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?

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

]]>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?

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

]]>As a Brewery or Pub Chain Owner you would like to take advantage of the economies of scale to bulk buy Quizzes for your pubs, and secure a nice discount.

So you pay your weekly fee, the supplier sends a quiz to all your pubs each week and . . .

“We’ve tried running quizzes here, but they don’t really work”,

“There’s a bloke who does our Quiz Nights and we’re not going to change”

“Our quizzes are the first Wednesday of the month – we don’t need one every week”

“No one in our pub liked the quiz you sent, so we stopped using them”

“Six rounds are too many for us – we use the best four rounds and throw the others”

You have got a lovely big discount, but most of the Quizzes end up in the bin. Which rather defeats the point.

There must be a better way.

**The Solution**

How about you buy a pool of Quiz Rounds (Note: Rounds not Quizzes) for use in your pubs; if they want to use them they do, and if they don’t they don’t. You only pay for what your pubs use.

Here’s how it works:

- You buy a number of Quiz Rounds that go into a pool.
- Any of your pubs that wants to use the free quizzes creates an account, and we link it to your account so they can access the quiz rounds.
- Each pub can then choose the number of rounds for their quiz night, the subjects for each round, and the difficulty for each round. They get exactly the quiz that they want, which might be completely different from every other pub.
- Every time a pub generates a quiz round the number in your pool decreases by one, so you are only paying for what the pubs actually use.
- When the number of rounds in your pool drops below your chosen limit, we’ll send you an e-mail to remind you to top up the pool.

**Get What You Want, Not What You’re Given**

The benefit to you is that you can supply your pubs with quizzes without paying for those that go in the bin.

The benefit to the pubs is they get to choose the frequency, length, subject and difficulty of the quiz they use, instead of the One Size Fits All approach that doesn’t really work.

]]>There are a couple of ways to legally pot the same ball in two consecutive shots as part of a beak in Snooker.

Can you work out how to pot the same ball in three consecutive shots as part of a beak?

One way to pot the same ball in two consecutive shots is when you are snookered on all the reds by a foul shot. You can nominate a colour as a free ball instead of a red, pot it, and then pot that colour after the red.

Another way is the pot the yellow as the colour after the last red, and then as the first of the colours.

The solution combines these two methods:

You are snookered on the last red by a foul shot, and you nominate the yellow as a free ball. You pot the yellow as a red AND pot the final red in this one shot. Incidentally scoring two points.

The yellow is respotted and you pot that as the colour after the red.

The yellow is again respotted and you pot that as the first of the colours.

The key is potting the red as well as the yellow on the first shot.

]]>There are a couple of ways to legally pot the same ball in two consecutive shots as part of a beak in Snooker.

Can you work out how to pot the same ball in three consecutive shots as part of a beak?

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

]]>As we've continued to have such great feedback we've decided to keep doing these!

The pubs in England are opening on 4th July, subject to Social Distancing that doesn't lend itself to Pub Quiz Nights, and there's no date set for Scotland, Wales and Northern Ireland at the time of writing.

So we're planing to run five more Virtual Quiz Nights in July, live streaming on our Facebook Page. These will be on 2nd, 9th, 16th, 23rd and 30th July at 7.00PM.

Details are under Events on our Facebook Page, or on various posts on Facebook or on our Twitter Page.

To access the Virtual Quiz Live Stream go to our Facebook Page just before 7.00PM and it will be there.

No Charge, No Prizes, About 40 minutes of fun, interesting and (occasionally) challenging questions.

]]>$$ \frac{1}{4} X \frac{8}{5} = \frac{18}{45} $$

To be completely correct you should cancel out the nines to leave two fifths, but the equation actually works even though the method used to produce it is decidedly dodgy.

This set us wondering whether there are any other equations where this incorrect method produces correct equations. Using single-digit, non-zero numbers (that is, 1 through to 9) on the left-hand side of the equation, how many correct equations can you find that follow this method?

