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I am after some new puzzles, if you have any please post here.
Ideally this thread should be for puzzles and hints only rather than solutions.
I'll start....
A watermelon weighs 100lbs and is 99% percent water. After a time drying in the sun it is 98% water.
How much water has it lost?
Ideally this thread should be for puzzles and hints only rather than solutions.
I'll start....
A watermelon weighs 100lbs and is 99% percent water. After a time drying in the sun it is 98% water.
How much water has it lost?
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In reply to lowersharpnose:
I like that one - it tricks the reader into making assumptions that the loss will be quite small. However...
I like that one - it tricks the reader into making assumptions that the loss will be quite small. However...
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In reply to lowersharpnose:
I always like the classic:
3 friends share the tab in their local pub. The barman tells them it is £30 so they each put in £10 each.
The barman then realises that the tab should have been £25
He gives each of the friends £1 back each, and pockets the other £2
The friends have now paid £9 each (£27) and the barman has £2 in his pocket
But where is the other £1???
I always like the classic:
3 friends share the tab in their local pub. The barman tells them it is £30 so they each put in £10 each.
The barman then realises that the tab should have been £25
He gives each of the friends £1 back each, and pockets the other £2
The friends have now paid £9 each (£27) and the barman has £2 in his pocket
But where is the other £1???
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In reply to lowersharpnose:
There's a room with a light in it. Outside, are three switches, A, B & C. From outside of the room you can't see if the light is lit. You have to find out which switch controls the light. You can only go inside the room once.
There's a room with a light in it. Outside, are three switches, A, B & C. From outside of the room you can't see if the light is lit. You have to find out which switch controls the light. You can only go inside the room once.
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In reply to lowersharpnose:
If you meet someone and they tell you “I have two children, and at least one of them is a boy”, what is the probability that the other child is a boy?
Instead, if the person says “I have two children and the eldest is a boy” does this change the probability that the other child is a boy? If so, how?
If you meet someone and they tell you “I have two children, and at least one of them is a boy”, what is the probability that the other child is a boy?
Instead, if the person says “I have two children and the eldest is a boy” does this change the probability that the other child is a boy? If so, how?
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In reply to lowersharpnose:
There are 60 random people in the town hall - what is the chance that two of them share the same birthday? (choose the closest answer)
a) 50%
b) 99%
c) 17%
d) 25%
There are 60 random people in the town hall - what is the chance that two of them share the same birthday? (choose the closest answer)
a) 50%
b) 99%
c) 17%
d) 25%
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In reply to lowersharpnose:
A man comes back to his house, turns the radio on, goes upstairs, turns the light on and shoots himself. What happened?
A man comes back to his house, turns the radio on, goes upstairs, turns the light on and shoots himself. What happened?
In reply to john arran:
Loved this one - neat variation on the Monty Hall problem and just as counter-intuitive.
Loved this one - neat variation on the Monty Hall problem and just as counter-intuitive.
In reply to lowersharpnose:
If you randomly select an Indian and a Chinese person, which one is more likely to score a century at cricket?
If you randomly select an Indian and a Chinese person, which one is more likely to score a century at cricket?
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In reply to lowersharpnose:
I keep turning my laptop round in frustration to try and read the answers!
Edit: found them
I keep turning my laptop round in frustration to try and read the answers!
Edit: found them
Post edited at 16:36
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In reply to lowersharpnose:
In the dead of night you're escaping the castle with a Knight, King and Queen. You have a 17 minute head start before the invading army catch you and you reach a rickety rope bridge. Only two can cross it at a time or it will collapse. One of them must be carrying a candle to light the way and you only have one candle. You can cross the bridge in a minute, the Knight is slower, taking 2 minutes, the Queen takes 5 but the elderly king takes 10. Carrying someone would snap the wood of the bridge but anyone can wait at either side. If the army reach the bridge before everyone's across they'll cut the rope. Can everyone cross the bridge and cut the rope before the army arrives?
In summary: Four people to cross taking 1, 2, 5 and 10 minutes respectively. Only two at a time and one must be holding the candle. You have 17 minutes and there are no tricks.
In the dead of night you're escaping the castle with a Knight, King and Queen. You have a 17 minute head start before the invading army catch you and you reach a rickety rope bridge. Only two can cross it at a time or it will collapse. One of them must be carrying a candle to light the way and you only have one candle. You can cross the bridge in a minute, the Knight is slower, taking 2 minutes, the Queen takes 5 but the elderly king takes 10. Carrying someone would snap the wood of the bridge but anyone can wait at either side. If the army reach the bridge before everyone's across they'll cut the rope. Can everyone cross the bridge and cut the rope before the army arrives?
In summary: Four people to cross taking 1, 2, 5 and 10 minutes respectively. Only two at a time and one must be holding the candle. You have 17 minutes and there are no tricks.
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In reply to lowersharpnose:
UKC Christmas Quiz is in preparation for the 25th, a few puzzles planned.
UKC Christmas Quiz is in preparation for the 25th, a few puzzles planned.
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In reply to Rampikino:
You have a 7 minute egg-timer and a 4 minute egg-timer. I want a nine minute egg. How long will I have to wait for my agg to be ready (once you have figured it out)?
You have a 7 minute egg-timer and a 4 minute egg-timer. I want a nine minute egg. How long will I have to wait for my agg to be ready (once you have figured it out)?
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In reply to lowersharpnose:
You have a 12 inch long stick. You can put 4 marks on it that allow you to assess every distance from 1 to 12 in one measure. Where do the marks go?
You have a 12 inch long stick. You can put 4 marks on it that allow you to assess every distance from 1 to 12 in one measure. Where do the marks go?
