For the benefit of overseas readers, or British readers in full-time employment, I should briefly explain the concept of daytime TV quiz phenomenon Pointless. The pinnacle of British public service broadcasting, it’s shown at 5.15pm every weekday on BBC One and is hosted by Alexander Armstrong of comedy double-act Armstrong & Miller, and Richard Osman of comedy double-act Armstrong & Osman. We shall investigate how we can use maths to analyse the show, improve our chances of winning it, and ultimately perhaps improve the show itself.

The aim of the game is in each round to give the most obscure correct answer to a given question. Each question ($Q$) has a large set of valid answers $A_Q$, questions perhaps asking contestants to name “Films starring Bruce Willis” or “Countries without an O in their name”. All the questions have been asked to 100 members of the public prior to the quiz (call this set $P$), and they each have 100 seconds to name as many examples as they can (giving rise to a set $A_p\subseteq A_Q$ for each $p\in P$. The contestant gets a point for every one of the 100 people who named their answer $a$:

\[ \mbox{score}(a) = \begin{cases}

| \{p\in P : a\in A_p \} | & \mbox{if}\ a\in A_Q \\

100 & \mbox{if}\ a\not\in A_Q.

\end{cases} \]

So an obvious answer like Die Hard or France will score a lot of points, and an obscure answer like Striking Distance or Central African Republic will score fewer points. Points are bad (hence the title) so it’s better to dredge up an obscure answer than stick with something safe. However an incorrect answer like Avatar or Mexico scores the maximum 100 points. At the end of the round the contestant with the most points is eliminated.

So far this sounds pretty simple; an obscure answer will beat an easy one, and you can decide what answer to give based on the scores of the other contestants so far, and your evaluation of how obscure your answer is and how confident of its correctness you are. There tend to be a few obvious answers scoring in the 60-100 range, a good number of ‘average’ answers worth 20-60, and a ‘long tail’ of lots of obscure answers with sub-20 scores. (In shocking ignorance of the increasing trend towards open data, *Pointless* never reveals the full set of scores for correct answers, so this assessment is necessarily slightly subjective.)

This distribution should be irrelevant though, since it’s only the relative order of the scores that matters. But this isn’t quite it, because the game is played in teams: four sets of two players ($T_i; i=1,\dots,4; |T_i|=2$). Both players in each team give an answer to the same question, and their scores are combined, with the highest-scoring team eliminated.

Now, when I say ‘combined’, what I of course mean is added — $\mathrm{score}(a_1,a_2) = \mathrm{score}(a_1) + \mathrm{score}(a_2)$. But it’s worth reflecting on the fact that this clearly isn’t the only way to combine two numbers (although it is the way that tends to predominate on game shows). Here the choice leads to some interesting results — and now the distribution of the answers along the 1-100 score range is crucial. An answer worth 80 and one worth 90 are both very obvious, and you might be hard-pressed to guess which answer would get which score. But there’s a world of difference between an answer worth 10, which will still be moderately well known, and an answer worth one point which is very obscure indeed. But Team A scoring 80+10=90 with an obvious answer and an alright one, beats Team B scoring 90+1=91 with an obvious answer and a brilliant one. Addition seems not to be properly rewarding what should be good gameplay.

This suggests that the best strategy to win at Pointless in general must be to play it safe: when trying to decide between an okay but safe answer, and one which is very good if right but possibly wrong, pick the safe answer, since its slightly higher score is unlikely to make the difference. This is compounded by the fact that the only aim in the round is not to come last, and a 100-scoring wrong answer is likely to be fatal (often there are no two correct answers that score over 100 in total). This is somewhat disappointing news, since it goes against the ethos of the show, which is to “plumb the depths of general knowledge”. The show is even named after the elusive ultra-obscure 0-point answers, and getting them is rewarded by a small addition to the jackpot prize (probably not enough to affect players’ motivations in the above). Now, it would be no good if all that could be got out of rigorous mathematical analysis of tea-time quiz formats was some admittedly well-founded sniping, so let’s see if we can use what we’ve learned to address these concerns.

From a format point of view tennis would be better if was best of 9 sets, first to 4 games in each set. Fairer, more exciting, more jeopardy

— Richard Osman (@richardosman) July 8, 2012

*I’m sure pointless co-host/producer Richard Osman will not mind me suggesting ways to improve the format of a popular British institution…*

## Pointless 2.0

The conundrum outlined above makes logical sense, because the scores in Pointless are essentially proportions: the percentage of the population that knows the answer you give. And the usual way to combine proportions is to multiply them rather than add them: $\mathrm{score}(a_1,a_2) = \mathrm{score}(a_1) \cdot \mathrm{score}(a_2)$. On this method, Team A scores 800 for their mediocre answers, and Team B just 90 for theirs. (Alternatively we could add the scores after taking their logarithms, but strangely, few quiz formats seem to go for this mechanism.) The team’s score could then be seen as replicating the proportion of the population that knew both of their answers, reflecting the joint skill of the team.

