such as:
2024 (and 2124, 2224 etc) has 7 such dates: 24 Jan, 12 Feb, 8 Mar, 6 Apr, 4 Jun, 3 Aug and 2 Dec. The years 2012, 2030, 2036, 2048, 2060 and 2072 get honourable mentions, with 6 dates each.
Why? All of these two digit years are numbers with lots of factors, relative to other numbers of similar size.
All of the honourable mentions except 2012 would have more such dates if our calendar had more months in each year, or more days in particular months.
There are 20 such years: 2037, 2041, 2043, 2047, 2053, 2058, 2059, 2061, 2062, 2067, 2071, 2073, 2074, 2079, 2082, 2083, 2086, 2089, 2094 and 2097.
Why? All of these two digit years are prime or have a factor more than the number of days in a month.
Notes:
– 2058 would not be in this list if it was a leap year i.e. 29 Feb 2058 would be a valid date.
– 2062 would not be in this list if 31 Feb was a valid date
There are only 16 days from 30 Jan 2030 to 15 Feb 2030.
Honourable mentions go to:
28 Jan 2028 to 14 Feb 2028 (17 days)
26 Jan 2026 to 13 Feb 2026 (18 days)
24 Jan 2024 to 12 Feb 2024 (19 days)
27 Feb 2054 to 18 Mar 2054 (19 days)
Note that the last of these dates is there only because February has only 28 days in that year.
There are 1097 days from 19 Mar 2057 to 20 Mar 2060.
Honourable mentions go to:
09 Sep 2081 to 28 Mar 2084 (931 days)
06 Dec 2072 to 25 Mar 2075 (839 days)
05 Dec 60 to 21 Mar 63 (836 days)
Each of these (and only these) date pairs surround two consecutive calendar years with no product dates (see 2. above).
The average gap between such dates is just over 171 days or about 5 months 20 days.
If we want to get the four digit year, we need something more than just month and day values, as the maximum product of month and day is only 31×12=372, a year well before our current year numbering system was developed in 525 by Dionysius Exiguus.
A natural addition is to combine the day and month with the hour to get a four digit year. Only 14 and 22 years this century have a total of 80 and 121 days with combinations of month day and hour like this, using a 12 and 24 hour clock respectively. Of these, 18 and 23 of those combinations occur this year (2016), as in the following table. For example, 24 Apr 2016 at 21h gives us 24 x 4 x 21 = 2016.
24 Apr 2016 at 21h 
24 Jul 2016 at 12h  28 Sep 2016 at 08h 
28 Apr 2016 at 18h  12 Aug 2016 at 21h  08 Dec 2016 at 21h 
16 Jun 2016 at 21h  14 Aug 2016 at 18h  12 Dec 2016 at 14h 
21 Jun 2016 at 16h  18 Aug 2016 at 14h  14 Dec 2016 at 12h 
24 Jun 2016 at 14h  21 Aug 2016 at 12h  21 Dec 2016 at 08h 
28 Jun 2016 at 12h  28 Aug 2016 at 09h  24 Dec 2016 at 07h 
16 Jul 2016 at 18h  14 Sep 2016 at 16h  28 Dec 2016 at 06h 
18 Jul 2016 at 16h  16 Sep 2016 at 14h 
There are 18 dates in 2016 using a 12 hour clock, being the nine in green italics above (both AM and PM).
Why is this year so prolific? Well, 2016 = 2^{5} x 3^{2} x 7 i.e. 8 prime factors – way above the average prime factor count of just over 2 for a number its size. So the separate factor count (including 1 and 2016) is 6 x 3 x 2 = 36. No other number between 2000 and 2099 has as many factors.
We could even combine month, day, hour and minute to get the four digit year. Only 34 years this century have a total of 3808 and 2644 combinations of month day hour and minute like this, using a 24 and 12 hour clock respectively. 1046 and 689 of them occur this year (2016).
In fact, as I’m publishing this on 19 Feb 2016, the next few occasions of this are on Sun 21 Feb, as follows:
21Jan2016 02:48  21 x 1 x 2 x 48 = 2016 
21Jan2016 03:32  21 x 1 x 2 x 48 = 2016 
21Jan2016 04:24  21 x 1 x 2 x 48 = 2016 
21Jan2016 06:16  21 x 1 x 2 x 48 = 2016 
21Jan2016 08:12  21 x 1 x 2 x 48 = 2016 
21Jan2016 12:08  21 x 1 x 2 x 48 = 2016 
21Jan2016 16:06  21 x 1 x 2 x 48 = 2016 
All but the last one can be used with a 12 hour clock, but suffixed with each of AM and PM in turn.
Lastly, we could go completely crazy and take this to its logical conclusion, combining month, day, hour, minute and second to get the four digit year. Only 35 years this century have a total of 25848 and 15672 combinations like this, using a 24 and 12 hour clock respectively, of which 8164 and 4797 of them occur this year (2016).
Notes:
The table below sets out both year number factors and event counts for 21^{st} century years. There are some explanatory notes at the foot of the table.
Year 
Factor count^{1} 
Prime 
Event counts 

