# Blog

## Archive

Show me a random blog post**2019**

**2018**

**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 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 books cross stitch**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):

- 5 6 8 9 10 100
- 5 6 7 8 10 100
- 4 6 7 8 9 100
- 3 6 7 8 10 100
- 3 5 7 8 9 100
- 2 5 6 8 9 100
- 2 6 7 8 9 100
- 5 6 8 9 75 100
- 3 6 8 10 75 100
- 2 6 9 10 75 100

By contrast, the following combination allows no totals between 101 and 999 to be reached:

- 1 1 2 2 3 3

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 numbers | Probability of being able to reach target |

0 | 0.964463439 |

1 | 0.983830962 |

2 | 0.993277819 |

3 | 0.985770510 |

4 | 0.859709475 |

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 | 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.

**2016-07-20**

Matthew

**2016-07-20**

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

**Add a Comment**

**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:

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.

### Similar posts

Countdown probability, pt. 2 | Pointless 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.

**Add a Comment**

**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.

**Add a Comment**

**
**

**© Matthew Scroggs 2018**

2016-07-20