But we can do better still. There’s no need to approximate the proportion that knew both answers. The question was actually put to 100 people, so we can award the team points according to the number of the 100 people who gave both of the team’s responses in their 100 seconds:

\[\mathrm{score}(a_1,a_2) = \begin{cases}

\left| \{p\in P : A_Q\cap\{a_1,a_2\}\subseteq A_p\} \right| & \mbox{if}\ A_Q\cap\{a_1,a_2\}\neq\emptyset \\

100 & \mbox{otherwise.}

\end{cases}\]

This would add an extra layer of subtlety to the game, since you can try and pick a pair of answers that don’t ‘go together’ (scuttling the popular tactic of taking inspiration from previous answers, for example picking a sequel to a film that scored well for your teammate). Here’s how it would work:

## Paul’s Humble Proposition for Improving the Scoring System for the TV Quiz Pointless

The first player in the team gives an answer in the normal way, and the tower descends as usual to reveal their score. The second player gives their answer, and the tower starts at the team’s current score, representing the subset of the 100 people who gave the first player’s answer. The tower descends until it reaches the number of people who gave both answers, and this is the team’s final score. The team with the highest score is eliminated.

Here is a brief summary of the reasons that my method is superior to the current system:

- Does not reward two average answers better than one good and one slightly-less-average answer;
- Removes the unsatisfactory situation where a team’s second answer is irrelevant because the team’s score is more than 100 ahead of all teams yet to play;
- Introduces a second level of tactics in choosing answers with a low intersection;
- Maintains the conceptual link between a team’s score and the 100 people;
- Increases the cachet of a ‘pointless’ answer since a team scoring one is unbeatable. (Though clearly ties where each team gives a 0-point answer would be broken by the value of the other answer.)

## Winning Money

After the first two rounds, the two remaining teams can confer between themselves and give a single joint answer, so the above doesn’t apply. The winning team goes forward to play the jackpot round, in which a cash prize can be won. The prize starts at £1000, rising by £250 whenever a pointless answer is given, and an extra £1000 is added at the start of each show, so the jackpot keeps building as it is not won.

This prize mechanism is essentially lifted from the BBC’s previous daytime quiz hit Eggheads, where a team of contestants tries to beat the eponymous resident panel of top quizzers. There the conceit makes some sense, as the prize will build if the Eggheads are ‘on form’, and it lessens the sense that they are robbing the competing quiz plebs of cash, instead merely redistributing it only to the most deserving ((A concern that seems not to trouble anyone at The Chase)). On Pointless this device only serves to make the prize basically random, since each game is independent.

But there’s an added wrinkle. Any team on the show that doesn’t make it to the jackpot round on their first appearance competes again on the next show. Of course, this means if you lose your first show and the winners don’t claim the jackpot (they usually don’t) the prize money comes with you to the next show and you have another crack at it. And if the prize was at its lowest, £1000, in your first game, it will have doubled for your second appearance (and even if the team that beats you does win the money, your jackpot has not lowered).

This leads to the interesting scenario that, if you find yourself in the last two teams looking at a jackpot of £1000, and you rate yourselves as having a better than 50% chance of winning a given show, you can actually increase your expected takings by throwing the match and returning to have a go at a bigger prize. (Of course this would be very unsportsmanlike and we do not recommend you actually do this, unless you are playing against us.)

## A puzzle

Finally, here is a puzzle I wrote some time ago as a sort of tribute to the show:

As explained above, four teams compete in each episode and any team which does not win plays again on the next episode (but do not return a third time). So on any episode, between zero and three of the teams competing may be ‘returning teams’. There are nine shows left in the series, and the producers would like there to be no teams due to return on the first episode of the next series, as this creates logistical difficulties. Richard Osman is preparing the questions for the last nine episodes.

Richard is able to rig the questions for each episode in one of two ways, either **(i)** filling the questions with facts that were mentioned in passing in the previous show, so that a returning team will definitely win (if there is one); or **(ii)** filling the show with questions whose correct answers were contradicted by a secret campaign of misinformation in the previous show, ensuring that a new team wins.

Unfortunately, he cannot remember how many returning pairs will compete in the next episode, and must rig all nine episodes in advance of their being filmed. Can he rig the questions in such a way as to guarantee that no teams are awaiting their second go when the series ends?

Blimey,your clever.are you related to Sheldon cooper? I like the maths but I don’t think Richard will.enjoy enjoying pointless,I will.

Nice article and nice puzzle. I think I have a solution. It’s based on the fact that whenever you have four new teams, it doesn’t matter how Richard has fixed the show, a new team will definitely win.

If N and R mean that Richard fixes the episode for a new or returning team to win respectively, then I think that the sequence R,N,R,N,R ensures that, no matter where you start, you will reach an episode with precisely 1 new team (I think this is the minimal sequence of fixes to guarantee a particular line-up). You then have to fix four more episodes so that you reach an episode with 4 new teams. Actually, it turns out that from an episode with 1 new team, by fixing four episodes you can get to any line-up EXCEPT an episode with 4 new teams. However, you can resolve this by not fixing any of the first three episodes, then fixing the next six as R,N,R,N,R,N.

If you have to fix every episode, then one solution would be: NNNRNRNRN