md =yy^{3} 
mdh 
mdhm =yyyy^{5} 
mdhms =yyyy^{6} 
hms 

12h  24h  12h  24h  12h  24h  12h  24h  
2001 
8  3^{1}23^{1}29^{1}  1  1  8  8  36  27  4  4  
2002  16  2^{1}7^{1}11^{1}13^{1}  2  4  6  64  54  288  204  16 
16 
2003 
1  Prime  2  
2004  12  2^{2}3^{1}167^{1}  3  
2005 
4  5^{1}401^{1}  2  
2006  8  2^{1}17^{1}59^{1}  4  4  4  28  20  8 
6 

2007 
6  3^{2}223^{1}  2  
2008 
8  2^{3}251^{1}  4  
2009  6  7^{2}41^{1}  3  6  3  28  14  8 
4 

2010 
16  2^{1}3^{1}5^{1}67^{1}  4  
2011  1  Prime  2  
2012 
6  2^{2}503^{1}  6  
2013  8  3^{1}11^{1}61^{1}  1  
2014 
8  2^{1}19^{1}53^{1}  3  4  4  28  20  8  6  
2015  8  5^{1}13^{1}31^{1}  3  1  8  8  36  27  4 
4 

2016 
36  2^{5}3^{2}7^{1}  4  18  23  1046  689  8164  4797  168  121 
2017  1  Prime  1  
2018 
4  2^{1}1009^{1}  5  
2019  4  3^{1}673^{1}  1  
2020 
12  2^{2}5^{1}101^{1}  5  
2021  4  43^{1}47^{1}  3  4  2  4 
2 

2022 
8  2^{1}3^{1}337^{1}  3  
2023  6  7^{1}17^{2}  1  1  4  6  18  18  2 
3 

2024 
16  2^{3}11^{1}23^{1}  7  4  6  86  71  460  318  32  26 
2025  15  3^{4}5^{2}  2  3  2  116  76  704  413  28 
20 
2026 
4  2^{1}1013^{1}  2  
2027  1  Prime  3  
2028 
18  2^{2}3^{1}13^{2}  4  2  46  48  334  266  24  20  
2029  1  Prime  1  
2030 
16  2^{1}5^{1}7^{1}29^{1}  6  4  3  72  40  328  173  28  16 
2031  4  3^{1}677^{1}  1  
2032 
10  2^{4}127^{1}  3  
2033  4  19^{1}107^{1}  2  
2034 
12  2^{1}3^{2}113^{1}  1  
2035  8  5^{1}11^{1}37^{1}  2  12  6  52  26  12 
6 

2036 
6  2^{2}509^{1}  6  
2037  8  3^{1}7^{1}97^{1}  
2038 
4  2^{1}1019^{1}  1  
2039  1  Prime  1  
2040 
32  2^{3}3^{1}5^{1}17^{1}  5  4  10  316  244  2240  1469  80  62 
2041  4  13^{1}157^{1}  
2042 
4  2^{1}1021^{1}  4  
2043  6  3^{2}227^{1}  
2044 
12  2^{2}7^{1}73^{1}  3  
2045  4  5^{1}409^{1}  3  
2046 
16  2^{1}3^{1}11^{1}31^{1}  1  1  42  25  236  127  16  10  
2047  4  23^{1}89^{1}  
2048 
12  2^{11}  6  1  66  50  660  412  12  10  
2049  4  3^{1}683^{1}  1  
2050 
12  2^{1}5^{2}41^{1}  3  22  11  124  62  16  8  
2051  4  7^{1}293^{1}  1  
2052 
24  2^{2}3^{3}19^{1}  2  4  7  228  168  1524  970  72  52 
2053  1  Prime  
2054 
8  2^{1}13^{1}79^{1}  4  
2055  8  3^{1}5^{1}137^{1}  2  
2056 
8  2^{3}257^{1}  4  
2057  6  11^{2}17^{1}  1  2  2  12  9  36  24  4 
3 
2058 
16  2^{1}3^{1}7^{3}  1  2  78  57  480  294  28  22  
2059  4  29^{1}71^{1}  
2060 
12  2^{2}5^{1}103^{1}  6  
2061  6  3^{2}229^{1}  
2062 
4  2^{1}1031^{1}  
2063 
1  Prime  3  
2064  20  2^{4}3^{1}43^{1}  2  68  36  500  256  28 
16 

