mscroggs.co.uk
mscroggs.co.uk

subscribe

Blog

 2018-05-02 
Every morning just before 7am, one of the Today Programme's presenters reads out a puzzle. Yesterday, it was this puzzle:
In a given month, the probability of a certain daily paper either running a story about inappropriate behaviour at a party conference or running one about somebody's pet being retrieved from a domestic appliance is exactly half the probability of the same paper containing a photo of a Tory MP jogging. The probability of no such photo appearing is the same as that of there being a story about inappropriate behaviour at a party conference. The probability of the paper running a story about somebody's pet being retrieved from a domestic appliance is a quarter that of its containing a photo of a Tory MP jogging. What are the probabilities the paper will (a) run the conference story, (b) run the pet story, (c) contain the jogging photo?
I'm not the only one to notice that some of Radio 4's daily puzzles are not great. I think this puzzle is a great example of a terrible puzzle. You can already see the first problem with it: it's long and words and very hard to follow on the radio. But maybe this isn't so important, as you can read it here after it's been read out.
Once you've done this, you can re-write the puzzle as follows: there are three news stories (\(A\), \(B\) and \(C\)) that the newspaper might publish in a month. We are given the following information:
$$\mathbb{P}(A\text{ or }B)=\tfrac12\mathbb{P}(C)$$ $$1-\mathbb{P}(C)=\mathbb{P}(A)$$ $$\mathbb{P}(B)=\tfrac14\mathbb{P}(C)$$
To solve this puzzle, we need use the formula \(\mathbb{P}(A\text{ or }B)=\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A\text{ and }B)\). These Venn diagrams justify this formula:
Venn diagrams showing that \(\mathbb{P}(A\text{ or }B)=\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A\text{ and }B)\).
Using the information we were given in the question, we get:
\begin{align} \tfrac12\mathbb{P}(C)&=\mathbb{P}(A\text{ or }B)\\ &=1-\mathbb{P}(C)+\tfrac14\mathbb{P}(C)-\mathbb{P}(A\text{ and }B)\\ \mathbb{P}(C)&=\tfrac45(1-\mathbb{P}(A\text{ and }B)). \end{align}
At this point we have reached the second problem with this puzzle: there's no answer unless we make some extra assumptions, and the question doesn't make it clear what we can assume. But let's give the puzzle the benefit of the doubt and try some assumptions.

Assumption 1: The events are mutually exclusive

If we assume that the events \(A\) and \(B\) are mutually exclusive—or, in other words, only one of these two articles can be published, perhaps due to a lack of space—then we can use the fact that
$$\mathbb{P}(A\text{ and }B)=0.$$
This means that \(\mathbb{P}(C)=\tfrac45\), \(\mathbb{P}(A)=\tfrac15\), and \(\mathbb{P}(B)=\tfrac15\). There's a problem with this answer, though: the three probabilities add up to more than 1.
This wouldn't be a problem, except we assumed that only one of the articles \(A\) and \(B\) could be published. The probabilities adding up to more than 1 means that either \(A\) and \(C\) are not mutually exclusive or \(A\) and \(B\) are not mutually exclusive, so \(C\) could be published alongside \(A\) or \(B\). There seems to be nothing special about the three news stories to mean that only some combinations could be published together, so at this point I figured that this assumption was wrong and moved on.
Today, however, the answer was posted, and this answer was given (without an working out). So we have a third problem with this puzzle: the answer that was given is wrong, or at the very best is based on questionable assumptions.

Assumption 2: The events are independent

If we assume that the events are independent—so one article being published doesn't affect whether or not another is published—then we may use the fact that
$$\mathbb{P}(A\text{ and }B)=\mathbb{P}(A)\mathbb{P}(B).$$
If we let \(c=\mathbb{P}(C)\), then we get:
\begin{align} \tfrac12c&=\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A\text{ and }B)\\ &=\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A)\mathbb{P}(B)\\ &=1-c+\tfrac14c-\tfrac14(1-c)c\\ \tfrac14c^2-\tfrac32c+1&=0. \end{align}
You can use your favourite formula to solve this to find that \(c=3-\sqrt5\), and therefore \(\mathbb{P}(A)=\sqrt5-2\) and \(\mathbb{P}(B)=\tfrac34-\tfrac{\sqrt5}4\).
In this case, our assumption appears to be more reasonable—as over the course of a month the stories published by a paper probably don't have much of an effect on each other—but we have the fourth, and probably biggest problem with the puzzle: the question and answer are not interesting or surprising, and the method is a bit tedious.

Similar posts

World Cup stickers 2018, pt. 3
World Cup stickers 2018, pt. 2
A 20,000-to-1 baby?
World Cup stickers 2018

Comments

Comments in green were written by me. Comments in blue were not written by me.
@Stefan: It's possible that that was intended but it's not completely clear. If that was the intention, I stick to my point that it's odd that the other story can be published alongside these two.
Matthew
                 Reply
Doesn’t the word ‘either’ in the opening phrase make A and B explicity exclusive?
Stefan
                 Reply
 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>
To prove you are not a spam bot, please type "sixa-y" backwards in the box below (case sensitive):

Archive

Show me a random blog post
 2020 

Jul 2020

Happy ϕ√3 Approximation Day!

May 2020

A surprising fact about quadrilaterals
Interesting tautologies

Mar 2020

Log-scaled axes

Feb 2020

PhD thesis, chapter ∞
PhD thesis, chapter 5
PhD thesis, chapter 4
PhD thesis, chapter 3
Inverting a matrix
PhD thesis, chapter 2

Jan 2020

PhD thesis, chapter 1
Gaussian elimination
Matrix multiplication
Christmas (2019) is over
 2019 
▼ show ▼
 2018 
▼ show ▼
 2017 
▼ show ▼
 2016 
▼ show ▼
 2015 
▼ show ▼
 2014 
▼ show ▼
 2013 
▼ show ▼
 2012 
▼ show ▼

Tags

phd golden ratio london underground machine learning gaussian elimination inverse matrices curvature manchester map projections game of life cambridge estimation gerry anderson harriss spiral approximation matrix multiplication accuracy countdown graphs martin gardner raspberry pi wave scattering bempp news pac-man signorini conditions final fantasy data visualisation realhats games logic finite element method matrices london christmas card nine men's morris pythagoras latex twitter dataset draughts european cup advent calendar matt parker craft people maths interpolation rugby asteroids boundary element methods big internet math-off video games game show probability hexapawn mathsteroids captain scarlet pi folding tube maps weather station dragon curves oeis error bars golden spiral stickers preconditioning trigonometry bodmas exponential growth misleading statistics quadrilaterals go speed braiding graph theory ucl pi approximation day reddit simultaneous equations wool convergence sobolev spaces football national lottery sound electromagnetic field royal baby probability propositional calculus mathslogicbot determinants mathsjam hats python manchester science festival geogebra the aperiodical hannah fry noughts and crosses binary arithmetic talking maths in public weak imposition statistics programming data bubble bobble sorting javascript coins php ternary sport chebyshev puzzles radio 4 squares a gamut of games computational complexity polynomials world cup platonic solids pizza cutting logs rhombicuboctahedron cross stitch dates chalkdust magazine menace numerical analysis christmas folding paper geometry chess inline code reuleaux polygons palindromes triangles tmip royal institution matrix of minors tennis flexagons fractals matrix of cofactors books plastic ratio frobel light

Archive

Show me a random blog post
▼ show ▼
© Matthew Scroggs 2012–2020