Some Pointless Mathematics

I’m a big fan of the TV quiz show ‘Pointless’. If you’ve not seen it, stop reading this and go see it. If you have seen it then you’ll be familiar with the show’s set-up. Four teams compete each show to make it through to the final round, where they have a chance to win the jackpot. Only one team makes it through to the final round each show. However unlike in other quiz shows, Pointless teams get two attempts to make it through to the final round, meaning the same team can return and appear on two consecutive shows. However teams who make it to the final round on their first attempt cannot compete for a second time and are replaced by a new team on the next show. Ultimately what I’m trying to say is that, on any given episode of Pointless, the 4 competing teams are made up of a combination of New and Returning teams.

This begs the question: what is the probability associated with each combination of new and returning teams?

The probability of there being four returning teams is zero. This is a result of the fact that the team who makes it through to the final round cannot return to compete in the next show. Therefore a new team is always present as a replacement for the winners of the last show. Every other combination is possible though.

The relationships between these combinations can be shown using a type of system called a ‘Markov chain‘. A Markov chain is a mathematical system that undergoes transitions from one state to another among a finite number of states. It is described as memoryless in that the next state depends only on the current state and not any preceding state. For us, each state refers to the number of new and returning teams in a show.

As the transition is different depending on whether a New team or a Returning team makes it through to the final, we obtain a relatively simple yet interesting Markov chain as shown below:

Markov Chain
We have assumed that each team has the same probability of winning.

Note: State 1 is the case of 4 New teams, and we assign from left to right on the above markov chain, the numbers for the other states

From the above Markov chain, we can see a clear progression towards state 4 (2 New and 2 Returning teams). However this Markov chain only tells us the probability of transitions between states, not the probability of each individual state occuring. To obtain these probabilities, we construct what is known as a ‘Markov matrix’. Each row refers to a state with the entries in each row being the probability of transitioning into each different state, a better and more thorough explanation of Markov matrices can be found here. For now take my word that below is the relevant Markov matrix. .

Markov Matrix
There is a 0.75 probability of transitioning from state 2 to state 3. So the third entry in row 2 is 0.75.

If we were to multiply this matrix by itself an infinite number of times we would obtain the exact probability of each of these states. So an approximation can be found by multiplying this matrix by itself a large number of times. Armed with, I multiplied this matrix by itself 1000 times to obtain the following matrix:

Markov Matrix^1000
Each column now refers to the probability of each state.

From this matrix we can see that for any given show of Pointless,

There is a 2.9% chance of 4 New teams,

There is an 11.4% chance of 1 New team and 3 Returning teams,

There is a 34.3% chance of 3 New teams and 1 Returning team, and

There is a 51.4% chance of 2 New teams and 2 Returning teams.

However as is often the case with Mathematics, these results are only true in a perfect world. One in which Pointless is a never-ending facet of life. But as TV shows are filmed in series’ and each new series of Pointless starts with 4 New teams, we would assume that the actual figures will be skewed towards the top of the Markov chain. Despite this, these probabilities still provide a great estimate.

EDIT: This link is to another blog post with a different (and substantially more elegant) method to solving the same question.

2 thoughts on “Some Pointless Mathematics

  1. Some pretty nifty mathematics, but I just thought I’d point out that new series don’t actually start with 4 new pairs. The pairs from the previous series get to carry across their second appearance, just as the jackpot gets carried across :)

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