2065 
8  5^{1}7^{1}59^{1}  1  12  6  52  26  12  6  
2066  4  2^{1}1033^{1}  3  
2067 
8  3^{1}13^{1}53^{1}  4  4  28  20  8  6  
2068  12  2^{2}11^{1}47^{1}  1  22  13  124  68  16 
10 

2069 
1  Prime  1  
2070  24  2^{1}3^{2}5^{1}23^{1}  3  4  9  142  124  832  593  40 
36 
2071 
4  19^{1}109^{1}  
2072  16  2^{3}7^{1}37^{1}  6  34  20  236  130  20 
12 

2073 
4  3^{1}691^{1}  
2074  8  2^{1}17^{1}61^{1}  
2075 
6  5^{2}83^{1}  2  
2076  12  2^{2}3^{1}173^{1}  1  
2077 
4  31^{1}67^{1}  2  
2078  4  2^{1}1039^{1}  2  
2079 
16  3^{3}7^{1}11^{1}  8  6  130  76  616  338  28  18  
2080  24  2^{5}5^{1}13^{1}  4  4  8  256  192  1820  1171  60 
48 
2081 
1  Prime  2  
2082  8  2^{1}3^{1}347^{1}  
2083 
1  Prime  
2084  6  2^{2}521^{1}  5  
2085 
8  3^{1}5^{1}139^{1}  1  
2086  8  2^{1}7^{1}149^{1}  
2087 
1  Prime  1  
2088  24  2^{3}3^{2}29^{1}  3  8  5  226  121  1384  716  56 
32 
2089 
1  Prime  
2090  16  2^{1}5^{1}11^{1}19^{1}  5  4  5  72  58  328  224  28 
22 
2091 
8  3^{1}17^{1}41^{1}  1  4  4  28  20  8  6  
2092  6  2^{2}523^{1}  1  
2093 
8  7^{1}13^{1}23^{1}  1  2  8  12  36  36  4  6  
2094  8  2^{1}3^{1}349^{1}  
2095 
4  5^{1}419^{1}  1  
2096  10  2^{4}131^{1}  4  
2097 
6  3^{2}233^{1}  
2098  4  2^{1}1049^{1}  1  
2099 
1  Prime  2  
2100  36  2^{2}3^{1}5^{2}7^{1}  10  16  590  397  4056  2391  112 
85 

