UKC

A puzzle thread. Not too, hard. Certainly not requiring A level maths, probably just primary school maths with good thinking skills.

I'll start.

Solve this quickly...

You buy a bike and a pump for £100. The bike costs £90 more than the pump. How much was the pump?

Now check your answer.

In reply to lowersharpnose:

Would I win the by now traditional £5?

In reply to lowersharpnose:

A fiver

In reply to lowersharpnose:

Don't forget to show your working

In reply to Andy Hardy:

Yes you would, pump in the post.

More puzzles please.

In reply to lowersharpnose:

Apparently children find this easier than adults.
http://www.thepoke.co.uk/2015/03/25/number-parked-car/

In reply to marsbar:

Ahah. Thanks.

There is a surgeon who works at the Northern General who has a brother who is a surgeon at The Children's Hospital. But no surgeon at The Children's Hospital has a brother who is a surgeon at The Northern General.

What is going on?

...No answer necessary. I am just collecting puzzles, not answers.

In reply to lowersharpnose:

Obvious to me,but I won't spoil it.

In reply to lowersharpnose:

easy

In reply to lowersharpnose:

Everyday sexism

In reply to lowersharpnose:

This is probably stretching the limits of simple, but I watched this recently and found it quite interesting

An Impossible Bet
youtube.com/watch?v=eivGlBKlK6M&

And the solution:
youtube.com/watch?v=C5-I0bAuEUE&

In reply to lowersharpnose:

This inscription was found carved into a wooden post. What does it mean?

TOTI
EMUL
ESTO

In reply to lowersharpnose:

Make the number 24 'Countdown style' by using each of the numbers 1, 3, 4, and 6 once each, along with any of the arithmetic operations + - * /

In reply to lowersharpnose:

What is the next group of numbers in this sequence
1
11
21
1211
111221
312211

In reply to lowersharpnose:

Write out the expansion of the expression
(a-x)(b-x)(c-x) ... (z-x)

In reply to lowersharpnose:

Construct a regular pentagon by folding a rectangular strip of paper

In reply to lowersharpnose:

Construct a 3-4-5 triangle by folding a square of paper

In reply to Paul Robertson:

At some point you have (x-x) so your expansion =0

In reply to lowersharpnose:

Multiple Choice: If you choose an answer to this question at random, what is the chance you will be correct?
A) 25%
B) 50%
C) 60%
D) 25%

In reply to Paul Robertson:

Thanks all. Just what I am after.

In reply to lowersharpnose: Olden but golden...

Three friends order identical meals in a restaurant at £10 each. When they pay, the manager tells the waiter that those meals are on special, 3 meals for £25 so give the diners £5 back. The waiter is dishonest, and only gives them £1 back each and puts £2 in his pocket.

This means they have paid £9 each. £9 X 3 = £27. The waiter has £2 in his pocket. £27 +£2 = £29. Where is the missing £1?

In reply to lowersharpnose:
This one needs to be drawn to make it easier

Four prisoners are buried up to their necks in the ground. Three in a row , one behind the other, all facing the same direction (so only one at the back can see the other two, and the middle guy can only see the one in front, and the front guy can't see anyone), and one on the otherside of a brick wall (so out of sight of the other three, but they know he is there). All fo them are wearing a hat. There are two white hats and two black hats, and the prisoners know this.

The guard decides to play an evil game and tells them that one of them has to tell him which colour hat they are wearing, or he will shoot them all. They cannot guess, and he gives them 5 seconds for one of them to decide.

If it can be done, which prisoner can know with certainty which colour hat he is wearing?

In reply to Bjartur i Sumarhus:

Nice phrasing. I like the idea of not being told what prisoner has what hat on.

However, it will be either the middle guy or the one furthest from the wall ( in the case where the two guys in front of him have the same colour hats on)

In reply to Bjartur i Sumarhus:



I can't answer that, because I am now dead!

In reply to lowersharpnose:

Yes, if the guy at the back says nothing, then the middle guy realises that he must have a different colour hat to the guy in front of him...so he can know with certainty

In reply to Bjartur i Sumarhus:

http://spikedmath.com/comics/445-three-logicians-walk-into-a-bar.png

In reply to lowersharpnose:

Sort of similar:

There are 100 prisoners, buried in a row like before and each has a black or white hat put on them (with the total number of each type of hats unknown to the prisoners). Starting at the back of the row the guard asks each prisoner in turn what colour hat they have on, and shoots them if they get it wrong. The prisoners ahead can hear what all the previous answers are, and if the prisoner got shot.