Post edited at 19:59
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In reply to john arran:
Or instead, you know your new neighbours have two children. You see a boy playing in their garden. What is the probability that their other child is a boy?
Or, you meet someone and they tell you "I have two children and at least one of them is a boy born on a Tuesday". What is the probability that their other child is a boy.
> If you meet someone and they tell you “I have two children, and at least one of them is a boy”, what is the probability that the other child is a boy?
> Instead, if the person says “I have two children and the eldest is a boy” does this change the probability that the other child is a boy? If so, how?
Or instead, you know your new neighbours have two children. You see a boy playing in their garden. What is the probability that their other child is a boy?
Or, you meet someone and they tell you "I have two children and at least one of them is a boy born on a Tuesday". What is the probability that their other child is a boy.
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In reply to lowersharpnose:
There is a pile of twelve coins, all of equal size and a simple balance scales. Eleven are of equal weight. One is of a different weight. In three weighings find the unequal coin and determine if it is heavier or lighter.
There is a pile of twelve coins, all of equal size and a simple balance scales. Eleven are of equal weight. One is of a different weight. In three weighings find the unequal coin and determine if it is heavier or lighter.
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In reply to Robert Durran:
The latter is a real opinion-splitter! Since we're allowed hints on this thread it may be worth considering the difference in probability of having a boy born on a Tuesday if you have one son and one daughter compared to if you have two sons.
> Or instead, you know your new neighbours have two children. You see a boy playing in their garden. What is the probability that their other child is a boy?
> Or, you meet someone and they tell you "I have two children and at least one of them is a boy born on a Tuesday". What is the probability that their other child is a boy.
The latter is a real opinion-splitter! Since we're allowed hints on this thread it may be worth considering the difference in probability of having a boy born on a Tuesday if you have one son and one daughter compared to if you have two sons.
In reply to lowersharpnose:
I presume you mean the coin one. There is an interesting link here - I used to be a member of the University of East Anglia climbing club and we spent w/es in The Lakes, Wales, Peak even Scotland. This meant long hours in transit vans and we found lots of ways to pass the time. One w/e spent at the Rugby MC hut below Craig yr Ysfa was particularly grim weather wise and this puzzle occupied us for many hours in the van then the hut. It was eventually solved half way up Amphitheatre Buttress by myself and ANOther whose name I forget. The soaked bit of paper outlining the solution was nursed to base to be dried out and transcribed.
I presume you mean the coin one. There is an interesting link here - I used to be a member of the University of East Anglia climbing club and we spent w/es in The Lakes, Wales, Peak even Scotland. This meant long hours in transit vans and we found lots of ways to pass the time. One w/e spent at the Rugby MC hut below Craig yr Ysfa was particularly grim weather wise and this puzzle occupied us for many hours in the van then the hut. It was eventually solved half way up Amphitheatre Buttress by myself and ANOther whose name I forget. The soaked bit of paper outlining the solution was nursed to base to be dried out and transcribed.
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In reply to lowersharpnose:
Do you know the elegant solution which generalises to finding the maximum number of coins for which it is possible to find the odd one out and whether it is light or heavy with 3, 4, 5, 6.............weighings?
> That is one of the best and most satisfying puzzles I have ever done.
Do you know the elegant solution which generalises to finding the maximum number of coins for which it is possible to find the odd one out and whether it is light or heavy with 3, 4, 5, 6.............weighings?
In reply to john arran:
By which I presume you mean many people get it wrong!
> The latter is a real opinion-splitter!
By which I presume you mean many people get it wrong!
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In reply to hokkyokusei:
Can you trust the switches to show On or Off?
Can you trust the switches to show On or Off?
In reply to Phil Anderson:
Why is it a variation on the Monty Hall problem? Seems pretty unconnected to me.
> Loved this one - neat variation on the Monty Hall problem and just as counter-intuitive.
Why is it a variation on the Monty Hall problem? Seems pretty unconnected to me.
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In reply to Robert Durran:
I was trained as an engineer and any solution would do as long as it worked but I must have had a bit of a mathematical gene because I always searched for the 'elegant' solution if I could find it. I would really like to know the 'elegant' solution to the weighing problem.
I was trained as an engineer and any solution would do as long as it worked but I must have had a bit of a mathematical gene because I always searched for the 'elegant' solution if I could find it. I would really like to know the 'elegant' solution to the weighing problem.
In reply to keith-ratcliffe:
It would take me a while to reconstruct the details, but it is based on the idea that from three weighings for which the coins used are predetermined, you get a sequence such as RLB (correspoding to: right side down, left side down, balanced). There are 3^3 such sequences so it is possible to distinguish between a maximum of 27 different situations. But it turns out that when you allocate balls to the weighings, it is not possible to use three of the sequences, so we can distinguish between 24 different situations (ie 12 balls either light or heavy). It generalises to (3^n - 3)/2 for n weighings. So 39 balls for 4 weighings, 120 balls for 5 weighings etc. An ad hoc approach for more than three weighings would be a nightmare!
> I was trained as an engineer and any solution would do as long as it worked but I must have had a bit of a mathematical gene because I always searched for the 'elegant' solution if I could find it. I would really like to know the 'elegant' solution to the weighing problem.
It would take me a while to reconstruct the details, but it is based on the idea that from three weighings for which the coins used are predetermined, you get a sequence such as RLB (correspoding to: right side down, left side down, balanced). There are 3^3 such sequences so it is possible to distinguish between a maximum of 27 different situations. But it turns out that when you allocate balls to the weighings, it is not possible to use three of the sequences, so we can distinguish between 24 different situations (ie 12 balls either light or heavy). It generalises to (3^n - 3)/2 for n weighings. So 39 balls for 4 weighings, 120 balls for 5 weighings etc. An ad hoc approach for more than three weighings would be a nightmare!
