A King died and the elder of his two sons prepared to ascend to the throne. The family assembled to hear the reading of the old King's will, and found a little surprise in it.

The King had specified that he would be succeeded by the son who had the slowest horse. And as it happened both Princes owned just the one horse.

[Yes, we know that this isn't how a hereditary monarchy works, but this was a long time ago - don't spoil the puzzle!]

A race was duly held and the Prince who would become King selected. The question is, how did they organise the race?

As usual you can post the answers as a comment on this website, reply to the post on Facebook, or retweet or reply on Twitter @quizmastershop.

Answer at 9.00 on Monday

]]>Having one can add some interest to your Quiz Night, as the teams have to decide the round that they’ll use their Joker for double points.

There are two ways that you can do this:

- Announce all the Categories for the rounds before the quiz, and then get each team to declare their Joker Category; either before the quiz starts or before the first question for that round is asked.
- Allow the teams to declare their Joker after a round’s questions have been asked but before it’s marked.

Give it a go and see if your customers like it.

]]>A man has just moved to a small town with two barbers and he needs a haircut, so he has to choose which barbershop he will grace with his custom.

Now this is in the days before online reviews, so he can't search the internet and see which one has the most stars. So he has to do it the old way and see for himself. [We don't know why he doesn't ask a few well-groomed blokes down the pub which barber they use, but that would spoil the puzzle, so go with it :-)]

He goes to the first barber's premises and looks in through the window. What he sees is a barber with a very scruffy haircut - long bits, short bits, tufts, you name it.

He then moves on to the second barber and looks through the window at the proprietor. In contrast with his competitor he has an immaculate haircut - smooth, sleek, not a hair out of place.

The question is, which of the two barbers got his trade, and why?

The man deduced that each barber cut the other barber's hair, and so went to the scruffy barber's shop on the basis that he had given the tidy barber a good haircut.

]]>A man has just moved to a small town with two barbers and he needs a haircut, so he has to choose which barbershop he will grace with his custom.

Now this is in the days before online reviews, so he can't search the internet and see which one has the most stars. So he has to do it the old way and see for himself. [We don't know why he doesn't ask a few well-groomed blokes down the pub which barber they use, but that would spoil the puzzle, so go with it :-)]

He goes to the first barber's premises and looks in through the window. What he sees is a barber with a very scruffy haircut - long bits, short bits, tufts, you name it.

He then moves on to the second barber and looks through the window at the proprietor. In contrast with his competitor he has an immaculate haircut - smooth, sleek, not a hair out of place.

The question is, which of the two barbers got his trade, and why?

As usual you can post the answers as a comment on this website, reply to the post on Facebook, or retweet or reply on Twitter @quizmastershop.

Answer at 9.00 on Monday

]]>There are ten questions; five of them have numerical answers and the other five have one-word answers. You should add up the five numerical answer to produce a number, and you should take the first letters of the other five answers in the order that they appear to produce a word. When you put the number and the word together they form a word, phrase or saying that is still topical.

- In Scrabble how many points is the letter A worth?
- Which is the only natural number that can't be written in Roman numerals?
- What portmanteau word is used for the cross between a Cavalier King Charles Spaniel and Poodle?
- On how many hills is Rome supposedly built?
- Which board game's name is a portmanteau word from Yes in French and German?
- What score is nicknamed Snake Eyes when rolling two dice?
- Which sign of the zodiac is the last alphabetically?
- What nationality is Bono?
- Under what number were laxative pills listed in World War I, giving the bingo call Doctors Orders?
- Rhode Island is the smallest US state - which is the second smallest?

The answers are:

- One
- Zero
- Cavapoo or Cavadoodle (either works)
- Seven
- Ouija
- Two (It's two ones)
- Virgo
- Irish
- Nine
- Delaware

The numbers (1 + 0 + 7 + 2 + 9) add up to 19. The letters C, O, V, I and D spell COVID. And these make COVID-19.

]]>There are ten questions; five of them have numerical answers and the other five have one-word answers. You should add up the five numerical answer to produce a number, and you should take the first letters of the other five answers in the order that they appear to produce a word. When you put the number and the word together they form a word, phrase or saying that is still topical.