If the prisoners have the chance to decide on a strategy beforehand (and are all willing to cooperate!) what's the maximum that can stay alive on average?

In reply to marsbar:

What's the number under the car?

I failed maths GSCE (Twice)

Did very well in French though.

In reply to I can eat 50 eggs:

look at them upside down

In reply to lowersharpnose:

Find all pairs of natural numbers X and Y that satisfy the equation

x^3 - y^3 = 721

CB

In reply to Paul Robertson:


Is this correct? I'm stumped, so I decided if I could prove the opposite, and I think I've ruled it out. You need to get to 3*8, 2*12 or 4*6.

You can't make 8 from 1,4,6 +-/, similarly you can't make 4 from 1,3,6 or from 1,3,4. You can't make 2 and still be left with 12 from the others.

In reply to Philip:

I think I came to a similar conclusion. I don't think 24 can be done (at least in base 10) but would love to be proved wrong.

In reply to EdH:




This is a great problem, because the answer is the same however many different colours of hat there are. But it is interesting that, although the two colour version is relatively accessible to the layman, only those with a reasonably mathematical background ever get the general version (even though the two colour solution is just a special case of the general solution). A bit like generalising the proof that root 2 is irrational to all non-square numbers!

In reply to GrahamD:

It can be done. Just persevere.

In reply to Philip:

6?(1?3?4) = 24
Replace each ? with one of + - / *

In reply to Paul Robertson:

Cheers.

In reply to Paul Robertson:


I've done it but used brackets. Is that allowed?

In reply to Robert Durran:

Yes, brackets are allowed.

In reply to Paul Robertson:

13112221

In reply to ebygomm:

Only is repalcing "brother" with "sister" makes the puzzle easier.

In reply to cb294:

There ae 2 solutions. I wonder if 721 is the smallest number which gives more than 1 solution.......

In reply to Sink41:

That is brilliant. Very glad I gave up and watched the solution, because there is nko way I would have thought of it!

In reply to Robert Durran:

I wondered that too.
Playing with Hardy's famously uninteresting number 1729 you can get:
12^3 - 10^3 = 9^3 - 1^3 = 728

I haven't found anything smaller.

In reply to Paul Robertson:

Yes, I played with it as well; if there are two solutions then they generate a number which can be expressed in two different ways as the sum of two cubes.


Nor me!

In reply to Robert Durran:


Correct.


Don´t know, tried to play around to find a general approach to find the minimum, but gave up due to brain overheating. Been a while since I have been doing proper maths.

CB

Coincidentally the numerical answer to this is the same in UK currency as US, despite different currency units.

What is the largest amount of money you can have in coins and still not have change for a pound (or dollar) ?

In reply to Philip:

What does change for a dollar mean? (unless the answer is 99 cents)

In reply to Philip:

I'm struggling with this my two answers don't match and aren't actually that close.

In reply to Andy Hardy:


Thought about this too. One interpretation would be that you could have an unlimited number of £2 coins and still not be able to provide change for £1, but I guess this is not what was meant.

CB

In reply to cb294:

I'd not even thought of the £2 coin to be honest. I have two totals both over a pound/dollar which I think are correct but that don't match.

In reply to JackPalmieri:

Yes, I got £1.43 and $1.19, ignoring £1 and 2 coins.

I still can't get the countdown question though, unless I put 2 ** next to each other.

In reply to stevieb:


Not necessary. In fact you don't need * at all.

In reply to Paul Robertson:

Can you use an operator more than once?

In reply to Lusk:

Yes, same as on Countdown.

In reply to lowersharpnose:

What the largest difference possible, between the Hypotenuse and the shortest side of a right angled triangle.

Where every side is a whole number e.g. 3,4,5 triangle.

In reply to Paul Robertson:


Thought so, job done!

In reply to krikoman:

Three adjacent whole numbers? Otherwise there is no limit ,

CB

In reply to cb294:


Not three adjacent whole numbers. There has to be a limit though to make a right angled triangle, hasn't there?

It's the highest ratio Hypot to shortest side, we're after.

If it does go on for ever, what the greatest ration for a Hypot. less than 1000?