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In reply to john arran:
If we have a large number of randomly selected parents, each with two children and pick those with one or more boys, then the probability that the other is a boy is 1/3.
But that was *not* the problem posed.
You have to think about how the information was offered. If the speaker randomly chooses a true statement about his children, then, given he said ..one is a boy..
The options are :
B G (one way of making the statement)
G B (one way of making the statement)
B B (two ways of making the statement)
So, out of the four ways of making the statement, two arise from the B B combination, so the probability that the other is a boy is 1/2.
If we have a large number of randomly selected parents, each with two children and pick those with one or more boys, then the probability that the other is a boy is 1/3.
But that was *not* the problem posed.
You have to think about how the information was offered. If the speaker randomly chooses a true statement about his children, then, given he said ..one is a boy..
The options are :
B G (one way of making the statement)
G B (one way of making the statement)
B B (two ways of making the statement)
So, out of the four ways of making the statement, two arise from the B B combination, so the probability that the other is a boy is 1/2.
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In reply to lowersharpnose:
This interpretation makes the problem equivalent to my variation where a child appears in the garden at random.
The only properly correct response to all these problems is to ask precisely how the information given was generated.
> If the speaker randomly chooses a true statement about his children, then, given he said ..one is a boy...........
This interpretation makes the problem equivalent to my variation where a child appears in the garden at random.
The only properly correct response to all these problems is to ask precisely how the information given was generated.
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In reply to keith-ratcliffe:
Let's say they are all off, to begin with.
> Can you trust the switches to show On or Off?
Let's say they are all off, to begin with.
In reply to lowersharpnose:
Like I said - a real opinion-splitter
mainly because the question can be interpreted differently.
Like I said - a real opinion-splitter
mainly because the question can be interpreted differently.
In reply to Bjartur i Sumarhus:
No, no no! The question must be :
"(all other things being equal) the probability of a random person in a given country in which a higher proportion of people play cricket (eg.India) than in another given country (e.g.China), of scoring a century is higher than that of a random person in the latter country (where a lower proportion of people play cricket) ?”
is
a) True
b) Untrue
c) We cannot know
Please, no more....
> If you randomly select an Indian and a Chinese person, which one is more likely to score a century at cricket?
>No, no no! The question must be :
"(all other things being equal) the probability of a random person in a given country in which a higher proportion of people play cricket (eg.India) than in another given country (e.g.China), of scoring a century is higher than that of a random person in the latter country (where a lower proportion of people play cricket) ?”
is
a) True
b) Untrue
c) We cannot know
Please, no more....
Post edited at 22:34
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In reply to Robert Durran:
I have just written out a truth table to look at this and whilst I get that there are different numbers of permutations if you know that 1 boy was born on Tuesday, it feels like that spoof proof that 1 = 2. So to put my mind at rest please could you let me know the answer to the following questions
1. I have 2 children, 1 of the them is a left handed boy, born on a Tuesday. What is the probability my other child is also a boy?
2. I have 2 children, 1 of them is a brown-eyed left handed boy, born on a Tuesday. What is the probability my other child is also a boy?
If the additional info about the boy makes no difference to the probabilities, why should the day of birth? If the additional info about the boy makes a difference, how many additional pieces of info would you need to be 99% certain of my other childs gender?
I have just written out a truth table to look at this and whilst I get that there are different numbers of permutations if you know that 1 boy was born on Tuesday, it feels like that spoof proof that 1 = 2. So to put my mind at rest please could you let me know the answer to the following questions
1. I have 2 children, 1 of the them is a left handed boy, born on a Tuesday. What is the probability my other child is also a boy?
2. I have 2 children, 1 of them is a brown-eyed left handed boy, born on a Tuesday. What is the probability my other child is also a boy?
If the additional info about the boy makes no difference to the probabilities, why should the day of birth? If the additional info about the boy makes a difference, how many additional pieces of info would you need to be 99% certain of my other childs gender?
In reply to lowersharpnose:
I find it beyond belief that there are people mindlessly "disliking" posts in this thread rather than engaging in intelligent querying and discussion!
I find it beyond belief that there are people mindlessly "disliking" posts in this thread rather than engaging in intelligent querying and discussion!
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In reply to Andy Hardy:
Excellent questions! I'll reply properly when I have time later.
> I have just written out a truth table to look at this and whilst I get that there are different numbers of permutations if you know that 1 boy was born on Tuesday, it feels like that spoof proof that 1 = 2. So to put my mind at rest please could you let me know the answer to the following questions
> 1. I have 2 children, 1 of the them is a left handed boy, born on a Tuesday. What is the probability my other child is also a boy?
> 2. I have 2 children, 1 of them is a brown-eyed left handed boy, born on a Tuesday. What is the probability my other child is also a boy?
> If the additional info about the boy makes no difference to the probabilities, why should the day of birth? If the additional info about the boy makes a difference, how many additional pieces of info would you need to be 99% certain of my other childs gender?
Excellent questions! I'll reply properly when I have time later.
In reply to Robert Durran:
I don't think it's mindless as such; it looks rather like you have a stalker. The other disliked posts I spotted seem to have some credible explanation, however tenuous.
> I find it beyond belief that there are people mindlessly "disliking" posts in this thread rather than engaging in intelligent querying and discussion!
I don't think it's mindless as such; it looks rather like you have a stalker. The other disliked posts I spotted seem to have some credible explanation, however tenuous.
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In reply to john arran:
Its human nature, isn't it ? if someone rails against 'dislikes' then someone else will wind them up by giving them. As far as I can see they are pretty arbitary don't really add anything to the thread.