21^{st} Century event totals  213  80  121  3808  2644  25848  15672  994  724 
Notes for the table above:
Teachers may use these sort of dates and times as an extension activity for students, to get them thinking about factors – prime or otherwise
How might this happen? Surely things average out in the end? Well, no.
The EarthMoonSun system has three basic parameters used by civilisations in the past few thousand years to create their calendar:
Here I’m (very parochially) looking only at the Western calendars of the last 2000 years. (If you want to be less parochial, you might start with History of Calendars.)
The Julian calendar was instituted by Julius Caesar in 46BC. It had 365 days in each year, except for leap years, which had 366. This assumes a year length of exactly 365.25 days – pretty good, but about 11 minutes and 14 seconds too long. This meant that, by the 16th century, the year (and its seasons and so on) was about 12 days behind what the seasons would say.
So, in 1562, Pope Gregory introduced an adjustment to the Julian calendar, making years divisible by 100 but not by 400 to be nonleap years. This assumes a year length of 365.2425 days. The difference between the assumed and actual year length is then 26 seconds, and so will still accumulate, but only by one day in 3323 years. I think we can fairly leave it to our 54th century descendants to figure out what to do then.
Now, what about Friday the 13^{th}?
In the Gregorian calendar, there are 146097 days, which is exactly 20871 weeks. This means that each 400 years, New Years Day is the same day of the week. For example, in 2015, New Years Day was a Thursday, as will New Years Day be in 2415.
It also means that other days, like the 13^{th} of a month, are similarly unbalanced. In fact, the number of Friday the 13^{th}s per day of the week in each 400 years of the Gregorian calendar is as follows:
Day of week  13th of month 
Sunday  687 
Monday  685 
Tuesday  685 
Wednesday  687 
Thursday  684 
Friday  688 
Saturday  684 
So there you go, more bad things do happen on Friday the 13^{th} – because there are more of them!
Note: My apologies to Spanish or Italian readers, for whom I understand the equivalent culturally unlucky dates are Tuesday 13^{th} and Friday 17^{th} respectively.
About a year ago, the excellent XKCD site had a post with movie narrative charts. They show the power of graphical display to convey information in a way not possible with words. Visual representation of film character groupings can be understood much more naturally. Symbols are used for particular event or action types.
These charts, particularly for The Lord Of The Rings, inspired me to try a similar thing for the ‘time turner’ part of Harry Potter and the Prisoner of Azkaban.
I tried a couple of programs already on my PC, and ended up using the Microsoft Word drawing package, mainly because (a) it preserved lines and other chart objects as objects and (b) I more or less knew how to use it already. I have no idea how Randall Munroe did his XKCD charts freehand i.e. with no Undo button and no block move of sections of the chart as one part or another needed more space.
The PDF version of the chart is here. It contains one page with a version with explanatory notes and another version on another page without the notes. Use a program such as Adobe Reader if you want to print either or both pages as a poster. (There’s too much detail to display readably in less than A3 size at a minimum.)
The chart looks like this…
(click on the image to download the PDF version).
I learned a number of things creating this chart…
The film runs more like a stage play, having a succession of scenes with very localised action. Lord of the Rings orLawrence of Arabia it is not. One of those things I knew but had not realised, if you know what I mean.
Hermione is clearly the main character in this part of the film, and – I think – the film as a whole. She is the main action person, she has many of the plot defining lines. Maybe the film should have been called Hermione Grainger and the Prisoner of Azkaban, except for reasons that Jano Caro reflects on in an excellent article (from The Age) here.
I started out with the chart more or less just showing the effects of the time turners (Hermione and Harry) on the others (including their earlier selves). Only later did I add the quotes, saved character instances and wand usage. I think the chart is now just about as dense as it can reasonably be (if not already over the top), so it’s time to hand it over to others to look at.
I am definitely interested in feedback on this, including suggested improvements.
P.S. This is a copy of a post on my other blog https://idliketoknowmoreaboutit.wordpress.com/. I was in two minds about which blog to use, so now it’s on both.
You get a prize if you are the the first person in the line (from the front person, working backwards) to have a person ahead of them with the same birthday.
Where to stand to have the best chance of getting the prize?
This is the easiest of the three birthday problems, although I didn’t think so at the start.
To be more explicit in the problem definition, we say that, if you are the n^{th} person in the line, then
(a) the n1 people in front of you must all have different birth dates
(b) you must have the same birthday as one of those (n1) people
So, the trick is to find where the decreasing probability of (a) outweighs the increasing probability of (b).
Let’s say you are indeed the n^{th} person in the line.
Let’s also says that the probability that the (n1) people in front of you all have different birth dates is p.
(It’s actually some mildly complicated expression we figured out in Birthday Pairings.)
Then the probability of you being the winner is (assuming a simple 365 day year)
(n1) p ===== (1) 365
Now imagine you move back one space, to the (n+1)^{th} position. That extra person
– must have a different birth date to all those in front, but also
– provides an extra birth date for you to match with.
So now, the probability of you as the person being the winner is
(366n) n p ======= === (2) 365 365
So, the best position will be the smallest n for which expression (1) is greater than expression (2) i.e.
(n1) (366n) n p ===== > p ======= === 365 365 365
Cancelling common factors, then moving 365 to the left hand side, we find that
365 (n1) > (366n) n
so
365n – 365 > 366n – n^{2
}so
n^{2} > n + 365
This is first true for n = 20, so stand 20^{th} in line.
For someone with their birthday on 29 February, the equation is (approximately)
(n1) (1462n) n p ===== > p ======= === 1461 1461 1461
Simplifying this expression in the same way as for other dates above, we find that