In reply to krikoman:

I don't see why there is a limit. There are an infinite number of Pythagorean triples.

11^2 = 121

121 = 60 + 61

61^2 = 60^2 + 11^2

I am I missing something?

In reply to krikoman:

961 = 31^2

so,
481^2 = 480^2 + 31^2

481/31 = 15 and a bit

In reply to lowersharpnose:



There is no limit, you and CB294 are correct (of course).
But using whole numbers for all sides the highest, under 1000 is 43 924 925 with the ratio of 21.51162791

The previous being 41 840 841 Ratio = 20.51219512


What's the formula for finding the next?

In reply to krikoman:

I am not sure what 43 924 925 is?

It is not a square number.

And it is bigger than 1000^2.

In reply to lowersharpnose:


The three sides are 43, 924 and 925

In reply to Robert Durran:

Ah, thank you.

In reply to krikoman:

Ok, ratio not difference.

Makes more sense, but again I don´t think there is a limit (but then I am not a number theorist or even a mathematician).

CB

In reply to Robert Durran:


give us a clue !

In reply to GrahamD:

Say what you see.

In reply to GrahamD:
Think of the catchphrase for Catchphrase...

Edit ^beat me to it

In reply to cb294:




Correct, there is no limit. There's even a formula to give you the numbers!!

In reply to BStar:


errr - a cultural reference which unfortunately is lost on me. I haven't seen it

In reply to stevieb:



I pinched the puzzle from a US site, where they don't have common 1 and 2 dollar coins, so in the uK version you can't count £1 and £2 obviously as they aren't change for a £1.

The $1.19 is correct (3x25,4x10,4x1), but I thought it was also £1.19 in the UK (50p, 20p,20p,20p,5p,2p,2p)- how do you get £1.43



For the countdown one I was stuck until I realised the original phrasing didn't require all 4 operators to be used. I was expect it to be ((-A+B)*C)/D to use each, but the answer actually uses minus and divide (twice).

In reply to Philip:

50p, 20p, 20p, 20p, 20p, 5p, 2p, 2p, 2p, 2p

In reply to Philip:

Hi
Yes, as BStar has put for £1.43.

I've got the countdown one now thanks. I had worked out fractions were probably involved but it still took me a while to get the sum

In reply to stevieb:
Yes, finally got the countdown one as well.

Here is a related one:

Construct natural numbers from four 3´s, all mathematical operators allowed.

E.g.,

1 = 3/3+3-3

2 = 3/3 + 3/3

3 = 3^3 /(3*3)

4 = ....

Which is the smallest real number that cannot be constructed in this way?

What is the largest number you can generate like this (although I dont think this has a solution, as you can think of notations where you can add operator symbols ad infinitum)?

CB


edited for typos

In reply to Paul Robertson:

Which, if any, of these statements are true?

1) One of these statements is false.
2) Two of these statements are false.
3) Three of these statements are false.
4) Four of these statements are false.
5) Five of these statements are false.
6) Six of these statements are false.
7) Seven of these statements are false.
8) Eight of these statements are false.
9) Nine of these statements are false.
10) Ten of these statements are false.

In reply to lowersharpnose:

1-9 are true, as you did not say "One and only one of these statements is false".

CB

In reply to cb294:

I haven't got pencil and paper out yet, but I'm stuck at 11

In reply to cb294:

OK. It would read better as 'Exactly three of these statement are false.' etc.

In reply to Paul Robertson:

11 has several solutions. You need to generate two using two of the threes.

This will help you with lots of other numbers as well,

CB

In reply to Paul Robertson:
If you are allowed !, then it's easy.

EDITed to remove answer

In reply to lowersharpnose:

Then 9 is true,

CB

In reply to lowersharpnose:
Yes, that works, even without getting to 9+2. Again, all operators are allowed, exponents as well.

CB

In reply to Paul Robertson:





Easy :--- 2 (atomic number 22 ) (12 letter of the Greek alphabet) LL 2

In reply to mgco3:


I am none the wiser for that!

In reply to lowersharpnose:
Ok, factorials help a lot.
I'm stuck on 22 now

In reply to mgco3:


I think you have it, but your answer is more cryptic than the original question!

In reply to cb294:

Exponents.

So, would a legit solution to 15 be (3! + 3^2) * 3/3 ?

I know (3+3) * 3 - 3 works.