- In Scrabble how many points is the letter A worth?
- Which is the only natural number that can't be written in Roman numerals?
- What portmanteau word is used for the cross between a Cavalier King Charles Spaniel and Poodle?
- On how many hills is Rome supposedly built?
- Which board game's name is a portmanteau word from Yes in French and German?
- What score is nicknamed Snake Eyes when rolling two dice?
- Which sign of the zodiac is the last alphabetically?
- What nationality is Bono?
- Under what number were laxative pills listed in World War I, giving the bingo call Doctors Orders?
- Rhode Island is the smallest US state - which is the second smallest?

Can you solve it?

As usual you can post the answers as a comment on this website, reply to the post on Facebook, or retweet or reply on Twitter @quizmastershop.

Answer at 9.00 on Monday

]]>You will notice that each card, except those in the bottom row, is above two cards in the row below.

The challenge is to place the 15 cards such that the value on each card, except those in the bottom row, is the difference between the two cards below it.

Can you solve the puzzle?

The top card is 5,

Supported by 4 and 9,

The middle row is 7, 11 and 2,

Beneath those are 8, 1, 12 and 10,

With a bottom row of 6, 14, 15, 3 and 13.

]]>You will notice that each card, except those in the bottom row, is above two cards in the row below.

The challenge is to place the 15 cards such that the value on each card, except those in the bottom row, is the difference between the two cards below it.

Can you solve the puzzle?

Answer at 9.00 on Monday

]]>Tuesday 1st - Not Just An Udder Quiz (Online) 19.40

Wednesday 2nd - Robbie's Quarantine Quiz (Online) 20.00

Sunday 6th - The Inn on the Green 20.00

Monday 7th - Hemingways 20.00

Tuesday 8th - Not Just An Udder Quiz (Online) 19.40

Sunday 13th - The Inn on the Green 20.00

Monday 14th - Hemingways 20.00

Tuesday 15th - Not Just An Udder Quiz (Online) 19.40

Sunday 20th - The Inn on the Green 20.00

Monday 21st - Hemingways 20.00

Tuesday 22nd - Not Just An Udder Quiz (Online) 19.40

Sunday 27th - The Inn on the Green 20.00

Monday 28th - Hemingways 20.00

Tuesday 29th - Not Just An Udder Quiz (Online) 19.40

]]>

You have a coin that is biased so that the probability of a head is not the same as the probability of a tail. The probabilities of each are fixed but you don't know the probabilities.

How can you use this coin to produce an outcome where both alternatives have a 50% probability?

The solution should work irrespective of the degree of bias in the coin, although clearly not if it lands heads 100% of the time!

If the probability of a head is h and the probability of a tail is t, then the probability of a head followed by a tail is h*t and the probability of a tail followed by a head is t*h. They are the same.

So one person calls head-tail or tail-head and you toss the coin twice. If it lands head-head or tail-tail, repeat until the two are different and you have your result.

]]>You have a coin that is biased so that the probability of a head is not the same as the probability of a tail. The probabilities of each are fixed but you don't know the probabilities.

How can you use this coin to produce an outcome where both alternatives have a 50% probability?

The solution should work irrespective of the degree of bias in the coin, although clearly not if it lands heads 100% of the time!

Answer at 9.00 on Monday

]]>The problem is that all of the treasure chests have been labelled incorrectly.

By opening just one treasure chest how can you determine the contents of all three chests?

Open the treasure chest labelled "Gold and Silver Coins".

If it contains gold coins the treasure chest labelled "Gold Coins" must contain silver coins, and the treasure chest labelled "Silver Coins" must contain gold and silver coins.

Likewise, if it contains silver coins the treasure chest labelled "Silver Coins" must contain gold coins, and the treasure chest labelled "Gold Coins" must contain gold and silver coins.

ALL of the treasure chests are incorrectly labelled.

]]>The problem is that all of the treasure chests have been labelled incorrectly.

By opening just one treasure chest how can you determine the contents of all three chests?

Answer at 9.00 on Monday

]]>In that puzzle we asked the number of chocolates in a pile ten layers tall where the pile was a tetrahedron. That is with one chocolate at the top resting on three chocolates in the layer below, and so on.

And we followed this by asking if there was a formula for a pile of any height.