If the four digits are a, b, c and d, we are trying to solve the equation

$$ \frac{a}{b} X \frac{c}{d} = \frac{10a + c}{10b + d} $$

or

$$ ac(10b + d) = bd(10a + c) $$

Ignoring the trivial cases when a = b and c = d there are seven solutions, which can be inverted to make another seven, giving 14 in all. The first seven are:

$$ \frac{1}{2} X \frac{5}{4} = \frac{15}{24} $$

$$ \frac{1}{4} X \frac{8}{5} = \frac{18}{45} $$

$$ \frac{1}{6} X \frac{4}{3} = \frac{14}{63} $$

$$ \frac{1}{6} X \frac{6}{4} = \frac{16}{64} $$

$$ \frac{1}{9} X \frac{9}{5} = \frac{19}{95} $$

$$ \frac{2}{6} X \frac{6}{5} = \frac{26}{65} $$

$$ \frac{4}{9} X \frac{9}{8} = \frac{49}{98} $$

We'll let you invert them if you want!

]]>$$ \frac{1}{4} X \frac{8}{5} = \frac{18}{45} $$

To be completely correct you should cancel out the nines to leave two fifths, but the equation actually works even though the method used to produce it is decidedly dodgy.

This set us wondering whether there are any other equations where this incorrect method produces correct equations. Using single-digit, non-zero numbers (that is, 1 through to 9) on the left-hand side of the equation, how many correct equations can you find that follow this method?

Answer at 9.00 on Monday

]]>One day they came to visit and they sat in a neat line in three chairs while we talked. Being curious people here at Quiz Master Shop Towers (in both senses of the word curious), we wanted to know which girl was which. So we asked them each a question.

We asked the girl on the left "Who is sitting in the middle?" and she replied "Faith".

Then we asked the girl in the middle "Who are you?" and she replied "Charity".

Finally we asked the girl on the right "Who is sitting in the middle?" and she replied "Hope".

We had asked all three girls the name of the girl in the middle, and received three different answers.

So who was sitting where?

The key to solving this is finding Faith, who always tells the truth.

Faith can't be the girl on the left, as she would have had to be lying when she said the middle girl was Faith.

Faith can't be the girl in the middle, as she would have had to be lying when she said she was Charity.

So the girl on the right is Faith.

She says the middle girl is Hope, and that must be correct.

Leaving the girl on the left to be Charity.

]]>One day they came to visit and they sat in a neat line in three chairs while we talked. Being curious people here at Quiz Master Shop Towers (in both senses of the word curious), we wanted to know which girl was which. So we asked them each a question.

We asked the girl on the left "Who is sitting in the middle?" and she replied "Faith".

Then we asked the girl in the middle "Who are you?" and she replied "Charity".

Finally we asked the girl on the right "Who is sitting in the middle?" and she replied "Hope".

We had asked all three girls the name of the girl in the middle, and received three different answers.

So who was sitting where?

Answer at 9.00 on Monday

]]>You have ten sets of ten weights. With one of the sets of weights there was a manufacturing error, and each weight in the set was made ten grams overweight. This means that the set in question is 100 grams overweight in total.

Unfortunately you don't know which is the faulty set, and visually there is no way to determine if a weight is heavier than its counterpart from another set. But you do need to find the faulty set of weights.

Fortunately you have an accurate weighing machine available.

How can you find out which is the faulty set of weights in one weighing?

Select one weight from set one, two weights from set two, and so on. Work out what the total weight should be, and then weight them all.

If the weighing comes out ten grams overweight then the faulty set is set one. If it's 20 grams overweight it's set two that is faulty. And so on.

]]>You have ten sets of ten weights. With one of the sets of weights there was a manufacturing error, and each weight in the set was made ten grams overweight. This means that the set in question is 100 grams overweight in total.

Unfortunately you don't know which is the faulty set, and visually there is no way to determine if a weight is heavier than its counterpart from another set. But you do need to find the faulty set of weights.

Fortunately you have an accurate weighing machine available.

How can you find out which is the faulty set of weights in one weighing?

Answer at 9.00 on Monday

]]>This leaves buying the questions or writing your own.

With the best will in the world it will take you two to three hours to write 60 questions, and probably a lot longer. Three hours to write 60 questions is three minutes to think up and then check each question. We write questions every day, and would love to achieve that run rate!

Or you could buy 60 questions from Quiz Master Shop for £5.94. Three hours work to save less than six quid – and the National Living Wage is £8.72 per hour. Is your time really worth that little?]]>A cricket team needs seven runs to win the game, with only three balls of the game remaining. Both batsmen are on 94.

The fielding side does not concede any extras or overthrows, and yet the batting side wins the match, and both batsmen reach their 100s and are not out.

How?