1461 (n1) > (1462n) n
so
1461n – 1461 > 1462n – n^{2
}so
n^{2} > n + 1461
This is first true for n = 39, so stand 39^{th} in line.
Notes
I once used some of these and other mazes I’d made to form the complete text of a birthday card. Use them to make your own birthday and Christmas cards!
Use a name above, and then ‘s and room mazes to make a bedroom door sign.
Thanks for the idea Cindy!
America 
Best Teacher (well, they should get some recognition too!) 
DANCE 
Get well soon 
Happy Birthday 
Happy Birthday (italics) 
Happy new year (300 x 200) 
Merry Christmas (300 x 200) 
Random 
so have a good year (200 x 125) 
Most of these animal shape mazes have colouring to match the animal.
Bat  Dolphin  Frog  Lion 
Mouse  Rabbit  Rat  Tortoise 
Buildings (with building colour pattern) 
A simple maze (30 x 20), but also presented as wavy, circularly wavy, circular, wiggly, zigzag and centre zigzag (which is trickiest?) 
Guitar (very tricky to get the shapes looking right for this and the Harley) 
Harley 
Heart (for Valentine’s day) 
Jigsaw puzzle piece 
Jigsaw puzzle piece (inverted colouring) 
Maze within a maze (150 x 100) 
Maze within a maze (300 x 200) 
Pyramid #1 
Pyramid #2 
Skateboard (starts and ends on the wheels) 
Wheel 
These colour patterns were available. Parameters such as number of cycles or lines, lines thickness or angle and colour boundary could be configured. I could also use any image (preferably a simple one with bright strong colouring) as a colour template – see the animal shapes above.
Circle  Lines – wavy  Spiral – linear  
Explosion  Lines – zigzag  Star  
Flower  Patchwork  Sunrise  
Hypercircle 1  Random areas  Weave  
Hypercircle 2  Spiral – exp  Random per cell  
Lines – straight  Spiral – log  Single colour 
Have fun with them! Tell me what you do with them.
All text and images are copyright © Mark Michell 20012003, but may be used freely for nonprofit purposes. Write to me about any other uses of these mazes. It would be nice of you to say where you got the images if you do use them. I have deliberately not spoiled the images themselves with such notices.
Albert and Bernard just became friends with Cheryl, and they want to know when her birthday is. Cheryl gives them a list of 10 possible dates:
May 15  May 16  May 19  
June 17  June 18  
July 14  July 16  
August 14  August 15  August 17 
Cheryl then tells Albert and Bernard separately the month and the day of her birthday respectively, to which Albert and Bernard respond by saying:
Albert: I don’t know when Cheryl’s birthday is, but I know that Bernard does not know too.
Bernard: At first I didn’t know when Cheryl’s birthday is, but now I do know.
Albert: Then I also know when Cheryl’s birthday is.
So when is Cheryl’s birthday?
The key is to be clear at all times about what Albert and Bernard each know, and what we know as a result of hearing what they say.
Initially:
– Albert knows that the month of Cheryl’s birthday is either May, June, July or August.
– Bernard knows that the day of Cheryl’s birthday is either 14, 15, 16, 17, 18 or 19.
– each knows what the other knows.
So now our table of possible dates looks like this…
May 15  May 16  May 19  
June 17  June 18  
July 14  July 16  
August 14  August 15  August 17 
So now our table of possible dates looks like this…
May 15  May 16  May 19  
June 17  June 18  
July 14  July 16  
August 14  August 15  August 17 
So now our table of possible dates looks like this…
May 15  May 16  May 19  
June 17  June 18  
July 14^{th}  July 16  
August 14  August 15  August 17 
So Cheryl’s birthday is on July 16.
I have made a list of every possible Target puzzle. There are over 2500 Targets with word counts of 100 or more (not including obscure words). Why don’t we see Targets like these, especially on weekends? Contact your newspaper and tell them what you want!
For those new to these puzzles, each Target places letters into a 3×3 grid and asks the reader for a list of words made from those letters where
For example, for the letters
I  L  C 
O  Y  R 
S  U  U 
an ‘excellent’ level list of nonobscure words (i.e. those in SIL – see below) would include
cloy, cosily, cosy, CURIOUSLY, curly, lousy, lyric, oily, rosily, rosy, sourly, surly, usury, your, yours
but exclude plurals (or similar – see below) such as
coils, curs, curios, lyrics, roils
If we allow nonobscure and obscure words (from SOWPODS – see below), then we would also add
coly, cory, corylus, crusily, crusy, lory, loury, ricy, roily, ruly, scry, sluicy, soily, syli
My mother (Hi Mum!) is a very keen Target puzzler, frequently getting more than the (newspaper defined) ‘excellent’ number of words. She often complains that there only 20 or 30 words to find. Others (Hi Peter!) are quite happy with, say, 15 as the ultimate target. I think it depends on the optimum length of time the puzzler wants to spend on the puzzle.
So I started wondering how many (or how few) words there could be in one Target. Of course, it all depends on which words are in our large word list.
I have based all the following data on
(There are other lists, such as the Enhanced North American Benchmark Lexicon, a public domain list used in many games, such as Words With Friends. In size, this falls between the two lists above. There is also the extremely good Wordsmith anagram page, whose word counts seem to be similar to (i.e. within 5%10% of) those (The Age) ‘excellent’ targets.
Looking only at Target words (those with four to nine letters), SOWPODS has 153886 and SIL has 106346.
As mentioned above, the Target word lists (unfortunately for our purposes) exclude plurals ending in ‘S’.
Excluding aardvarks and acres is simple enough.
On the other hand, Targets often include words such as alas (which is also the plural of ala) but exclude taxis (which is also a singular noun) – perhaps because the plural alas and the singular taxis are both obscure. And what to do with series, which is both singular and plural? In general, it seems that if a word can be plural (i.e. has a singular form within a given word set), then it’s excluded.
Moreover, they also seem to exclude words such as abdicates and abates, which are not plurals, but the third person singular forms of the corresponding verbs.
In any case, as far as trying to accurately distinguish words which should be excluded from those which should not, well, it turned out to be quite a lot of work.
So, in all that follows, I estimate that I have ‘wrongly’ categorised perhaps 50 words (in or out of the list). Still, I am using my best guess as to the list of allowable words, and will not keep saying that I am doing so.
Using these quirky rules, I arrived at the following word counts:
Letters 
Full lists 
Used lists 