In reply to Paul Robertson:
to tie mules to

2 (TI - Titanium, atomic number 22) (Mu - 12th letter of the greek alphabet) LL's 2

I can understand the most complex of puzzles but haven't a clue when it comes to understanding women!!! But then, who does?

(Gets coat)

In reply to lowersharpnose:

Gargh, finally solved the countdown one; it took about 3 days! Good question.

In reply to lowersharpnose:

No, as you use a 2. The only numbers symbols to be used are four threes.

Exponents are allowed, though, so you could write 3! +3^3 / 3 =15

23 is the first difficult number.

CB

In reply to mgco3:

still cant get this

In reply to lowersharpnose:

Whilst rooting around in the loft with your grandmother you have found a set of her old kitchen scales. They work on the old-fashioned balance system. Unfortunately all the weights that were originally with them have been lost.

You want to be able to weigh any whole number of grammes up to 40g.

What is the minimum number of weights that you will have to buy to do this?

(An interesting pattern emerges when you get the right answer...)

In reply to Paul Robertson:

Cheers for the classic countdown, took about 15 minutes and I've done it before.

In reply to AdrianC:

6?
Won't list them just now.

In reply to Lusk:

Less than 6.

In reply to AdrianC:

Fewer...

I can do it with five weights, don´t see an obvious solution for doing it with four.

CB

In reply to AdrianC:
You can use 4 in some silly combination

2g ,4g, 8g, 16g maybe. This gives most even numbers if you weigh twice and add. Which if you know its less than one and more than another then the wieght is in between. However this only works for some numbers and is a bad solution. The other way round this would be to weigh to a whole number and then split it in half and add this to 20g to reach any value up to 40g. This wont work for some solutions is x has 2 as a factor. However by dividing and splitting more times it is still possible.

Using dividing and splitting you can prob get this down to less than 4 but i cant be bothered right now to work it out.

In reply to cb294:

OK, found the solution with four weights.

CB

In reply to Bjartur i Sumarhus:




I remember puzzling over this one over 30 years ago when I was a child in hospital. It took me ages to spot the sleight of hand involved, and taught me a valuable lesson about mis-direction.

In reply to Le Chevalier Mal Fet:

The restaurant has £25 in the till (which is correct), and the customers have paid £27. The customers' (not the restaurant's) missing £2 is in the waiter's pocket. All v confusing.

In reply to Le Chevalier Mal Fet:

Extra note. The waiter simply lied about the discount/how much he should give back to them. I.e £3 rather than £5. He kept the difference.

In reply to Gordon Stainforth:

That's the beauty of this puzzle: when it's explained it sounds perfectly obvious, but when it's explained in the original way again it still sounds just as baffling. Sometimes just explaining it doesn't really help understand what was wrong, just what is now right!

In reply to john arran:

The really fascinating, tricky bit is explaining precisely the way the original wording misled. I don't think I've quite nailed it as succinctly as I could. Language can be a very slippery thing.

In reply to Gordon Stainforth:

*Possible spoiler*


I think the key thing in the waiter problem is timing. There's a time in the story where the diners really have paid out £30. Later on, when £1 has been returned to each of them and the waiter has £2 the situation is different. They've paid out £27 and the waiter has £2 with £25 in the till. The trick in the question is taking pieces of correct data from different situations and implying that they should match up when there's no reason for them to do so.

In reply to cb294:

Four is correct (as is fewer, thank-you!) I still wonder about the numerical pattern in the weights you need and can't quite see why it's there. Any thoughts?

In reply to Matt Schwarz:

it reads to tie mules to ..

In reply to cb294:

cant work out less than 6. does 5 or 4 need the amount to be split then weighed more than once?

In reply to Matt Schwarz:

The trick is that is is a balance, i.e. you can add weights on either side. In terms of numbers, you need to find the smallest set from which you can generate all natural numbers up to 40 by addition or subtraction of its constituents.

CB

In reply to cb294:

Aye, I can get 5 in my head, have to get the pen & paper out to work out 4!
I'm assuming the item being weighed can't be split.

In reply to AdrianC:

Hard to say anything without giving the solution away, but I don´t really understand it either.

However, the pattern clearly continues, as the potential fifth weight that would allow you to maximally expand the weighing range would be 81...

