mscroggs.co.uk
mscroggs.co.uk

subscribe

Blog

 2013-12-15 
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
  1234
100\(\frac{1}{4}\)1
20\(\frac{1}{2}\)\(\frac{3}{4}\)0
3\(\frac{3}{4}\)\(\frac{1}{2}\)00
4\(\frac{1}{4}\)000
Call the probability of an episode having one, two, three or four new teams \(P_1\), \(P_2\), \(P_3\) and \(P_4\) respectively. After a few episodes, the following must be satisfied:
$$P_1=\frac{1}{4}P_3+P_4$$ $$P_2=\frac{1}{2}P_2+\frac{3}{4}P_3$$ $$P_3=\frac{3}{4}P_3+\frac{1}{2}P_4$$ $$P_4=\frac{1}{4}P_1$$
And the total probability must be one:
$$P_1+P_2+P_3+P_4=1$$
These simultaneous equations can be solved to find that:
$$P_1=\frac{4}{35}$$ $$P_2=\frac{18}{35}$$ $$P_3=\frac{12}{35}$$ $$P_4=\frac{1}{35}$$
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.
                        
(Click on one of these icons to react to this blog post)

You might also enjoy...

Comments

Comments in green were written by me. Comments in blue were not written by me.
 Add a Comment 


I will only use your email address to reply to your comment (if a reply is needed).

Allowed HTML tags: <br> <a> <small> <b> <i> <s> <sup> <sub> <u> <spoiler> <ul> <ol> <li> <logo>
To prove you are not a spam bot, please type "y-axis" in the box below (case sensitive):
 2013-07-24 
A news story on the BBC Website caught my eye this morning. It reported the following "uncanny coincidence" between a Northern Irish baby and a Royal baby:
But both new mothers share the name Catherine, the same birthday - 9 January - and now their sons also share the same birth date.
I decided to work out just how uncanny this is.
The Office for National Statistics states that 729,674 babies are born every year in the UK. This works out at 1,999 babies born each day, assuming that births are uniformly distributed, so there will be approximately 1,998 babies who share Price Nameless's birthday.
So, what is the chance of the mother of one of these babies having the same birthday as Princess Kate? To work this out I used a method similar to that which is used in the birthday "paradox", which tells us that in a group of 23 people there is a more than 50% chance of two people sharing a birthday, but that's another story.
First, we look at one of our 1,998 mothers. The chance that she shares Princess Kate's birthday is 1/365 (ignoring leap days). The chance that she does not share Princess Kate's Birthday is 364/365.
Next we work out the probability that none of our 1,998 mothers shares Princess Kate's birthday. As our mothers' birthdays are independent we can multiply the probabilities together to do this (this is why we are looking at the probability of not sharing a birthday instead of sharing a birthday). Our probability therefore is \(\left(\frac{364}{365}\right)^{1998} = 0.00416314317\).
Back to the original question, we wanted to know the probability that one of our mothers shares Princess Kate's birthday. To calculate this we do take 0.00416314317 away from 1. This gives 0.99583685682 or 99.6%.
There is a 99.6% chance that there is a resident of the UK who shares the same birthday as Princess Kate and had a child on the same day.
Uncanny.
But let's be fair. The mother in our story is also called Kate. So what are the chances of that? In fact, the same method can be followed, working with the probability of having neither the same birthday or name as Princess Kate.
I think it is safe to assume that this would still be considered news-worthy if our non-princess was called Katie, Cate, Cathryn, Katie-Rose or any other name which is commonly shortened to Kate, so I included a number of variations and used this fantastic tool to find the probability of a mother being called Kate. The data only goes back to 1996, but as the name is dropping in popularity, we can assume that before 1996 at least 1.5% of babies were called Kate. Disregarding males, we can estimate that 3% of mothers are called Kate.
If anyone would like the details of the rest of the calculation, please comment on this post and I will include it here. For anyone who trusts me and isn't curious, I eventually found that the probability of none of our 1,998 mothers share the same name and birthday as Princess Kate is 0.84855028964. So the probability of another Kate having a child on the same day and sharing Princess Kate's birthday is 0.15144971035 or 15.1%. Just over one in seven.
So this is as uncanny as anything else which has a probability of one in seven, such as the Royal baby being born on a Monday (uncanny!).
×3      ×3      ×3      ×3      ×3
(Click on one of these icons to react to this blog post)

You might also enjoy...

