However, for Harriet, it isn’t just the overall quantity that counts, it is the number of female puppies, because females earn her a premium. On average, 50 per cent of puppies are female. “What is the chance I’ve got more female puppies than Matilda?” wonders Harriet.

You might think she needs to use sophisticated probability maths to work this out, but in fact there is a way to figure out the solution with some relatively simple reasoning.

Answer next week

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#97 Cabinet reshuffle

Solution

The Ruritanian PM reshuffled the five senior ministers as follows:

Anerdine: Health to defence

Brinkman: Chancellor to education

Crass: Defence to chancellor

Dyer: Home secretary to health

Eejit: Education to home secretary

We were told Dyer goes to health, so only Crass or Eejit could take Dyer’s job (which isn’t chancellor).

From this starting point, you will find that the only cycle of moves that works is ACBED, resulting in the jobs listed above.

The Ruritanian prime minister is in a fix. Thanks to a series of incompetent policy decisions, all five of her senior ministers need to be axed from their posts. However, the PM cannot afford to sack them completely, because they will wreak havoc if moved to the back benches.

She has a solution: a reshuffle! She will move each of the five ministers to one of the other top posts, but no two of them will directly swap with each other. Anerdine will move to the department of the person who will become chancellor. Brinkman will replace the person who will be the new home secretary. Crass will take over the post being vacated by the person who will take Eejit’s job. Dyer will be become health secretary even though he has been lobbying to become chancellor. The current defence secretary will get the department of the person who is becoming the education secretary.

Can you figure out who currently has which job, and where they are moving to?

Answers next week

#96 Inside the box

Solution

Start at X and work outwards. The red lines that link nine dots arranged in a square follow the path of the classic four line/nine dot “out of the box” solution.

Members of the Rackemup rifle club are tired of Pat Perfect winning the annual shooting tournament every year. She always hits the bullseye. To spice up the contest, this year she has agreed to be blindfolded. Even so, during practice she still hits the target four times out of five. In the first round she competes against Kate, who, even when fully sighted, only hits two times in five.

To give Kate the best chance of a fluke win, the contest will involve just a single shot each. The two competitors fire, but only one of them hits the target. “I reckon it’s a 1 in 3 chance that’s yours Kate,” says the referee as he strides up to inspect whose bullet it was. But he is wrong about the odds. What is the chance that Kate is the victor?

Answer next week

#87 Poll position

Solution

There will always be a house from which, if you start the polling there, a Red majority can’t emerge. Imagine scoring +1 for every Blue voter and -1 for every Red voter. Start at any point and keep score as you go clockwise. Since there are equal Blues and Reds, when you get back to where you started, you will be at zero. However, there will be some house on the tour when your score is the most negative. If you start the poll at the neighbour to the right (clockwise) of this, your score on the tour will always be positive or zero. In other words, Red will never be in the majority.

A biased and manipulative pollster is doing a survey of your cul de sac, where half the residents vote Red and the other half Blue. He wants to use the poll to “prove” that Reds are in the majority. His plan is to pick a house to start and visit all the homes in a loop going clockwise, but to stop the poll at the first instance that Red takes the lead.

As it happens, you know how everyone on the street votes. Is there a house where you can get the pollster to start from where you can be certain his plan will be foiled. If so, what is an easy way to find it?

Answer next week

#86 Yam tomorrow

Solution

Abel got 10 yams, Babel seven, Cabel five and the monkey three. Suppose in the morning the sailors find three yams and each gets one. This means Cabel took (1.5 x 3) + 1 = 5.5 yams – but it must be a whole number. Raising the number to two each (six yams) means Cabel found 10 (and took three), Babel found 16 (and took five) and Abel found 25 (and took eight). The next solution with whole numbers would be 106 yams, but that exceeds the capacity of the crate.