CB

In reply to cb294:


Working in base 3 you can write out any natural number using 0s, 1s and 2s
When you do that you are representing a number as a sum of powers of 3
The problem is the 2s
If you could find a way of turning the 2s into -1s then you would have a recipe for representing any number as a sum of powers of 3 (possibly multiplied by -1)

Now note that (in base 3):
2 = 10-1
20 = 100-10
200 = 1000-100
etc.

So, putting all this information together for the example of 20 (decimal) and working in base 3:

20 (decimal) = 202 (base 3)
202 = 200 + 2 = (1000-100)+(10-1) = (1000+10)-(100+1)

26 (decimal) = 222 (base 3)
222 = 200 + 20 + 2 = (1000-100)+(100-10)+(10-1) = 1000-1

Does that help?

In reply to Paul Robertson:

Tie an overhand knot in the strip of paper and gently tighten it pressing it flat as you go.

In reply to Paul Robertson:

Thanks! I found the solution intuitively, by expanding from the smallest two, then three weights.

As I understand your argument, you can construct any natural number from a sum of powers of another natural number (proof? Appears obvious, though. Writing numbers out in binary, decimal or hex does just this).

Since 2x3^n =3^(n+1) - 3^n, any power of three thus needs only appear 1, 0, or -1 times (i.e., the weight can be put on the other side of the balance). For powers of two you would also only ever need 0 or 1 weights, but more different powers.

CB

In reply to cb294:
Yes.
The number system is called "balanced ternary". It was used in the design of a mechanical calculator built in the 19th century.
http://www.mortati.com/glusker/fowler/ternary.htm

In reply to Paul Robertson:

Interesting link, thanks!

CB

In reply to Paul Robertson:


A friend gave me a very similar / essentially identical one:

Make 27 using 1, 3, 4 and 9 following the same rules.

In reply to lowersharpnose:

Albert and Bernard just became friends with Cheryl, and they want to know when her birthday is. Cheryl gives them a list of 10 possible dates.

May 15 May 16 May 19

June 17 June 18

July 14 July 16

August 14 August 15 August 17

Cheryl then tells Albert and Bernard separately the month and the day of her birthday respectively.

Albert: I don’t know when Cheryl’s birthday is, but I know that Bernard does not know too.

Bernard: At first I don’t know when Cheryl’s birthday is, but I know now.

Albert: Then I also know when Cheryl’s birthday is.

So when is Cheryl’s birthday?

In reply to lowersharpnose:

Let me introduce you to my 3 children. Here is Tom, here is Fred - what's the probability of my other child being a boy?
(no funny business - Tom and Fred are male)

In reply to hang_about:



Quite low, otherwise you would likely have said "Let me introduce you to my 3 boys"

In reply to hang_about:

1/4?

which UKC thinks is all uppercase!

In reply to Bjartur i Sumarhus:

Wondered about this one, too. I almost think that there is no clean solution. Here is my reasoning:

A knows the month and concludes that the day alone cannot be enough for B to know C´s birthday.

Thus, May and June are excluded, otherwise B could have known right away (if given a unique day like 18th or 19th).

B thus concludes that May and June are out. This is now sufficient for him to know the exact birthday from the day alone.

Thus, B was not told the 14th, as this could still be July and August. However, B would at that point know the month regardless of whether he was told 16th (July), 15th or 17th (August).

I don´t see how this information is now sufficient for A to exclude two of these three dates,

What am I missing?

CB

In reply to Andy Hardy:


Unless, of course, one of them was born on a Tuesday

In reply to cb294:



You're thinking back to front We are told that A can work it out therefore he must be able to exclude two of the three days. He knows the month - if it was August he couldn't know whether it was the 15th or 17th it so the fact that he can work it out implies it must be July 16th.

In reply to tom_in_edinburgh:

Just realized this too. Totally got off track, the task is not to know whether A could deduce the date, but rather whether an outside observer can deduce the date from the fact that first B and then A could.

CB

In reply to hang_about:

1/2.

Disabuse me.

In reply to lowersharpnose:



There are at least two boys.
The birth order is either BBB, BBG, BGB or GBB
In only one of these four equally likely possibilities are there three boys.

In reply to Robert Durran:




Robert, I don't really see how you can conclude this; you seem to be assuming that the father has the strategy of deliberately revealing any boys before any girls (rather like Monty Hall deliberating revealing a goat). But we're not given this information.