Its human nature, isn't it ? if someone rails against 'dislikes' then someone else will wind them up by giving them. As far as I can see they are pretty arbitary don't really add anything to the thread.
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In reply to GrahamD:
It seems that has now happened - more dislikes have appeared. A bit pathetic. I'd still be genuinely interested to know why I got dislikes for a couple of posts earlier. Disagreement? Unclear explanation? Different interpretation of problem?
> Its human nature, isn't it ? if someone rails against 'dislikes' then someone else will wind them up by giving them.
It seems that has now happened - more dislikes have appeared. A bit pathetic. I'd still be genuinely interested to know why I got dislikes for a couple of posts earlier. Disagreement? Unclear explanation? Different interpretation of problem?
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In reply to Robert Durran:
I find it beyond belief that there are people mindlessly "disliking" posts in this thread rather than engaging in intelligent querying and discussion!
Because disliking a post is easier than engaging in rational discussion?
Feck 'em.
I find it beyond belief that there are people mindlessly "disliking" posts in this thread rather than engaging in intelligent querying and discussion!
Because disliking a post is easier than engaging in rational discussion?
Feck 'em.
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In reply to lowersharpnose:
How many gifts are there in total in the well known song "The Twelve days of Christmas"?
Generalise to n days.
How many gifts are there in total in the well known song "The Twelve days of Christmas"?
Generalise to n days.
1
In reply to Andy Hardy:
To clarify, the interpretation of the problem is the standard: What proportion of all the two child families with at least one boy born on a Tuesday consist of two boys?
The answer is 13/27=0.481. It at first seems unintuitive that this should be different from the answer without the Tuesday bit, which is 1/3=0.333 (Only BB out of the equally likely BB, BG, GB families consists of 2 boys). The obvious question is whether it is significant that the answer is only a little bit les than 1/2.
A table quickly gives the answer as 27/55=0.491, which is even closer to 1/2.
We need to know the proportion of children with brown eyes. If it is, say, 1 in 5, then the answer is 139/279=0.498
The less likely it is for a given boy to meet all the conditions, the closer the answer gets to 1/2. In fact, if the probability of a given boy meeting all the conditions is 1/n, then a bit of thought about the table shows that the answer is (2n-1)/(4n-1) = 1/2 - 1/(8n-2). For the original Tuesday n=7, include left handedness and n=14. For a left handed boy born on a Tuesday on the first of January, n=2 x 7 x 365 = 5110 and the probability is 0.499976.
Here is how to make it more intuitive that it does make a difference:
A BB family has had two shots at having a boy born on a Tuesday, so nearly* twice as many BB families will have a boy born on a Tuesday than will BG or GB families. So say we have 70 BG, 70 GB and 70 BB families. About 10 of each of the BG and GB families will have a boy born on a Tuesday but nearly 20 of the BB families will. So of the nearly 40 families with a boy born on a Tuesday, nearly 20 (nearly 1/2) will consist of two boys.
* "nearly" because a rare family will have two boys born on a Tuesday, in a sense getting more than their fair share and depriving another family of a boy born on a Tuesday. Adding extra information makes such families rarer and rarer and so makes the probability closer and closer to 1/2.
The probability will never exceed 1/2 however much additional information is given!
> I have just written out a truth table to look at this and whilst I get that there are different numbers of permutations if you know that 1 boy was born on Tuesday, it feels like that spoof proof that 1 = 2. So to put my mind at rest please could you let me know the answer to the following questions.
To clarify, the interpretation of the problem is the standard: What proportion of all the two child families with at least one boy born on a Tuesday consist of two boys?
The answer is 13/27=0.481. It at first seems unintuitive that this should be different from the answer without the Tuesday bit, which is 1/3=0.333 (Only BB out of the equally likely BB, BG, GB families consists of 2 boys). The obvious question is whether it is significant that the answer is only a little bit les than 1/2.
> 1. I have 2 children, 1 of the them is a left handed boy, born on a Tuesday. What is the probability my other child is also a boy?
A table quickly gives the answer as 27/55=0.491, which is even closer to 1/2.
> 2. I have 2 children, 1 of them is a brown-eyed left handed boy, born on a Tuesday. What is the probability my other child is also a boy?
We need to know the proportion of children with brown eyes. If it is, say, 1 in 5, then the answer is 139/279=0.498
The less likely it is for a given boy to meet all the conditions, the closer the answer gets to 1/2. In fact, if the probability of a given boy meeting all the conditions is 1/n, then a bit of thought about the table shows that the answer is (2n-1)/(4n-1) = 1/2 - 1/(8n-2). For the original Tuesday n=7, include left handedness and n=14. For a left handed boy born on a Tuesday on the first of January, n=2 x 7 x 365 = 5110 and the probability is 0.499976.
> If the additional info about the boy makes no difference to the probabilities, why should the day of birth?
Here is how to make it more intuitive that it does make a difference:
A BB family has had two shots at having a boy born on a Tuesday, so nearly* twice as many BB families will have a boy born on a Tuesday than will BG or GB families. So say we have 70 BG, 70 GB and 70 BB families. About 10 of each of the BG and GB families will have a boy born on a Tuesday but nearly 20 of the BB families will. So of the nearly 40 families with a boy born on a Tuesday, nearly 20 (nearly 1/2) will consist of two boys.
* "nearly" because a rare family will have two boys born on a Tuesday, in a sense getting more than their fair share and depriving another family of a boy born on a Tuesday. Adding extra information makes such families rarer and rarer and so makes the probability closer and closer to 1/2.
> If the additional info about the boy makes a difference, how many additional pieces of info would you need to be 99% certain of my other childs gender?