This would be easy if there were another over after the current over, as one batsman hits a six, and then the other batsman is on strike in the next over and also hits six.

But with only three balls left how do both batsman get on strike without scoring the seven runs to win the game first. Here's how . . .

Off the first ball the batsman scores four runs, so he is now on 98 and three are needed to win.

Off the second ball the same batsman hits the ball into the outfield, and the batsmen run three, thinking that they have won the match. Unfortunately an alert umpire spots that they have run one short, and the scores are level. At least the first batsman can celebrate his 100.

Crucially for this puzzle the second batsman is now on strike, and off the final ball hits a six to win the match, and reach his 100 as well.

]]>A cricket team needs seven runs to win the game, with only three balls of the game remaining. Both batsmen are on 94.

The fielding side does not concede any extras or overthrows, and yet the batting side wins the match, and both batsmen reach their 100s and are not out.

How?

Answer at 9.00 on Monday

]]>The earliest the UK pubs will be opening will be July (and we're no confident that will happen), so we're running four more Virtual Quiz Nights in June, live streaming on our Facebook Page. These will be on 4th, 11th, 18th and 25th June at 7.00PM.

Details are under Events on our Facebook Page, or on various posts on Facebook or on our Twitter Page.

To access the Virtual Quiz Live Stream go to our Facebook Page just before 7.00PM and it will be there.

No Charge, No Prizes, About 40 minutes of fun, interesting and (occasionally) challenging questions.

]]>From what we could gather they had been shooting at a four-ringed target, with seven points for a bullseye, five points for an inner, three points for a magpie, and one point for an outer.

Very oddly, archer A had not hit the bullseye, but had hit all the other rings; archer B had not registered an inner, but had hit all the other rings; archer C had got everything but a magpie; and archer D had not hit an outer, but had registered scores in all the other rings. No one had missed with any arrow.

This seemed very odd, but stranger still they had all scored 30 points.

Intrigued by all this we spent a couple of pints trying to work out the archers' scoring, and we'll leave you with this question: how many bullseyes were scored by the archer who got three magpies?

Archer A, who got no bullseyes, must have shot at least eight arrows, and archer D, with no outers, could have fired no more than eight arrows. So we know they all shot eight arrows.

For archer A we have

i + m + o = 8

5i + 3m + o = 30

With possible scores of five inners (25), one magpie (3) and two outers (2) or four inners (20), three magpies (9) and one outer (1).

Archer B could have got three bullseyes (21), two magpies (6) and three outers (3) or two bullseyes (14), five magpies (15) and one outer (1).

And archer D could only have got one bullseye (7), one inner (5) and six magpies (18).

Archer C got no magpies so we don't need to bother about the scores.

From this, only archer A could have got three magpies, and got no bullseyes.

]]>From what we could gather they had been shooting at a four-ringed target, with seven points for a bullseye, five points for an inner, three points for a magpie, and one point for an outer.

Very oddly, archer A had not hit the bullseye, but had hit all the other rings; archer B had not registered an inner, but had hit all the other rings; archer C had got everything but a magpie; and archer D had not hit an outer, but had registered scores in all the other rings. No one had missed with any arrow.

This seemed very odd, but stranger still they had all scored 30 points.

Intrigued by all this we spent a couple of pints trying to work out the archers' scoring, and we'll leave you with this question: how many bullseyes were scored by the archer who got three magpies?

Answer at 9.00 on Monday

]]>At Middleton Cricket Club there are two prestigious awards given to players, based on their performances in the triangular tournament. There is one trophy given to the player with the best batting average, and another for the player with the best bowling average. These trophies go back almost as far as the tournament itself, and the players are very keen to win them.

[For readers who are not completely up to speed on batting and bowling averages, here is a quick explanation.

A player's batting average is the number of runs they have scored divided by the number of times they are out. If a player scores 100 runs and is out four times the batting average is 25. The higher the batting average, the better.

Similarly, a player's bowling average is the number of runs scored off their bowling divided by the number of batsmen they get out. If a player has 100 runs scored off their bowling and gets five batsmen out the bowling average is 20. The lower the bowling average, the better.]

This season the trophy for the best batting average is closely fought between Sam Slogger and Neil Nurdler, with the winner in doubt until the final match. The players' records against both Easton and Weston have been displayed on boards throughout the season, so both players know how the other is performing.