Put another way, the four options BBB, BBG, BGB, GBB are not equally likely possibilities, once we are given the information that two randomly selected children are boys. As common sense suggests, this information increases the probability of BBB to 1/2.

P(BBB | BB) = P(BB | BBB) * P(BBB)/P(BB) = 1/2
P(BBG | BB) = P(BB | BBG) * P(BBG)/P(BB) = 1/6.

In reply to Robert Durran:

To make sense of the question, we have to assume that the parent is asking a random question (but truthful) question about a random selection of his own children.

If he has three boys, then he could have asked that question three ways - it would have worked with any of the three picks of two from three.

Tom & Fred
Tom & Henry
Fred & Henry

If he has two boys and a girl, there is only one way he could ask that question(if Henrietta had been one of the two, he could not have asked that question)

Three ways of getting one quarter (your BBB) is the same as one way of getting the three-quarters (your other combinations). Hence 1/2.

I am a bit rusty with my probability terminology.

In reply to crossdressingrodney:


We are not told that he selected two children at random. In the absence of information about how the children were selected, I interpreted it (I thought in the conventional way) as simply giving the information that at least two of the children are boys (ie what proportion of three children families with at least two boys will in fact have three boys). I do agree that the way in which the children are selected effects the answer though (something few people are aware of). If the problem is posed as "The family which just moved in next door has three children and at least two of them are boys. What is the probability that all three are boys", then your response should be "how did you come across this information". If the response is that the parents said they had at least two boys then the answer is 1/4, but if the response is that two boys appeared in the garden, then the answer is 1/2.

In reply to Robert Durran:

If the response is that the parents said they had at least two boys then the answer is 1/4

I don't think it is. If there are two boys and a girl, then the statement ' We have at least two boys' can only be offered about the one pair of boys. If there are three boys then the statement can be offered about any of the three possible pairs of boys. This brings the odds back to evens.

1/4 is the right answer to a question like : Considering families with three children, that have either two or three boys. What are the odds that a family picked at random will have three boys? Answer 1/4.

But that is not the problem posed.

In reply to Robert Durran:

I think we agree on the maths. I had a mental image of children hurtling around, and the father introducing the first two that happened to fly by -- at least that is how I introduce my children...

In reply to lowersharpnose:


the problem as you originally posed it was, as I explained, ambiguous, but I think the conventional interpretation in the absence of further information is "what proportion of three children families with at least two boys do in fact have three boys".

In reply to Robert Durran:

Sorry to have disappeared - hospital visits then the joy of Principal Component Analysis.
Robert's interpretation is what I meant. Random selection of children. I have none - which is probably a good thing.
Off to sleep and dream of eigenvalues.

In reply to hang_about:

The answer to the question you did pose is 1/2.

You have ignored the fact that there are three ways of picking two boys from three boys and only one way of picking two boys from two boys and a girl.

In your scenario, the parent of Fred and Tom either has a third child that is a boy or a girl, say Henry or Helen.

If there are three boys, then there are *three* ways the parent can make a statement like, 'I have two boys, what is the chance the third is a boy'

He could point out Fred & Tom or Fred & Henry or Tom & Henry.

The weighted probability is 3 * 1/8 = 3/8

If he has two boys and a girl, then there is only one way he could make the same statement, Fred & Tom. The odds of having two boys and a girl is 3/8. So the probability that he could make the statement this was is 1 * 3/8 = 3/8

Which is the same.

In reply to hang_about:


That was not my interpretation. My interpretation was that all you know is that at least two of the children are boys but you do not know how they were selected.

My interpretation is equivalent to a game where three coins are tossed together until at least two of them are heads and you win if, in fact, all three are heads. Your chances of winning are 1/4.

In reply to lowersharpnose:

Can someone explain the birthday to me I'm baffled

In reply to Pesda potato:
A says he does´t know the birthday, but he is certain the neither does B. Thus, the month cannot be May or June, because in this case it could be possible that B was told either the 18th or 19th, which are unique.

From this, B concludes that A must have been told July or August. As long as B was not told the 14th, this is enough to identify the birthday (August 15th or 17th, or July 16th). As B says he knows the date, 14th of either month is excluded.

A then claims he also knows the date. Had A been told August, he would have no means of knowing whether B was told 15th or 17th. He therefore must have been told July, as only the 16th is left.

