# Blog

## Archive

Show me a random blog post**2018**

### Sep 2018

Runge's Phenomenon### Jul 2018

World Cup stickers 2018, pt. 3Mathsteroids

### Jun 2018

World Cup stickers 2018, pt. 2### May 2018

A bad Puzzle for Today### Apr 2018

Building MENACEs for other games### Mar 2018

A 20,000-to-1 baby?World Cup stickers 2018

### Jan 2018

*Origins of World War I*

Christmas (2017) is over

**2017**

**2016**

**2015**

**2014**

**2013**

**2012**

## Tags

folding paper folding tube maps london underground platonic solids london rhombicuboctahedron raspberry pi weather station programming python php inline code news royal baby probability game show probability christmas flexagons frobel coins reuleaux polygons countdown football world cup sport stickers tennis braiding craft wool electromagnetic field people maths trigonometry logic propositional calculus twitter mathslogicbot oeis matt parker pac-man graph theory video games games chalkdust magazine menace machine learning javascript martin gardner noughts and crosses reddit national lottery rugby puzzles game of life dragon curves fractals pythagoras geometry triangles european cup dates palindromes chalkdust christmas card ternary bubble bobble asteroids final fantasy curvature binary arithmetic bodmas statistics error bars estimation accuracy misleading statistics pizza cutting captain scarlet gerry anderson light sound speed manchester science festival manchester dataset a gamut of games hexapawn nine men's morris draughts chess go radio 4 data map projections aperiodical big internet math-off sorting polynomials approximation interpolation chebyshev**2013-12-15**

## 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:

N^{o} of new teams today | |||||

N
^{o}of new teams tomorrow | 1 | 2 | 3 | 4 | |

1 | 0 | 0 | \(\frac{1}{4}\) | 1 | |

2 | 0 | \(\frac{1}{2}\) | \(\frac{3}{4}\) | 0 | |

3 | \(\frac{3}{4}\) | \(\frac{1}{2}\) | 0 | 0 | |

4 | \(\frac{1}{4}\) | 0 | 0 | 0 |

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.

### Similar posts

Countdown probability, pt. 2 | Countdown probability | World Cup stickers 2018, pt. 3 | World Cup stickers 2018, pt. 2 |

### Comments

Comments in green were written by me. Comments in blue were not written by me.

**© Matthew Scroggs 2018**

Add a Comment