SOWPODS 
SIL 
SOWPODS 
SIL 

4 
5454 
3130 
4461 
2354 
5 
12478 
6919 
8873 
4623 
6 
22157 
11492 
16205 
8236 
7 
32909 
16882 
22580 
11562 
8 
40161 
19461 
27381 
13226 
9 
40727 
16693 
26846 
10291 
Total 
153886 
74577 
106346 
50292 
In both lists, roughly one in every three words is a plural ending in ‘S’.
Each distinct letter in each nine letter word makes a distinct Target. For SOWPODS, the total number of Targets is 187749, an average of 6.99 for each distinct set of nine letters; the corresponding figures for SIL are 71262 and 6.92.
Some summary statistics of word counts:
SOWPODS  SIL  
Maximum  515  284 
Average  90  51 
Median  77  43 
Mode  54  25 
Minimum  1  1 
A comparison of word counts for particular targets suggests that the top (‘excellent’) targets in the newspapers I read tend to be about 40%45% of those from the SOWPODS word list.
I know this post is stretching the ‘maths’ theme of this blog, so here is a sop to that.
(You can skip this bit if you like, unless you are indeed a SSMS, in which case read on…)
More detailed statistical measures of word counts are as follows:
SOWPODS  SIL  
Maximum  515  284 
Minimum  1  1 
Range  514  283 
Average  90  51 
St Dev  58  33 
Q1  48  27 
Median  77  43 
Q3  118  67 
IQR=Q3Q1  70  40 
Notes:
1. Skewness
The distribution of word counts is positively skewed. As would be expected for a positively skewed distribution,
Mode < Median < Mean.
2. Outliers
Of the 187749 SOWPODS Targets, 6073 or 3,2% have word counts greater than 223=118+1.5×70. These word counts are – formally speaking – outliers, having counts above the upper fence. As the low fence is less than zero, there are no low outliers. That said, we would not normally use the notion of outliers with such large sample sizes, at least not in terms of box plot displays of outliers.
What follows is a selection of Targets which are extreme in some sense, having the most or least of a particular property.
Beneath each Target are specified
central letter : SIL word count / SOWPODS word count
if the letters are rearranged to allow each indicated central letter. SIL counts are not provided if there is no SIL nine letter word for those letters.
The following are the Targets with some of the largest SOWPODS word counts.
E: – / 515 
E: 260 / 493 
E: – / 481 

E: – / 469 
E: – / 467 
E: – / 467 
See below for solutions. One of the nine letter words is in SIL, though I think it’s a less common word than two or three of the other words here.
There are 7 Targets whose only SOWPODS word is the nine letter word:
X: – / 1 
X: – / 1 
V: – / 1 

X: – / 1 
Q: – / 1 
Q: – / 1 

X: 1 / 1 
See below for solutions. None except the last are common words (i.e. in SIL).
There are also 15 SIL Targets which have only the nine letter word in the list:
X: 1 / 1 
J: 1 / 2 
C: 1 / 2 