CB

How many people do you need, for there to be a greater than 50% probability that at least 2 of those people share the same birthday? (Assume no one is born in a leap year)

In reply to lowersharpnose:



Mathematics aside the sex ratio in humans is not 50% (more boys born overall) and even more importantly, not independent: The probability of a second child having the same sex as the first is higher than 50%.

It is not known for certain why this has evolved in this way in humans. The baby boy excess may potentially compensate a higher death rate in young males, but it is even less understood how this sex ratio distortion may be achieved.

However, such phenomena are seen all over the animal kingdom, and in some cases understood in molecular detail. In flies there is a phenomenon called meiotic drive that introduces a bias as to which sperm (X or Y containing) mature more effectively, the bias is thus introduced in the fathers.

A similar mechanism in humans could explain the observed excess of same sex siblings.

Thus, the answer will be a little bit above 1/2.

Consider yourself disabused....

CB

In reply to lowersharpnose:

Saw another video recently, beaker ball balance problem that may interest people on here:

youtube.com/watch?v=QD3hbVG1yxM&

In reply to Robert Durran:



I have a dim memory of this one, and that the naming of the boys matters....

You had BBB as the only way to get three boys looking at it in terms of birth order.

But, since we have names for them, we can say there are six ways of having three boys, from a birth order perspective:

Tom, Fred, Third Boy
Tom, Third Boy, Fred
etc, etc...

My memory is that when we do this the answer to the original question comes out at 1/2.

However, I'd need to check this!

In reply to John Gillott:

I deliberately avoided using terms such as 'third boy' as it implies birth order. If children are introduced in age order then the probability of the third child being a boy is 1/2 (OK slightly more boys but that's country dependent and I think planes on treadmills are also involved). If you don't know anything about the birth order it becomes a population problem.

In reply to hang_about:

Yes, I know. But you did name them. So, I'm using 'Third Boy' to mean the boy that isn't named Tom or Fred. In this sense the boys are not interchangeable.

In reply to John Gillott:




I think this is equivalent to picking 2 of the three children at random and finding that they are boys. I'm not sure that naming them actually changes anything except as a means of keeping track of them.*

It is still undoubtedly true that 1/4 of all three children families with at least two boys will have three boys

*If I dimly remember rightly intrinsically indistinguishable particles such as electrons behave differently and it is meaningless to label them so the statistics works differently to, say, "identical" marbles But quantum mechanics is weird.........

In reply to Robert Durran:

I think your birth order approach is fine, but given the way the question was posed we get 1/2 as the answer when we are introduced to two named children, Tom and Fred. This was the original puzzle:

'Let me introduce you to my 3 children. Here is Tom, here is Fred - what's the probability of my other child being a boy? (no funny business - Tom and Fred are male)'

Since we have met the two boys and we know their names, individually that is, they are not exchangeable. There are six ways this specific three child family could have been created if the third, unseen, child is a boy, and six if it is a girl.

In reply to John Gillott:



I am afraid that I cannot see how knowing the names of the two boys effects the problem. We still do not know how the two children were selected. At random? As many boys as possible? The two eldest children? In the absence of this knowledge, the problem is ambiguous and the only thing we know for certain is that the family includes at least two boys.

In reply to Robert Durran:

You and your wife have produced two boys. If you have a third child , what is the chance it will be a boy?

In reply to lowersharpnose:


1/2 (assuming that the sex of each child is independent and the probability of any one child being a boy is 1/2)

In reply to lowersharpnose:

You have two jugs, one of water, one of wine.
Scoop a large ladle of the wine, pour it into the water, and mix it in.
Now take a ladleful of that mixture, and pour it back into to wine jug.

Now is there more wine in the water, or more water in the wine, or equally contaminated/diluted?

In reply to Robert Durran:

Ok, so you get 1/2 as an answer to "you and your wife have produced two boys. If you have a third child , what is the chance it will be a boy?"

What about "My wife and I have three children. Fred and Tom and another child. What are the odds the other child is a boy?"

(I am wondering how similar I can make the wording and still have you giving different answers).

In reply to Robert Durran:

We are focusing on the fact that Tom and Fred are boys, but we could just say, as the original problem did, here are two of my kids, Tom and Fred, and I'm now going to introduce you to my third kid. Or, if you prefer, let the kids themselves randomise it in a way: 'hey, look, the kids are coming, that's Tom in the lead, then there's Fred, who is coming next?'