The probability will never exceed 1/2 however much additional information is given!
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In reply to lowersharpnose:
My local hospital only does Caesarians on a Tuesday so all my kids are Tuesdays! not relevant I know. Can someone please post some answers as my brain is hurting!
My local hospital only does Caesarians on a Tuesday so all my kids are Tuesdays! not relevant I know. Can someone please post some answers as my brain is hurting!
In reply to keith-ratcliffe:
Beat me to it, was just about to post a version with snooker balls. It is an absolute classic!
CB
Beat me to it, was just about to post a version with snooker balls. It is an absolute classic!
CB
In reply to Robert Durran:
Maybe it's just the way I internalise these sorts of problems, but to me it seems similar in the way the odds change counter-intuitively once a little more information is given.
Saying "at least one of them is a boy" is similar to opening a non-prize door in the Monty Hall Problem.
Given the connection (as I saw ir) I then solved it in the same way, by listing the available options given at each stage.
Maybe it's just the way I internalise these sorts of problems, but to me it seems similar in the way the odds change counter-intuitively once a little more information is given.
Saying "at least one of them is a boy" is similar to opening a non-prize door in the Monty Hall Problem.
Given the connection (as I saw ir) I then solved it in the same way, by listing the available options given at each stage.
1
A young boy and his father are on their way home from soccer practice when a distracted driver crosses the center line and hits them head on. The father dies at the scene of this horrible car accident, but the boy is still alive when the emergency medical technicians arrive.
The boy is transported to the hospital in an ambulance and he is taken immediately into surgery. But the surgeon steps out of the operating room and says, "Call Dr. Baker stat to the operating room. I can't operate on this boy - he is my son!"
The question: Who is the surgeon?
Thought provoking if you don't get it straight away.
The boy is transported to the hospital in an ambulance and he is taken immediately into surgery. But the surgeon steps out of the operating room and says, "Call Dr. Baker stat to the operating room. I can't operate on this boy - he is my son!"
The question: Who is the surgeon?
Thought provoking if you don't get it straight away.
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In reply to lowersharpnose:
Mary's mum has four children. The first child is called April, the second May, the third June.
What is the name of the fourth child?
Mary's mum has four children. The first child is called April, the second May, the third June.
What is the name of the fourth child?
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In reply to john arran:
July was too obvious, so I picked another month at random.
> You need to show your reasoning!
July was too obvious, so I picked another month at random.
In reply to Robert Durran:
in which case ... GOTCHA!
> July was too obvious, so I picked another month at random.
in which case ... GOTCHA!
Post edited at 13:46
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In reply to john arran:
Mary obviously.
> Mary's mum has four children. The first child is called April, the second May, the third June.
> What is the name of the fourth child?
Mary obviously.
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In reply to Pete Dangerous: Plays on people's innately sexist views regarding high powered careers. A good little puzzle.
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In reply to Robert Durran: I thought the obvious solution to the weighing problem was an elegant one. You have lost me!
In reply to blackmountainbiker:
What's the obvious solution!
> I thought the obvious solution to the weighing problem was an elegant one. You have lost me!
What's the obvious solution!
In reply to blackmountainbiker:
What is the light bulb problem?
> This is very much like the light bulb problem, I like it a lot.
What is the light bulb problem?
In reply to Robert Durran:
You might have described it but I couldn't follow you. My solution: You split the coins into 2 piles of 6 and put them on the balance scales. The light pile contain the lighter coin so you retain that pile and discard the other 6. Then split your 6 into 2 piles of 3 and put them on the balance scales. The lighter pile is retained, the heavier discarded. The, this is the best bit, select 2 coins at random from your remaining 3 and put 1 on each side of the scale. Either they will balance, in which case your remaining coin is the lightest or they won't balance and you will find which is lightest.
You might have described it but I couldn't follow you. My solution: You split the coins into 2 piles of 6 and put them on the balance scales. The light pile contain the lighter coin so you retain that pile and discard the other 6. Then split your 6 into 2 piles of 3 and put them on the balance scales. The lighter pile is retained, the heavier discarded. The, this is the best bit, select 2 coins at random from your remaining 3 and put 1 on each side of the scale. Either they will balance, in which case your remaining coin is the lightest or they won't balance and you will find which is lightest.
In reply to Robert Durran:
The one where you have switches and one light bulb in a room you can't see. Requires similar reasoning skills.
The one where you have switches and one light bulb in a room you can't see. Requires similar reasoning skills.
In reply to blackmountainbiker:
You don't know if the odd coin is heavier or lighter though.
> ... My solution: You split the coins into 2 piles of 6 and put them on the balance scales. The light pile contain the lighter coin so you retain that pile and discard the other 6.
You don't know if the odd coin is heavier or lighter though.
3
In reply to lowersharpnose:
A man is heading towards the centre of a field. He knows, with certainty, that when he gets there, he is going to die. Why?
A man is heading towards the centre of a field. He knows, with certainty, that when he gets there, he is going to die. Why?
In reply to EddInaBox: yes you do. There is only one lighter coin so if the 2 coins balance it must be lighter, if they don't you know which is lighter.
In reply to lowersharpnose:
Not really in the spirit of some of the others on here, but this one kept me busy for a long time a while ago. I've not seen it elsewhere. It was a real situation.
I put my favourite 100 music tracks onto a memory stick (or similar) and set the music player in my car to play the tracks randomly. (It was a multi-day trip across Australia). On average, how many tracks (including repeated tracks) would have to play before I had heard all 100 tracks at least once?
The question arose because I had put a particular recent favourite (it was Adele's "Someone Like You") onto the stick and after several days it had still not played.