Comments

Comments in green were written by me. Comments in blue were not written by me.
 Add a Comment 


I will only use your email address to reply to your comment (if a reply is needed).

Allowed HTML tags: <br> <a> <small> <b> <i> <s> <sup> <sub> <u> <spoiler> <ul> <ol> <li> <logo>
To prove you are not a spam bot, please type "cosine" in the box below (case sensitive):

Archive

Show me a random blog post
 2026 

May 2026

World Cup stickers 2026

Apr 2026

A new puzzle every day
Mixing Wordle with other games

Feb 2026

Christmas (2025) is over
 2025 

Dec 2025

Christmas card 2025

Nov 2025

Christmas (2025) is coming!

Sep 2025

The partridge puzzle

Aug 2025

TMiP 2025 puzzle hunt

Jun 2025

A nonogram alphabet

Mar 2025

How to write a crossnumber

Jan 2025

Christmas (2024) is over
Friendly squares
 2024 

Dec 2024

A regular expression Christmas puzzle
Christmas card 2024

Nov 2024

Christmas (2024) is coming!

Feb 2024

Zines, pt. 2

Jan 2024

Christmas (2023) is over
 2023 
▼ show ▼
 2022 
▼ show ▼
 2021 
▼ show ▼
 2020 
▼ show ▼
 2019 
▼ show ▼
 2018 
▼ show ▼
 2017 
▼ show ▼
 2016 
▼ show ▼
 2015 
▼ show ▼
 2014 
▼ show ▼
 2013 
▼ show ▼
 2012 
▼ show ▼

Tags

finite group asteroids sound final fantasy crossnumber phd crosswords wool fonts numerical analysis propositional calculus national lottery gerry anderson pac-man weather station accuracy pascal's triangle royal institution guest posts crochet nonograms partridge puzzle error bars coins talking maths in public turtles polynomials mean anscombe's quartet noughts and crosses rhombicuboctahedron sobolev spaces databet crossnumbers games palindromes youtube sport christmas matrix of cofactors dragon curves news game of life reuleaux polygons errors runge's phenomenon bluesky video games statistics nine men's morris inverse matrices logic mathsjam quadrilaterals platonic solids hyperbolic surfaces arithmetic thirteen advent calendar signorini conditions fence posts convergence folding paper captain scarlet weak imposition pokémon numbers chebyshev puzzles countdown tennis 24 hour maths approximation golden ratio finite element method big internet math-off recursion braiding hannah fry draughts preconditioning probability radio 4 rust golden spiral pokémon wordle oeis tmip geogebra wordle programming wave scattering bots pi approximation day hats people maths martin gardner cambridge coventry mathsteroids friendly squares regular expressions manchester science festival php menace football books ucl simultaneous equations chalkdust magazine bodmas trigonometry realhats newcastle stickers hexapawn alphabets light manchester plastic ratio data visualisation javascript computational complexity logo mathslogicbot fractals map projections speed matrix multiplication stirling numbers estimation game show probability kenilworth harriss spiral binary the aperiodical european cup correlation edinburgh inline code gaussian elimination machine learning graph theory pizza cutting python triangles a gamut of games electromagnetic field london rugby boundary element methods pythagoras data gather town latex datasaurus dozen royal baby frobel logs geometry exponential growth matrix of minors live stream zines dates pi ternary reddit london underground go folding tube maps bempp matt parker kings standard deviation raspberry pi arrangement puzzles sorting graphs chess squares cross stitch flexagons warwick christmas card world cup dataset misleading statistics bubble bobble matrices dinosaurs curvature interpolation tetris determinants craft

Archive

Show me a random blog post
▼ show ▼
© Matthew Scroggs 2012–2026