This week we have changed the pile to a Pyramid. That is with one chocolate at the top resting on four chocolates in the layer below, and so on.

So how many chocolates are in a pile ten layers tall?

And following on from this, is there a formula to calculate the number of chocolates in a pile of any height?

Each of the layers in the pile (or pyramid) is a square, and so the numbers of chocolates in each of the layers are successive square numbers. That is, 1, 4, 9, 16 and so on.

So taking the two parts of the puzzle in reverse order, to calculate the number of chocolates in a pyramid of n layers, we need to sum the first n square numbers. And there is a formula for that:

$$ \frac{n(n + 1)(2n + 1)}{6} $$

We can see it is correct when n = 1

$$ \frac{1 x 2 x 3}{6} = 1 $$

and when n = 2

$$ \frac{2 x 3 x 5}{6} = 5 $$

So to answer the first part last, when n = 10

$$ \frac{10 x 11 x 21}{6} = 385 $$

]]>

In that puzzle we asked the number of chocolates in a pile ten layers tall where the pile was a tetrahedron. That is with one chocolate at the top resting on three chocolates in the layer below, and so on.

And we followed this by asking if there was a formula for a pile of any height.

This week we have changed the pile to a Pyramid. That is with one chocolate at the top resting on four chocolates in the layer below, and so on.

So how many chocolates are in a pile ten layers tall?

And following on from this, is there a formula to calculate the number of chocolates in a pile of any height?

Answer at 9.00 on Monday

]]>However, if you look a little closer at many of the companies on the internet that are selling quizzes, they actually sell quiz rounds. So this isn't such an outlandish claim. Many companies do sell their weekly or daily (or some other time interval) quiz, which comes as a complete entity, while others allow you to buy individual rounds on different topics.

Some companies write quizzes and some companies write quiz rounds.

Quiz Master Shop writes neither quizzes nor quiz rounds, and this is the thing that sets us apart.

Quiz Master Shop writes questions.

All of our questions are in a database (over 14,000 and still rising) and our clever software does the rest. But why does that matter to you?

Well, if you're a brewery or pub chain that wants quizzes for your pubs, instead of getting your supplier's periodic offering, each of your pubs can download the quiz they want. And this can, and most likely will, be different from all your other pubs.

And if you're a pub, club, PTA or whatever, you can choose the number of rounds that you want, the subject for each of those rounds, and the difficulty for each of those subjects.

In both cases our software does the rest, and you get the quizzes that you want.

Get What You Want, Not What You're Given

]]>- In which sitcom did Windsor Davies play Sergeant Major Williams?
- What is the first line of Bohemian Rhapsody?
- What does NIMBY stand for?
- Which song from The Wizard of Oz won the Best Original Song Oscar in 1939?
- What four-letter word means to prepare land for crops?
- What is the title of T H White's novel about King Arthur's childhood, which Disney made into a film?
- What foodstuff are people said to chew when engaged in smalltalk?
- Which legendary horsewoman was the wife of Leofric, Earl of Mercia?
- According to the nursery rhyme, what does Little Tommy Tucker do for his supper?
- Write the sentence suggested by the answers to the nine questions above in the order that they were asked.

The answers are

- It Ain't Half Hot Mum
- Is this the real life?
- Not In My Back Yard
- Over The Rainbow
- Till
- The Sword in the Stone
- Fat
- Lady Godiva
- Sings

And then taking the first word of each answer in order

It Is Not Over Till The Fat Lady Sings

]]>- In which sitcom did Windsor Davies play Sergeant Major Williams?
- What is the first line of Bohemian Rhapsody?
- What does NIMBY stand for?
- Which song from The Wizard of Oz won the Best Original Song Oscar in 1939?
- What four-letter word means to prepare land for crops?
- What is the title of T H White's novel about King Arthur's childhood, which Disney made into a film?
- What foodstuff are people said to chew when engaged in smalltalk?
- Which legendary horsewoman was the wife of Leofric, Earl of Mercia?
- According to the nursery rhyme, what does Little Tommy Tucker do for his supper?
- Write the sentence suggested by the answers to the nine questions above in the order that they were asked.

Answer at 9.00 on Monday

]]>If we assume that the pile is a tetrahedron (that is with one chocolate at the top resting on three chocolates in the layer below, and so on) how many chocolates are in a pile ten layers tall?

