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 2014-04-11 

Countdown probability, pt. 2

As well as letters games, the contestants on Countdown also take part in numbers games. Six numbers are chosen from the large numbers (25,50,75,100) and small numbers (1-10, two cards for each number) and a total between 101 and 999 (inclusive) is chosen by CECIL. The contestants then use the six numbers, with multiplication, addition, subtraction and division, to get as close to the target number as possible.
The best way to win the numbers game is to get the target exactly. This got me wondering: is there a combination of numbers which allows you to get every total between 101 and 999? And which combination of large and small numbers should be picked to give the highest chance of being able to get the target?
To work this out, I got my computer to go through every possible combination of numbers, trying every combination of operations. (I had to leave this running overnight as there are a lot of combinations!)

Getting every total

There are 61 combinations of numbers which allow every total to be obtained. These include the following (click to see how each total can be made):
By contrast, the following combination allows no totals between 101 and 999 to be reached:
The number of attainable targets for each set of numbers can be found here.

Probability of being able to reach the target

Some combinations of numbers are more likely than others. For example, 1 2 25 50 75 100 is four times as likely as 1 1 25 50 75 100, as (ignoring re-orderings) in the first combination, there are two choices for the 1 tile and 2 tile, but in the second combination there is only one choice for each 1 tile. Different ordering of tiles can be ignored as each combination with the same number of large tiles will have the same number of orderings.
By taking into account the relative probability of each combination, the following probabilities can be found:
Number of large numbersProbability of being able to reach target
00.964463439
10.983830962
20.993277819
30.985770510
40.859709475
So, in order to maximise the probability of being able to reach the target, two large numbers should be chosen.
However, as this will mean that your opponent will also be able to reach the target, a better strategy might be to pick no large numbers or four large numbers and get closer to the target than your opponent, especially if you have practised pulling off answers like this.
Edit: Numbers corrected.
Edit: The code used to calculate the numbers in this post can now be found here.

Similar posts

Countdown probability
Pointless probability
A bad Puzzle for Today
A 20,000-to-1 baby?

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 2016-07-20 
I've pushed a version of the code to https://github.com/mscroggs/countdown-numbers-game
Matthew
 2016-07-20 
Sadly, I lost the code I used when I had laptop problems. However, I can remember what it did, so I shall recreate it and put it on GitHub.
Matthew
 2016-07-20 
If you could, I'd love to have the code you used to do this exhaustive search?

I'm a fan of the game myself (but then I'm French, so to me it's the original, "Des chiffres et des lettres"), but for the numbers game, this is pretty much irrelevant to the language and country :)
Francis Galiegue
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 2014-04-06 

Countdown probability

On Countdown, contestants have to make words from nine letters. The contestants take turns to choose how many vowels and consonants they would like. This got me wondering which was the best combination to pick in order to get a nine letter word.
Assuming the letters in countdown are still distributed like this, the probability of getting combinations of letters can be calculated. As the probability throughout the game is dependent on which letters have been picked, I have worked out the probability of getting a nine letter word on the first letters game.

The probability of YODELLING

YODELLING has three vowels and six consonants. There are 6 (3!) ways in which the vowels could be ordered and 720 (6!) ways in which the consonants can be ordered, although each is repeated at there are two Ls, so there are 360 distinct ways to order the consonants. The probability of each of these is:
$$\frac{21\times 13\times 13\times 6\times 3\times 5\times 4\times 8\times 1}{67\times 66\times 65\times 74\times 73\times 72\times 71\times 70\times 69}$$
So the probability of getting YODELLING is:
$$\frac{6\times 360\times 21\times 13\times 13\times 6\times 3\times 5\times 4\times 8\times 1}{67\times 66\times 65\times 74\times 73\times 72\times 71\times 70\times 69} = 0.000000575874154$$

The probability of any nine letter word

I got my computer to find the probability of every nine letter word and found the following probabilities:
ConsonantsVowelsProbability of nine letter word
090
180
270
360.000546
450.019724
540.076895
630.051417
720.005662
810.000033
900
So the best way to get a nine letter word in the first letters game is to pick five consonants and four vowels.

Similar posts

Countdown probability, pt. 2
Pointless probability
A bad Puzzle for Today
A 20,000-to-1 baby?

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 2014-03-29 

New machine unfriendly £1 coin, pt. 2

Following my last post, I wrote to my MP (click to enlarge):
Today I received this reply (click to enlarge):
I'm excited about hearing what the Treasury has to say about it...

Similar posts

The end of coins of constant width
New machine unfriendly £1 coin
A bad Puzzle for Today
A 20,000-to-1 baby?

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 2014-03-19 

New machine unfriendly £1 coin

Vending machines identify coins by measuring their width. Circular coins have the same width in every direction, so designers of vending machines do not need to worry about incorrectly rotated coins causing a blockage or being misidentified. But what about seven-sided 20p and 50p coins?
Perhaps surprisingly, 20p and 50p coins also have a constant width, as show by this video. In fact, the sides of any regular shape with an odd number of sides can be curved to give the shape a constant width.
3, 5, 7 and 9 sided shapes of constant width.
Today, a new 12-sided £1 coin was unveiled. One reason for the number of sides was to make the coin easily identified by touch. However, as only polygons with an odd number of sides can be made into shapes of constant width, this new coin will have a different width when measured corner to corner or side to side. This could lead to vending machines not recognising coins unless a new mechanism is added to correctly align the coin before measuring.
Perhaps an 11-sided or 13-sided design would be a better idea, as this would be easily distinguishable from other coins by touch which being a constant width to allow machines to identify it.

Similar posts

The end of coins of constant width
New machine unfriendly £1 coin, pt. 2
A bad Puzzle for Today
A 20,000-to-1 baby?

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 2013-12-23 

Assorted christmaths

Here is a collection of Christmas relates mathematical activities.

Flexagons

I first encountered flexagons sometime around October 2012. Soon after, we used this template to make them at school with year 11 classes who had just taken GCSE papers as a fun but mathematical activity. The students loved them. This lead me to adapt the template for Christmas:
And here is an uncoloured version of the template on that site if you'd like to colour it yourself and a blank one if you'd like to make your own patterns:
The excitement of flexagons does not end there. There are templates around for six faced flexagons and while writing this piece, I found this page with templates for a great number of flexagons. In addition, there is a fantastic article by Martin Gardner and a two part video by Vi Hart.

Fröbel stars

I discovered the Fröbel star while searching for a picture to be the Wikipedia Maths Portal picture of the month for December 2013. I quickly found these very good instructions for making the star, although it proved very fiddly to make with paper I had cut myself. I bought some 5mm quilling paper which made their construction much easier. With a piece of thread through the middle, Fröbel starts make brilliant tree decorations.

Similar posts

Design your own flexagon
Electromagnetic Field talk
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Christmas card 2017

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