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 50mm quilling paper which made their construction much easier. With a piece of thread through the middle, Fröbel starts make brilliant tree decorations.

Pointless Probability

Last week, I was watching Pointless and began wondering how likely it is that a show features four new teams.

On the show, teams are given two chances to get to the final—if they are knocked out before the final round on their first appearance, then they return the following episode. In all the following, I assumed that there was an equal chance of all teams winning.

If there are four new teams on a episode, then one of these will win and not return and the other three will return. Therefore the next episode will have one new team (with probability 1). If there are three new teams on an episode: one of the new teams could win, meaning two teams return and two new teams on the next episode (with probability 3/4); or the returning team could win, meaning that there would only one new team on the next episode. These probabilities, and those for other numbers of teams are shown in the table below:

	No of new teams today
Noof new teams tomorrow		1	2	3	4
	1	0	0	1 4	1
	2	0	1 2	3 4	0
	3	3 4	1 2	0	0
	4	1 4	0	0	0

Call the probability of an episode having one, two, three or four new teams P₁, P₂, P₃ and P₄ respectively. After a few episodes, the following must be satisfied:

P1 =	1	P3 + P4
	4
P2 =	1	P2 +	3	P3
	2		4
P3 =	3	P3 +	1	P4
	4		2
P4 =	1	P1
	4

And the total probability must be one:

P1 + P2 + P3 + P4 = 1

These simultaneous equations can be solved to find that:

P1 =	4
	35
P2 =	18
	35
P3 =	12
	35
P4 =	1
	35

So the probability that all the teams on an episode of Pointless are new is one in 35, meaning that once in every 35 episodes we should expect to see all new teams.

Edit: This blog answered the same question in a slightly different way before I got here.