And following on from this, is there a formula to calculate the number of chocolates in a pile of any height?

Each of the layers in the pile (or tetrahedron) is a triangle, and so the numbers of chocolates in each of the layers are successive triangular numbers. That is, 1, 3, 6, 10 and so on.

So taking the two parts of the puzzle in reverse order, to calculate the number of chocolates in a tetrahedron of n layers, we need to sum the first n triangular numbers. And there is a formula for that:

$$ \frac{n(n + 1)(n + 2)}{6} $$

We can see it is correct when n = 1

$$ \frac{1 x 2 x 3}{6} = 1 $$

and when n = 2

$$ \frac{2 x 3 x 4}{6} = 4 $$

So to answer the first part last, when n = 10

$$ \frac{10 x 11 x 12}{6} = 220 $$

]]>If we assume that the pile is a tetrahedron (that is with one chocolate at the top resting on three chocolates in the layer below, and so on) how many chocolates are in a pile ten layers tall?

Answer at 9.00 on Monday

]]>There is a heart to the left of the spade. The six is on the left of the diamond. There is a diamond on the left of the spade. And the eight is to the right of the seven.

You are then offered a bet - correctly identify the three cards and win £1, get it wrong and lose £1

Could work out the positions of the three cards and win the bet?

As you know there is both a heart and a diamond to the left of the spade, the spade must be on the right.

The six is to the left of the diamond, so the diamond is in the middle and the leftmost card is the six.

This means the heart is on the left.

As the eight is to the right of the seven, the seven is in the middle and the eight is the right-hand card.

Left to right the cards are six of hearts, seven of diamonds and eight of spades.

Did you win £1?

]]>There is a heart to the left of the spade. The six is on the left of the diamond. There is a diamond on the left of the spade. And the eight is to the right of the seven.

You are then offered a bet - correctly identify the three cards and win £1, get it wrong and lose £1

Could work out the positions of the three cards and win the bet?

Answer at 9.00 on Monday

]]>This set us wondering, and we ran a few polls on Social Media. The results were fascinating with an almost even split between those who do give half points and those that don’t. In fact our Quizmasters are in three camps:

- The aforementioned who feel that an answer is either correct or it’s not, with no half measures.
- Those that give half a point if the answer is half right. For example, name the two Flowerpot Men – half a point for each, or if the answer is Antigua and Barbuda – half a point for Antigua.
- And those that give half a point to teams at the bottom end of the leader board, to encourage them, knowing that it won’t affect the result.

Having been to many, many quizzes between us, we don’t think that there is any hard and fast advice that we can give, apart from be consistent. As long as you have your approach, you should stick to it.

But do remember that you want all the teams to come back next time, even those that aren’t getting many questions right. So maybe some judiciously awarded half points might make them feel a bit happier about the quiz – something to think about perhaps.

]]>The game was simplicity itself; one player made an evens bet of one pound on whether the top two cards in the pack were the same colour or different colours. The other player then turned over the top two cards, and the pound was won and lost on the basis. Play continued through the pack in this way, and then the two players swapped roles.

Our friend explained that he was going to bet on the cards being different, as once the top card had been turned there were 25 of the same colour and 26 of the other colour.

His friend returned, play started, and it being such a basic game we rather lost interest. Until we heard our friend make an exasperated remark, and we enquired as to the problem.

He explained that he had lost every hand so far. Naturally we sympathised, and then asked if he was going to continue with his strategy. He looked even more perplexed, and replied that it had reached the point where both bets had the same probability.

How much money had he lost?

Let there be r red cards left and b black cards. The probability of turning two red cards is

$$ \frac{r(r - 1)}{(r + b)(r + b -1)} $$

and of turning two black cards

$$ \frac{b(b - 1)}{(r + b)(r + b - 1)} $$

and so the probability of two cards of the same colour is the sum of these

$$ \frac{r(r - 1) + b(b - 1)}{(r + b)(r + b - 1)} $$

Likewise the probability of two cards of different colours is

$$ \frac{2rb}{(r + b)(r + b - 1)} $$

As the two probabilities are the same we have

$$ r^2 - r + b^2 - b = 2rb $$

$$ (r - b)^2 = r + b $$

Now r + b must be even as the cards are removed in pairs and so it must be an even square less than 52. That is, 4, 16 or 36.