Z: 1 / 3 
Z: 1 / 3 
Q: 1 / 3 

Q: 1 / 3 
Z: 1 / 3 
Z: 1 / 3 

U: 1 / 4 
M: 1 / 4 
Z: 1 / 5 

Z: 1 / 5 
Z: 1 / 6 

In my humble opinion, some of the nine letter words above are rather obscure – but not according to SIL, so there they are.
The following are the Targets with four (the maximum) or three 9 letter words in SIL.
Beneath each Target, the numbers in parentheses are respectively the SIL and SOWPODS counts for nine letter words.
Note: In each case, the selected central letter gives the highest total word counts for those nine letters (as shown below). The central letter is not otherwise relevant for this category.
153 / 250 (4 / 4) 
110 / 210 (3 / 5) 
135 / 232 (3 / 4) 

164 / 375 (3 / 4) 
73 / 120 (3 / 3) 
109 / 211 (3 / 3) 

83 / 133 (3 / 3) 
75 / 142 (3 / 3) 
151 / 274 (3 / 3) 

162 / 275 (3 / 3) 
116 / 236 (3 / 3) 
77 / 138 (3 / 3) 

155 / 257 (3 / 3) 
54 / 100 (3 / 3) 
68 / 118 (3 / 3) 
This is a very restrictive condition, so the word counts are quite low.
Each Target below has exception counts in parentheses, i.e. the SIL and SOWPODS word counts for words beginning with letters other than the central letter.
SIL top 3:
25 / 39 (0 / 2) 
19 / 25 (0 / 2) 
16 / 26 (0 / 4) 
SOWPODS top 3:
– / 26 ( / 0) 
– / 22 (0 / 0) 
– / 18 ( / 0) 
Note that the word counts for SIL and SOWPODS are not that different.
This is the opposite to the previous category. With up to eight choices for the initial letter in any given Target, the word counts are also much higher.
Each Target below also has exception counts in parentheses, i.e. the SIL and SOWPODS word counts for words which do begin with the central letter.
SIL top 3:
133 / 223 (0 / 3) 
129 / 186 (0 / 3) 
124 / 180 (0 / 2) 
Interestingly, 37 of the top 38 SIL entries here have ‘E’ as the central letter.
SOWPODS top 3:
96 / 171 (0 / 0) 
89 / 156 (0 / 0) 
86 / 149 (0 / 0) 
This takes the view from the other end. That is, which words appear in most Targets? Note that SIL words are in bold italics.
Word(s) 
SIL 
SOWPODS 
rein, rine 
8869 
19961 
rite, tier, tire, trie 
8052 
19625 
nite, tine 
8423 
18734 
aine, eina 
– 
18386 
aret, rate, tare, tear 
7268 
18249 
site, stie 
7926 
18222 
rise, seir, sire 
7178 
17161 
sien, sine 
6699 
17111 
gien 
– 
16810 
ilea 
– 
16102 
Conversely, some words are not found in any Targets.
Word length 
SIL  SOWPODS  
Words in no Targets 
Total 
% 
Words in no Targets 
Total 
% 

4 
15 
4459 
0.30% 
15 
4459 
0.30% 
5 
120 
8953 
1.30% 
120 
8953 
1.30% 
6 
717 
16529 
4.50% 
717 
16529 
4.50% 
7 
3198 
22538 
14.20% 
3198 
22538 
14.20% 
8 
12121 
27328 
44.30% 
12121 
27328 
44.30% 
The four letter words not found in any Targets are:
SIL
buzz, foxy, hadj, hajj, jamb, java, jeez, jiff, jinx, john, jowl, juju, qoph, wavy, waxy
SOWPODS
foxy, fozy, hajj, jaxy, jazy, jeff, jiff, juju, juku, jynx, waxy, yuky, yuzu, ziff, zizz
Note that each of the words in the SOWPODS list contains at least one of ‘J’, ‘Y’ or ‘Z’ (so jazy had no chance :))
The word angriest and its anagrams (rangiest and the SOWPODSonly angstier, astringe, ganister, reasting, stearing, and tasering) are remarkable in at least two ways.
Firstly, there are more anagrams (excluding plurals ending in s) for this eight letter group than any other.
Secondly, the letters of these words make at least one nine letter word with the addition of each of 9 (SIL) /18 (SOWPODS) different letters, for a total of 11/29 nine letter words:
Added letter  Words  
SIL  SOWPODS  
B  BREASTING  
C  RECASTING  
D  ASTRINGED  
E  RESEATING STINGAREE 

G  RESTAGING  
I  GRAINIEST GRANITISE 
SERIATING 
K  STREAKING  
L  GNARLIEST RESLATING 

M  MASTERING STREAMING 

N  GANNISTER  
O  ORANGIEST  
P  REPASTING TRAPESING 

R  ARRESTING  ASTRINGER RASTERING SERRATING 
S  ASSERTING  
T  RESTATING  ASTERTING RETASTING 
U  SIGNATURE  
W  WASTERING  
Y  ESTRAYING  
Total  11  18 
As above, SIL words are in bold italics.
STRAPLINE 
GNARLIEST 
TRIANGLED 