Martin
Not really in the spirit of some of the others on here, but this one kept me busy for a long time a while ago. I've not seen it elsewhere. It was a real situation.
I put my favourite 100 music tracks onto a memory stick (or similar) and set the music player in my car to play the tracks randomly. (It was a multi-day trip across Australia). On average, how many tracks (including repeated tracks) would have to play before I had heard all 100 tracks at least once?
The question arose because I had put a particular recent favourite (it was Adele's "Someone Like You") onto the stick and after several days it had still not played.
Martin
1
In reply to Martin Hore:
The chance of not hearing one particular track is quite easy. You can simply multiply .99 (chance of not hearing it once) by itself until you get less than .5 (chance of 'probably' having heard it by now). Turns out that after 70 tracks you should probably have heard that track.
Generalising to all tracks is harder - I'll need to think more when I'm not at work!
The chance of not hearing one particular track is quite easy. You can simply multiply .99 (chance of not hearing it once) by itself until you get less than .5 (chance of 'probably' having heard it by now). Turns out that after 70 tracks you should probably have heard that track.
Generalising to all tracks is harder - I'll need to think more when I'm not at work!
In reply to john arran:
But the player is probably not completely random in its track selection, it probably remembers the last few tracks so that it doesn't play the same track twice in quick succession.
But the player is probably not completely random in its track selection, it probably remembers the last few tracks so that it doesn't play the same track twice in quick succession.
In reply to EddInaBox:
That's true in reality, but the question stated they were set to play randomly.
> But the player is probably not completely random in its track selection, it probably remembers the last few tracks so that it doesn't play the same track twice in quick succession.
That's true in reality, but the question stated they were set to play randomly.
In reply to john arran:
Ok, I think I have it.
If x different tracks have been played so far, the chance of the next track being a new one is (100-x)/100
This will remain the case until the next track is actually new, whereupon x will increment.
If something has a probability p then it will take on average 1/p attempts for it to occur.
The average number of attempts to play a new track will therefore be 1/((100-x)/100), or rather 100/(100-x)
X starts at 0 and reaches 99 before there are no new tracks to play, so the whole equation is:
100/(100-0) + 100/(100-1) + 100/(100-2) + ... + 100/(100-99)
Excel tells me that this comes out at 518.7377518, so you will need to wait for 519 tracks before it is probable that every track has been played.
Ok, I think I have it.
If x different tracks have been played so far, the chance of the next track being a new one is (100-x)/100
This will remain the case until the next track is actually new, whereupon x will increment.
If something has a probability p then it will take on average 1/p attempts for it to occur.
The average number of attempts to play a new track will therefore be 1/((100-x)/100), or rather 100/(100-x)
X starts at 0 and reaches 99 before there are no new tracks to play, so the whole equation is:
100/(100-0) + 100/(100-1) + 100/(100-2) + ... + 100/(100-99)
Excel tells me that this comes out at 518.7377518, so you will need to wait for 519 tracks before it is probable that every track has been played.
In reply to john arran:
... which, assuming an average song length of 3.5 minutes, makes for just over 30 hours of music before everything is likely to have played. Your favourite tune, however, would most likely have been played within the first 5 hours, so you were unlucky.
... which, assuming an average song length of 3.5 minutes, makes for just over 30 hours of music before everything is likely to have played. Your favourite tune, however, would most likely have been played within the first 5 hours, so you were unlucky.
In reply to EddInaBox:
Interestingly my player did quite regularly seem to play the same track twice in succession. I guess if it is completely random it should do this once every 100 tracks, so assuming John is right (he gets the same answer as I did) then this will happen several times before all the tracks are played.
What I tended to notice though was occasions when the same track was repeated in a set of say 5. Intuitively that doesn't seem very random, but in fact it will happen quite often.
I believe that because of this people complained to whichever company it was that their random generator "wasn't random enough" so they changed the algorithm so that no track was repeated until all had been played. So not random at all really. And customers were happy again.
Martin
> But the player is probably not completely random in its track selection, it probably remembers the last few tracks so that it doesn't play the same track twice in quick succession.
Interestingly my player did quite regularly seem to play the same track twice in succession. I guess if it is completely random it should do this once every 100 tracks, so assuming John is right (he gets the same answer as I did) then this will happen several times before all the tracks are played.
What I tended to notice though was occasions when the same track was repeated in a set of say 5. Intuitively that doesn't seem very random, but in fact it will happen quite often.
I believe that because of this people complained to whichever company it was that their random generator "wasn't random enough" so they changed the algorithm so that no track was repeated until all had been played. So not random at all really. And customers were happy again.
Martin
1
In reply to john arran:
Hi John
Glad I kept you busy for a few minutes. It took me a great deal longer to work it out, but I eventually got the same answer with the same logic.
Unfortunately, although I was a maths teacher a long time ago (35 years ago), I didn't remember the standard rule that if something has a probability p then it will take on average 1/p attempts for it to occur. So I had to work this out from first principles along the way.
In reality I had a fair few more than 100 tracks so it could, and did, take quite a bit longer.
If anyone's ever driven from Arapiles to Sydney via Broken Hill you'll know how boring bits of the road can get so I wasn't short on thinking time. (Not bad navigation by the way. we really did want to go via Broken Hill!)
Martin
Hi John
Glad I kept you busy for a few minutes. It took me a great deal longer to work it out, but I eventually got the same answer with the same logic.
Unfortunately, although I was a maths teacher a long time ago (35 years ago), I didn't remember the standard rule that if something has a probability p then it will take on average 1/p attempts for it to occur. So I had to work this out from first principles along the way.
In reality I had a fair few more than 100 tracks so it could, and did, take quite a bit longer.