If r + b is 36 then r - b is 6, and so r is 21 and b is 15.

If r + b is 16 then r - b is 4, and so r is 10 and b is 6.

If r + b is 4 then r - b is 2, and so r is 3 and b is 1.

However both r and b must be even, as either two red cards or two black cards have been removed on each hand. So the only answer is ten red cards and six black cards (or the other way round) are left in the pack.

There are 36 cards already removed from the pack, which is 18 hands. Our friend is £18 down - no wonder he's exasperated!

]]>The game was simplicity itself; one player made an evens bet of one pound on whether the top two cards in the pack were the same colour or different colours. The other player then turned over the top two cards, and the pound was won and lost on the basis. Play continued through the pack in this way, and then the two players swapped roles.

Our friend explained that he was going to bet on the cards being different, as once the top card had been turned there were 25 of the same colour and 26 of the other colour.

His friend returned, play started, and it being such a basic game we rather lost interest. Until we heard our friend make an exasperated remark, and we enquired as to the problem.

He explained that he had lost every hand so far. Naturally we sympathised, and then asked if he was going to continue with his strategy. He looked even more perplexed, and replied that it had reached the point where both bets had the same probability.

How much money had he lost?

Answer at 9.00 on Monday

]]>Click to see the Sample Rounds or Buy Your Quiz and get started.

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The gist of the disagreement was that one of the pair was boasting of his encyclopaedic knowledge of music, and the other did not believe these boasts and thus was getting very angry.

The one who was boasting then made an astonishing claim: give me any name, any name at all, and I will sing you a song with that name in the lyrics - for a fee, of course.

You could have heard a pin drop.

"How much?" came the reply, "and what happens if you can't?"

So the pair agreed on a bet of £1,000 on whether the braggart would be able to sing a song with the (yet to be stated) name in the lyrics.

Not a person in the pub could take their eyes of the scene as the name of Nehemiah Kalashnikov Ring-Ding the Younger was given, although a few people started to smirk.

And then to the amazement of everyone the singer won the bet. How?

Very simply, the song was Happy Birthday.

]]>The gist of the disagreement was that one of the pair was boasting of his encyclopaedic knowledge of music, and the other did not believe these boasts and thus was getting very angry.

The one who was boasting then made an astonishing claim: give me any name, any name at all, and I will sing you a song with that name in the lyrics - for a fee, of course.

You could have heard a pin drop.

"How much?" came the reply, "and what happens if you can't?"

So the pair agreed on a bet of £1,000 on whether the braggart would be able to sing a song with the (yet to be stated) name in the lyrics.

Not a person in the pub could take their eyes of the scene as the name of Nehemiah Kalashnikov Ring-Ding the Younger was given, although a few people started to smirk.

And then to the amazement of everyone the singer won the bet. How?

Answer at 9.00 on Monday

]]>So the police concoct a "cunning plan".

They offer both Bill and Ben the following plea bargain: If you testify that your accomplice was involved in the burglary and he remains silent, you will go free and he will go to prison for ten years; if you both remain silent, you will both go to jail for one year for fighting; if you both testify, you will both be incarcerated for five years.

Bill and Ben have no way to communicate, and so no way to agree a plan.

What should they do?

This is quite a well known puzzle, known as the Prisoners' Dilemma, and there is no real correct answer.

The best collective course of action is for both Bill and Ben to stay silent. In this way they both get a year inside and then walk away.

Individually keeping silent has a big risk, because if your accomplice testifies against you a ten-year stretch awaits you, while he walks free.

So the insurance of talking means you could walk free, and in the worst case your sentence would be reduced to five years.

But the two prisoners are still better off keeping quiet, if they can trust the other.

Tricky!

]]>So the police concoct a "cunning plan".

They offer both Bill and Ben the following plea bargain: If you testify that your accomplice was involved in the burglary and he remains silent, you will go free and he will go to prison for ten years; if you both remain silent, you will both go to jail for one year for fighting; if you both testify, you will both be incarcerated for five years.

Bill and Ben have no way to communicate, and so no way to agree a plan.

What should they do?

Answer at 9.00 on Monday

]]>