LATERISED 
MISLEARNT 
RESINATED 
SIL one word Targets
STRONGBOX 
JUDGMATIC 
WITTICISM 

HYBRIDIZE 
MERCERIZE 
QUERULOUS 

QUIBBLING 
SPOROZOON 
VICTIMIZE 

CHIHUHUA 
KIBBUTZIM 
PELLETIZE 

PROTOZOON  GHETTOIZE  VIBRAHARP 
SOWPODS one word Targets
EXSUCCOUS  GOGGLEBOX  IMPLUVIUM  
LIXIVIOUS  MONOCOQUE  SILIQUOUS  
STRONGBOX 
CRATERING 
ASPERSING 
EARTHIEST 

EARTHLING HALTERING HEARTLING LATHERING 
ALLOTYPIC 
AUCTIONED 

BURBLIEST 
CATTINESS 
CONTAINER CRENATION NARCOTINE 

CORSETING 
ENDEARING ENGRAINED GRENADINE 
ENLISTING LISTENING TINSELING 

LUCRATIVE 
MONOGRAPH NOMOGRAPH GRAMOPHON 
REPRISING RESPIRING SPRINGIER 
SIL top 3:
COMPELLER 
FLUMMOXED 
HOPEFULLY 
SOWPODS top 3:
COMPLEXLY 
JOLTINGLY 
CROWBERRY 
VAPORISED 
RESHAPING 
REDEPOSIT 
and
OUTFABLED 
AMBULATED 
ADOPTABLE 
Although some students are already getting started…
(The calm before the storm.)
I’m not quite sure what it’s going to be like, but it looks even bigger than I imagined.
I don’t actually know what I intend to do after school (except maybe a gap year).
Actually, why am I doing year 12 anyway?
(That’s OK. I tell students, only halfjokingly, that I still don’t know what I’m going to do when I grow up.)
How do teachers expect us to do homework as well as revise for SACs?
There’s just too much to do.
(Yep. Too much work is still not enough at times. Just do what you can.)
Now I know enough about the subjects to know how big it’s going to be and there’s just so much work, so much to get through.
It just looks too big for me live up to my own and others’ expectations.
And I still don’t know why I’m doing it.
(And yet most students push into it with a good deal of effort and persistence, even with subjects they’re not enjoying much. I, for one, am inspired by their application to the task.)
SACs, homework, SACs, homework…
Will it ever end?
(Head down, bum up. Just remember that it will end in less than 6 months.)
More SACs, the GAT, then at least the midyear break.
(The last time for a break for a few days, at least until the exams are over, so it is important to get some rest now.)
We’re nearing the end of coursework, but there’s still a lot to do in such a short time. It feels so rushed.
There’s still such a lot to do, so much I don’t really get yet.
I’ve done some test exams – a sobering experience. The questions are so much harder than those in the textbooks.
It just shows me how far there is to go. Time is running short.
(Just remember – the exams are in November, not now, so you will be much more prepared for them by then – no, really.)
Counting the days now. I still don’t feel prepared.
(Yes, you can count the days. They’re going to zoom past very quickly now.)
We’re finished all the coursework. No time left. Just revising and revising now.
(Revising will make it all start to come together. Things you haven’t thought about for weeks, months even, will just make more sense than they did before.)
Aaagh! My head is exploding.
(As that wise sage Dory once said, “Just keep swimming. Just keep swimming.”)
Will I be able to answer the questions? How will I go?
(1. The aim in doing the exam isn’t to get 100%. You’re there to show them what you can do. This may mean getting 90%, 75% or 60%. All your marks contribute to the end result.
2. Expect to leave some questions unanswered, due to the difficulty of the question, or just time pressure. Remember that the examiners have to distinguish between those getting 95% and those getting 100%, so the hardest questions will be answered correctly by only 5% of those doing the exam.)
Finished! More relief than joy.
Well, maybe not quite the ATAR I wanted, but I’ll do something with it.
But I still don’t know what I want to do.
(1. Even if you don’t have a 5 year plan, when making final course selections, follow the subjects you like. You’re more likely to end up doing things you want to do.
2. And lastly, just remember that in 2 years from now, probably less, no one will care what your ATAR was. It will have been a key for you to open some doors. And if the doors you were aiming for remain shut, you may be surprised and even delighted by what other opportunities arise.)