If anyone's ever driven from Arapiles to Sydney via Broken Hill you'll know how boring bits of the road can get so I wasn't short on thinking time. (Not bad navigation by the way. we really did want to go via Broken Hill!)
Martin
In reply to blackmountainbiker:
Known in computer science as a binary search algorithm.
> My solution: You split the coins into 2 piles of 6 and put them on the balance scales...
Known in computer science as a binary search algorithm.
In reply to cb294:
Alright, here's one about snooker. How can you legitimately pot the yellow with each of three successive shots in the same break?
> Beat me to it, was just about to post a version with snooker balls. It is an absolute classic!
Alright, here's one about snooker. How can you legitimately pot the yellow with each of three successive shots in the same break?
In reply to Martin Hore:
I didn't remember the 1/p rule either and I was just about to concede that I didn't have time to spend trying to extrapolate from combinatorial logic with fewer tracks, when that principle occurred to me like a Eureka! moment. It's very satisfying when a complete rethink of approach gives a very much simpler answer.
I didn't remember the 1/p rule either and I was just about to concede that I didn't have time to spend trying to extrapolate from combinatorial logic with fewer tracks, when that principle occurred to me like a Eureka! moment. It's very satisfying when a complete rethink of approach gives a very much simpler answer.
In reply to Oceanrower:
You haven't answered the question. How can you specifically pot three successive yellows in successive shots in the same break?
You haven't answered the question. How can you specifically pot three successive yellows in successive shots in the same break?
In reply to lowersharpnose:
What is the chance that the six lottery balls out of 50 are drawn in ascending order?
What is the chance that the six lottery balls out of 50 are drawn in ascending order?
In reply to John Gillott:
That's strange. I worked out an answer I thought was correct via a reasonably easy way that didn't really need a calculator, then decided to look online for how that kind of problem would ordinarily be solved and found the way suggested was far more computationally difficult but produced the same result. I won't say the result or the method here yet in case others are intrigued too, but suffice to say it isn't as hard as it might seem!
> What is the chance that the six lottery balls out of 50 are drawn in ascending order?
That's strange. I worked out an answer I thought was correct via a reasonably easy way that didn't really need a calculator, then decided to look online for how that kind of problem would ordinarily be solved and found the way suggested was far more computationally difficult but produced the same result. I won't say the result or the method here yet in case others are intrigued too, but suffice to say it isn't as hard as it might seem!
In reply to john arran:
Exactly, there's an overkill method, and a very simple one (which is quite satisfying).
Exactly, there's an overkill method, and a very simple one (which is quite satisfying).
In reply to Martin Hore:
You're confusing time and probability too.
If all the other tracks are 2 hours long, then you'll likely wait a long time for your favourite to come around again.
Most "random" track players are actually "shuffle" so they play all you tracks eventually but just in a random order. If this is the case you won't hear your favourite again until it's played ALL the other tracks.
You're confusing time and probability too.
If all the other tracks are 2 hours long, then you'll likely wait a long time for your favourite to come around again.
Most "random" track players are actually "shuffle" so they play all you tracks eventually but just in a random order. If this is the case you won't hear your favourite again until it's played ALL the other tracks.
In reply to John Gillott:
Seemed fairly obvious to me. First ball is any except the top 5 and then there is just one possibility for each of the succeeding balls.. Have I missed something?
Martin
Seemed fairly obvious to me. First ball is any except the top 5 and then there is just one possibility for each of the succeeding balls.. Have I missed something?
Martin
In reply to Rylstone_Cowboy:
I didn't want to give it away but if you insist.
After a foul, if you can't see a red you have a free shot on a colour, so you pot a yellow as a red.
Follow that with a yellow and, cos there's no more reds left your next ball will be another yellow.
Like I said, much the same way. Using a colour as an extra red.
I didn't want to give it away but if you insist.
After a foul, if you can't see a red you have a free shot on a colour, so you pot a yellow as a red.
Follow that with a yellow and, cos there's no more reds left your next ball will be another yellow.
Like I said, much the same way. Using a colour as an extra red.
Post edited at 12:10
In reply to Martin Hore:
There are many, many, possibilities. 1,2,3,4,5,6 will do, as will 1, 3, 5, 7, 9, 50, as will 31, 39, 44, 46, 47, 49, as will.... well, you get the idea. The only restriction is that the balls come out in ascending order.
> Seemed fairly obvious to me. First ball is any except the top 5 and then there is just one possibility for each of the succeeding balls.. Have I missed something?
> Martin
There are many, many, possibilities. 1,2,3,4,5,6 will do, as will 1, 3, 5, 7, 9, 50, as will 31, 39, 44, 46, 47, 49, as will.... well, you get the idea. The only restriction is that the balls come out in ascending order.
In reply to Oceanrower:
Thats what I thought, but for this to work there can't be any reds left and so there wouldn't be a free ball would there ?
> I didn't want to give it away but if you insist.
> After a foul, if you can't see a red you have a free shot on a colour, so you pot a yellow as a red.
> Follow that with a yellow and, cos there's no more reds left your next ball will be another yellow.
> Like I said, much the same way. Using a colour as an extra red.
Thats what I thought, but for this to work there can't be any reds left and so there wouldn't be a free ball would there ?
1
In reply to Oceanrower:
You can't see a red because there aren't any left, that's not a foul.
> After a foul, if you can't see a red you have a free shot on a colour, so you pot a yellow as a red.
> Follow that with a yellow and, cos there's no more reds left your next ball will be another yellow.
> Like I said, much the same way. Using a colour as an extra red.
You can't see a red because there aren't any left, that's not a foul.
2
In reply to Tony the Blade:
There has to be one red left, which you can't see.
