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:

21×13×13×6×3×5×4×8×1

67×66×65×74×73×72×71×70×69

So the probability of getting YODELLING is:

6×360×21×13×13×6×3×5×4×8×1	= 0.000000575874154
67×66×65×74×73×72×71×70×69

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:

Consonants	Vowels	Probability of nine letter word
0	9	0
1	8	0
2	7	0
3	6	0.000546
4	5	0.019724
5	4	0.076895
6	3	0.051417
7	2	0.005662
8	1	0.000033
9	0	0

So the best way to get a nine letter word in the first letters game is to pick five consonants and four vowels.

Sunday Afternoon Maths VII

Here's this week's collection. Answers & extensions here. Why not discuss the problems on Twitter using #SundayAfternoonMaths or on Reddit.

Ninety Nine

Source: UKMT Senior Maths Challenge 2013

In a ‘ninety nine’ shop, all items cost a number of pounds and 99 pence. Susanna spent £65.76. How many items did she buy?

Reverse Bases

Find two digits a and b such that ab in base 10 is equal to ba in base 4?

Find two digits c and d such that cd in base 10 is equal to dc in base 7?

Find two digits e and f such that ef in base 9 is equal to fe in base 5?

Sunday Afternoon Maths VI Answers & Extensions

This post contains the answers to this week's Sunday Afternoon Maths and some extension problems based around the originals.

Ellipses

The area of an ellipse is πab where a and b are the distances from the centre of the ellipse to the closest and furthest points on the ellipse.

In the first ellipse, a=5cm and b=4cm, so the area is 20πcm². In the second ellipse, a=5cm and b=3cm, so the area is 15πcm². Hence, the first ellipse has the larger area.

Extension

How far apart should the pins be placed to give the ellipse with the largest area?

Triangle Numbers

T_n = ¹/₂n(n+1), so:
T_n+T_n+1 = ¹/₂n(n+1) + ¹/₂(n+1)(n+2)
    = (n+1)²

So, we are looking for n such that (n+1)²+(n+3)²=(n+5)². This is true when n=5 (6²+8²=10²).

Extension

Find n such that T_n+T_n+1+T_n+1+T_n+2=T_n+2+T_n+3.

Ticking Clock

The second hand will always be pointing at one of the 60 graduations. If the minute and hour hand are 120° away from the second hand they must also be pointing at one of the graduations. The minute hand will only be pointing at a graduation at zero seconds past the minute, so the second hand must be pointing at 0. Therefore the hand are either pointing at: hour: 4, minute: 8, second: 0; or hour: 8, minute: 4, second: 0. Neither of these are real times, so it is not possible.

Extension

If the second hand moves continuously instead of moving every secone, will there be a time when the hands of the clock are all 120° apart?

Sunday Afternoon Maths VI

Here's this week's collection. Answers & extensions tomorrow. Why not discuss the problems on Twitter using #SundayAfternoonMaths.

Ellipses

Source: Maths Medicine

A piece of string 10cm long is tied to two pins 6cm apart. The string is used to draw an ellipse. The pins are then moved 2cm further apart and a second ellipse is drawn. Which ellipse has the larger area?

Triangle Numbers

Source: ATM Mathematics Teaching 239

Let T_n be the n^th triangle number. Find n such that T_n+T_n+1+T_n+2+T_n+3=T_n+4+T_n+5.

Ticking Clock

Is there a time of day when the hands of an analogue clock (one with a second hand that moves every second instead of moving continuously) will all be 120° apart?

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...

Sunday Afternoon Maths V Answers & Extensions

This post contains the answers to this week's Sunday Afternoon Maths and some extension problems based around the originals.

Two Lines

Let A have the equation y = mx + c. B will have the equation y = cx + m.
Therefore, mx + c = cx + m.
Which rearranges to x(m - c) = m - c.
So x = 1.
Substituting back in, we find y=m+c.
The co-ordinates of the point of intersection are (1,m+c).

Extension

Let a, b and c be three distinct numbers. What can you say about the points of intersection of the parabolas:
y = ax² + bx + c,
y = bx² + cx + a,
and y = cx² + ax + b?

Odd Sums

They are all equal to one third.

The sum of the first n odd numbers is n² (this can be proved by induction). This means that (sum of the first n odd numbers) ÷ (sum of the next n odd numbers) is ⁿ²/_(2n)²-n² = ⁿ²/_4n²-n² = ⁿ²/_3n² = ¹/₃

Extension

What is (sum of the first n odd numbers) ÷ (sum of the first n even numbers)?

x^xxxxx... Again

y = x^xxxxx... so y = x^y = e^{y ln x}.
By the chain rule and the product rule, ^dx/_dy = e^{y ln x}(^dx/_dy ln x + ^y/_x).
Rearranging, we get ^dx/_dy = ^{yey ln x}/_{x(1-e^{y ln x} ln x)}.
This simplifies to ^dx/_dy = ^{xxxxxx...xxxxxx...}/_{x(1-x^xxxxx... ln x)}

Extension

What would the graph of y = x^xxxxx... look like?

Folding Tube Maps

Once the map is folded, it will look like this:

For the final tetrahedron to be regular, the red lengths must be equal. Let each red length be 2 (this will get rid of halves in the upcoming calculations). By drawing a vertical line in we can work out the width and height of the rectangle:

The width of the rectangle is 3 (one and a half red lengths). Using Pythagoras' Theorem in the blue triangle, we find that the height of the rectangle is √3. Therefore, the ratio of the rectangle is √3:3 or 1:√3.

Extension

If the ratio of the rectangle is 1:a, what is the ratio of the lengths of the sides of the tetrahedron?

Sunday Afternoon Maths V

Here's this week's collection. Answers & extensions here. Why not discuss the problems on Twitter using #SundayAfternoonMaths.

Two Lines

Let A and B be two straight lines such that the gradient of A is the y-intercept of B and the y-intercept of A is the gradient of B (the gradient and y-intercept of A are not the same). What are the co-ordinates of the point where the lines meet?

Odd Sums

What is (1+3) ÷ (5+7)?

What is (1+3+5) ÷ (7+9+11)?

What is (1+3+5+7) ÷ (9+11+13+15)?

What is (1+3+5+7+9) ÷ (11+13+15+17+19)?

What is (sum of the first n odd numbers) ÷ (sum of the next n odd numbers)?

x^xxxxx... Again

Let y = x^xxxxx... [x to the power of (x to the power of (x to the power of (x to the power of (...)))) with an infinite number of xs]. What is ^dy/_dx?

Folding Tube Maps

Back in 2012, I posted instructions for folding a tetrahedron from tube maps. When tube maps are used, the sides of the tetrahedron are not quite equal. What ratio would the rectangular maps need to be in to give a regular tetrahedron?