After the final game has finished the boards are brought up to date, and the two players' records are as follows:

v Easton | |||

Runs | Outs | Average | |

Sam Slogger | 325 | 10 | 32.50 |

Neil Nurdler | 121 | 4 | 30.25 |

v Weston | |||

Runs | Outs | Average | |

Sam Slogger | 288 | 5 | 57.60 |

Neil Nurdler | 535 | 10 | 53.50 |

Sam Slogger is over the moon. He has the highest batting average for the season against both Easton and Weston, and immediately buys all of his team mates a drink in celebration.

But what happens when the prizes are given out at the end of the evening that leaves a bitter taste in Sam Slogger's mouth?

You would think that if a player has the best batting average against both the other teams that they must have the best batting average for the season. Sam Slogger certainly thought so, and he was wrong too.

The overall batting averages are:

Season | |||

Runs | Outs | Average | |

Neil Nurdler | 656 | 14 | 46.86 |

Sam Slogger | 613 | 15 | 40.87 |

Neil Nurdler has scored more runs and been out fewer times than Sam Slogger, and so must have the higher average.

This is an example of Simpson's paradox, where a trend that is evident in individual data sets reverses in when the data sets are combined.

In this particular case, it would appear from the batting averages that Easton is a stronger team than Weston. Sam Slogger played more games against Easton and Neil Nurdler more games against Weston, which is probably why the paradox happens here.

]]>At Middleton Cricket Club there are two prestigious awards given to players, based on their performances in the triangular tournament. There is one trophy given to the player with the best batting average, and another for the player with the best bowling average. These trophies go back almost as far as the tournament itself, and the players are very keen to win them.

[For readers who are not completely up to speed on batting and bowling averages, here is a quick explanation.

A player's batting average is the number of runs they have scored divided by the number of times they are out. If a player scores 100 runs and is out four times the batting average is 25. The higher the batting average, the better.

Similarly, a player's bowling average is the number of runs scored off their bowling divided by the number of batsmen they get out. If a player has 100 runs scored off their bowling and gets five batsmen out the bowling average is 20. The lower the bowling average, the better.]

This season the trophy for the best batting average is closely fought between Sam Slogger and Neil Nurdler, with the winner in doubt until the final match. The players' records against both Easton and Weston have been displayed on boards throughout the season, so both players know how the other is performing.

After the final game has finished the boards are brought up to date, and the two players' records are as follows:

v Easton | |||

Runs | Outs | Average | |

Sam Slogger | 325 | 10 | 32.50 |

Neil Nurdler | 121 | 4 | 30.25 |

v Weston | |||

Runs | Outs | Average | |

Sam Slogger | 288 | 5 | 57.60 |

Neil Nurdler | 535 | 10 | 53.50 |

Sam Slogger is over the moon. He has the highest batting average for the season against both Easton and Weston, and immediately buys all of his team mates a drink in celebration.

But what happens when the prizes are given out at the end of the evening that leaves a bitter taste in Sam Slogger's mouth?

Answer at 9.00 on Monday

]]>It first came to prominence in 1990 as a reader's question to Marilyn vos Savant's column in Parade magazine:

This is loosely based on an actual game show where this was the problem given to contestants.

So, you are on the game show you are asked to pick a door. You do so and the host opens one of the other two doors to reveal a goat. You can stick with your original door or change to the other unopened door. Do you stick? Do you change? Or does it make no difference?

This is known as the Monty Hall Problem after the original host of the game show.

In the three situations below assume that you always pick door one.

- If the car is behind door one the host can choose which door to open to reveal a goat, and if you switch you change from the car to the other goat and lose the car.
- If the car is behind door two the host has to open door three to reveal a goat, and if you switch from door one to door two you win the car.
- If the car is behind door three the host has to open door two to reveal a goat, and if you switch from door one to door two you win the car.

These are the only three possible situations, and in two of them you win by switching. So your odds of winning by switching are 2/3 and of losing are 1/3; you should always switch.

Put another way, the only way you can lose by switching is if you picked the car in the first place, and the odds of picking the correct door from three doors is 1/3.

When this answer was published in the magazine many people refused to believe it correct. Thousand's of them wrote in to complain.

But it is correct, however counter intuitive it might seem.

]]>

It first came to prominence in 1990 as a reader's question to Marilyn vos Savant's column in Parade magazine:

This is loosely based on an actual game show where this was the problem given to contestants.