Nominate the yellow as a free (red) ball, and pot both the yellow and the red in the same shot (2 points)
For potting the equivalent of a red, you nominate the yellow as the following colour.
Then you start on the colours with yellow.
There has to be one red left, which you can't see.
Nominate the yellow as a free (red) ball, and pot both the yellow and the red in the same shot (2 points)
For potting the equivalent of a red, you nominate the yellow as the following colour.
Then you start on the colours with yellow.
1
In reply to lowersharpnose:
A monk sets off up a mountain to pray. There's only one way up, a narrow track just one person wide. This winds around the mountain in an irregular way, steep in places traversing in others, and with a 4a move that's not easy in a robe and sandals, though luckily in a gully and not exposed. Sorry, getting carried away.
He sets of at dawn, and only reaches the summit at sunset. He spends a few days on the summit meditating (not much else to do up there.) Then sets off on the descent, again starting at dawn, though it takes him less time than the climb up to return to his starting point.
Will he occupy the same spot on the path at exactly the same time of day on the way down, as he did on the way up? Can you prove this?
[Oh and I automatically dislike any post I read that dislikes dislikes. I do this because it feels right.]
A monk sets off up a mountain to pray. There's only one way up, a narrow track just one person wide. This winds around the mountain in an irregular way, steep in places traversing in others, and with a 4a move that's not easy in a robe and sandals, though luckily in a gully and not exposed. Sorry, getting carried away.
He sets of at dawn, and only reaches the summit at sunset. He spends a few days on the summit meditating (not much else to do up there.) Then sets off on the descent, again starting at dawn, though it takes him less time than the climb up to return to his starting point.
Will he occupy the same spot on the path at exactly the same time of day on the way down, as he did on the way up? Can you prove this?
[Oh and I automatically dislike any post I read that dislikes dislikes. I do this because it feels right.]
Post edited at 17:02
1
In reply to johnjohn:
Think about two monks doing this on the same day................ OK, dawn will be at a different time a few days later but 'm not sure that makes any difference.
Martin
Think about two monks doing this on the same day................ OK, dawn will be at a different time a few days later but 'm not sure that makes any difference.
Martin
In reply to John Gillott:
Back to the drawing board then. I read it as consecutive ascending order. I like the problem though so I'll try again. It will be much harder of course.
Martin
Back to the drawing board then. I read it as consecutive ascending order. I like the problem though so I'll try again. It will be much harder of course.
Martin
In reply to johnjohn:
If monky's mountain is in the northern hemisphere and it's after mid-summer, he could be about halfway around midday.
No.
> Will he occupy the same spot on the path at exactly the same time of day on the way down, as he did on the way up?
If monky's mountain is in the northern hemisphere and it's after mid-summer, he could be about halfway around midday.
> Can you prove this?
No.
In reply to Martin Hore:
Was that the answer? (If not, that was a clue.)
(@lusk - you're not asked necessarily to specify the time that he occupies the same place on the path.)
Was that the answer? (If not, that was a clue.)
(@lusk - you're not asked necessarily to specify the time that he occupies the same place on the path.)
Post edited at 17:32
In reply to john arran:
Your solution looks good to me. I did it by setting up a recurrence relation for the expected waiting time. Same answer.
Your solution looks good to me. I did it by setting up a recurrence relation for the expected waiting time. Same answer.
In reply to John Gillott:
Seems almost trivial. Don't think I'm missing something......
> What is the chance that the six lottery balls out of 50 are drawn in ascending order?
Seems almost trivial. Don't think I'm missing something......
In reply to Robert Durran:
It is, when seen in the right way.
> Seems almost trivial. Don't think I'm missing something......
It is, when seen in the right way.
In reply to Robert Durran:
Here's another nice one, which can be cracked by formal methods at roughly undergrad level, but which can also be tackled by an 11 year old if they spot the method:
Drop two markers at random on a line of finite length, creating three parts. What is the chance that the length of the right hand part is at least twice that of the middle one?
Here's another nice one, which can be cracked by formal methods at roughly undergrad level, but which can also be tackled by an 11 year old if they spot the method:
Drop two markers at random on a line of finite length, creating three parts. What is the chance that the length of the right hand part is at least twice that of the middle one?
In reply to Fredt:
Well done. Spot on.
> There has to be one red left, which you can't see.
> Nominate the yellow as a free (red) ball, and pot both the yellow and the red in the same shot (2 points)
> For potting the equivalent of a red, you nominate the yellow as the following colour.
> Then you start on the colours with yellow.
Well done. Spot on.
In reply to lowersharpnose:
This following one isn't at all hard, but it can be interesting.
Two blue socks and two red socks in a draw, loose. Pick two at random - what is the chance of getting a correct pair?
Now answer the following, quickly:
Two identical pairs of running socks in a draw, labelled Left and Right, loose. Pick two at random - what is the chance of getting a correct pair?
This following one isn't at all hard, but it can be interesting.
Two blue socks and two red socks in a draw, loose. Pick two at random - what is the chance of getting a correct pair?
Now answer the following, quickly:
Two identical pairs of running socks in a draw, labelled Left and Right, loose. Pick two at random - what is the chance of getting a correct pair?
In reply to John Gillott:
The trivial solution leaves me slightly uneasy......... Must be correct though....... I think!
> Drop two markers at random on a line of finite length, creating three parts. What is the chance that the length of the right hand part is at least twice that of the middle one?
The trivial solution leaves me slightly uneasy......... Must be correct though....... I think!
In reply to Robert Durran:
My thoughts entirely. I thought Monty Hall was about to make an appearance for a moment but he never showed up.
My thoughts entirely. I thought Monty Hall was about to make an appearance for a moment but he never showed up.
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