So, you are on the game show you are asked to pick a door. You do so and the host opens one of the other two doors to reveal a goat. You can stick with your original door or change to the other unopened door. Do you stick? Do you change? Or does it make no difference?

Answer at 9.00 on Monday

]]>A grandmother and grandfather have two grandchildren who were born in the same year. The odd, and seemingly impossible situation, is that the first grandchild to be born has a birthday the day after the second grandchild to be born.

How is this possible?

The explanation is that the two grandchildren were born in different countries, just after midnight and then just before midnight.

For example, the first grandchild is born in Australia in the early hours of the 2nd May. It is still afternoon on the 1st May in Europe, and the second grandchild is born sometime before midnight.

Thus the firstborn's birthday is 2nd May and the second to be born's birthday is 1st May.

As an aside, one of our colleague's thinks they remember this actually happening, and being reported in a newspaper!

]]>A grandmother and grandfather have two grandchildren who were born in the same year. The odd, and seemingly impossible situation, is that the first grandchild to be born has a birthday the day after the second grandchild to be born.

How is this possible?

Answer at 9.00 on Monday

]]>The problem is to make a series of equations containing the numbers two, zero, two and zero (2020), in any order, and any number of common mathematical symbols. The first equation must total one, the second equation must total two, and so on . . . as far as you can go. We think that 30 is a good target, but go as far as you can.

In this puzzle you can combine numbers to make 20 or 22 (or 200 or 2020) so 20 + 20 = 40 is a valid equation. And you can raise to the power of 2 or 0, so 200 ^ 2 = 40,000 is also valid. You must use all four numbers each time - you can't use three!

We asked how long a sequence of equations you could produce - here is a list to 32, some from us and some from our solvers:

NB We've added some brackets that aren't really necessary (if you are familiar with BODMAS), but they make things clearer.

20 / 20 = 1

(2 / 2) + 0! + 0 = 2

2 + 2 - 0! + 0 = 3

2.0 + 2.0 = 4

2 + 2 + 0! + 0 = 5

2 + 2 + 0! + 0! = 6

(2 + 0!)! + 2 - 0! = 7

(2 + 0!)! + 2 + 0 = 8

(2 + 0!)! + 2 + 0! = 9

(2 / 0.2) + 0 = 10

(2 / 0.2) + 0! = 11

(2 + 0!)! + (2 + 0!)! = 12

(2 + 0!)! * 2 + 0! = 13

((2 + 0!)! + 0!) * 2 = 14

(2 + 0!) / 0.2 = 15

(2 + 0! + 0!) ^ 2 = 16

20 - (2 + 0!) = 17

20 - 2 + 0 = 18

20 - 2 + 0! = 19

22 - (0! + 0!) = 20

22 - (0! + 0) = 21

22 + 0 + 0 = 22

22 + 0! + 0 = 23

22 + 0! +0! = 24

((2 + 0!)! - 0!) ^ 2 = 25

(2 + 2)! + 0! + 0! = 26

(2 + 0!) ^ (2 + 0!) = 27

((2 + 0!)! / .2 (recurring)) + 0! = 28

(((2 + 0!)!) / .2) - 0! = 29

((2 + 0!)!) / 0.2 = 30

((2 + 0!)!) / .2) + 0! = 31

2 ^ ((2 + 0!)! - 0!) = 32

]]>We've had such great feedback it would be criminal not to keep doing these!

Details are under Events on our Facebook Page, or on various posts on Facebook or on our Twitter Page.

To access the Virtual Quiz Live Stream go to our Facebook Page just before 7.00PM and it will be there.

No Charge, No Prizes, About 30 minutes of fun, interesting and (occasionally) challenging questions.

]]>The problem is to make a series of equations containing the numbers two, zero, two and zero (2020), in any order, and any number of common mathematical symbols. The first equation must total one, the second equation must total two, and so on . . . as far as you can go. We think that 30 is a good target, but go as far as you can.

In this puzzle you can combines number to make 20 or 22 (or 200 or 2020) so 20 + 20 = 40 is a valid equation. And you can raise to the power of 2 or 0, so 200 ^ 2 = 40,000 is also valid. You must use all four numbers each time - you can't use three!

So how long a sequence of equations can you produce?

Answer at 9.00 on Monday

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

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