By specifying that a particular child is Tom and another particular child is Fred, that is them tagged and taken out of the equation. We can then ask anything we like about the third child. Will they be a red head or blonde? Dressed or naked? Human or alien? Boy or girl? We have no reason to presume anything so we go with 50:50 on the latter question.

Of course, if they were introduced in a way determined by the sex of children then you might be right, but that would be an odd puzzle. Most other things, including birth order, leave it at 50:50 for the sex of the last child.

In reply to Robert Durran:


We can also ignore the names and do a population sampling approach to get 1/2.

Of all the three child families containing at least two boys, 1/8 will have three boys and 3/8 will have two boys and one girl (the former has to be BBB, the latter can be BBG, BGB or GBB, focusing on birth order).

So, if we see two boys run out of a house that we know contains three children, we know that there was a 3/8 chance that the house had two boys and a girl in it to start with, and a 1/8 chance that it had three boys in it to begin with.

But, the puzzle doesn't ask for that answer. It asks for the odds on the sex of the child left in the house. To get to that we have to make a further comparison, which involves answering these questions:

1. Of houses containing three boys, what is the chance that the first two children out will be boys?

2. Of houses containing two boys and a girl, what is the chance that the first two children out will be boys?

The answer to 1. is that it is certain, the answer to 2. is 1/3.

We now notice that while it is three times more likely that households containing three children, at least two of whom are boys, will contain one girl rather than no girls, it is three times less likely that the first two children to run out of such a house will both be boys. Therefore, if we see two boys run out of a house that contains three children, at least two of whom are boys, there is a 50:50 chance that the remaining child is a boy.

In reply to John Gillott:







But isn't all this is just saying that the two children were chosen at random to which I agree the answer is 1/2. But I don't think there is anything in the original question which says they were chosen at random.

In reply to Robert Durran:

Well, non-random selection procedures could, presumably, be non-random in all sorts of ways. I'm going to assert that the non-random ways, looking at the collection of all possible non-random ways, so taking a population approach, all aggregate to 1/2 as well, so that when we do the grand aggregation of the random and the non random we still get 1/2. You can't argue with that!

In reply to John Gillott:

Do these two questions have the same answer (I'm trying to get to the bottom of the name thing).

I have 2 children. I have a son. What is the probability I have two sons?

Here is John, one of my two children. What is the probability I have two sons?

And does replacing "What is the probability I have two sons?" with "What is the probability my other child is a boy?" change either question?

In reply to John Gillott:


Except to say that it is a bit meaningless?!

In reply to lowersharpnose:


I would argue that it is, strictly speaking, ambiguous because I don't know how you selected the children to tell me about, but, in the absence of any other information I would say the best answer I can give is 1/4 (all I know is that you have at least 2 boys and a quarter of such families have 3 boys - I am not buying into the idea that knowing their names changes anything!).

In reply to Robert Durran:

My view is that the name thing can be important if it fixes the individual child and / or if it gives us some information. In the three children example, that the children have names is less important or is just another way of saying we have some definite information about particular children, that something has happened (two children of a particular sex have left the house or are in front of us). This affects the probabilities.

In this case you have just presented, my answers are:

1. 'I have 2 children. I have a son. What is the probability I have two sons?'

1/3, because the possibilities are BG, GB and BB

2. 'Here is John, one of my two children. What is the probability I have two sons?'

1/2, because the possibilities are that you had John then another child, or another child then John. Either way it is 50:50 whether the other child is male or female.

Changing the wording doesn't have any effect.

In reply to Robert Durran:

Making them even closer...
A) I have 2 children. I have a son. What is the probability I have two sons?

B) I have 2 children. I have a son, John. What is the probability I have two sons?

I think A and B are functionally the same.

In reply to Robert Durran:


Well, yes, I was being tongue in cheek.

But are you saying the original question was badly framed? If it is a non-random selection in a way we do no know, then surely the only sensible answer is 'give us more information' or 'any chance you like since it is a fix'

In reply to lowersharpnose:

A friend has just offered this variation:

I have one sibling. Of the two of us, I am the oldest boy.
What is the probability that the sibling is a girl?

In reply to John Gillott:


Yes, you should ask for more information. As I think I said earlier the "textbook" answer is 1/4, but in real life the selection might often be random (family moves in next door and the first two children you see playing in the garden are boys) and the answer is 1/2. The lesson is always to make sure you know how you the information arose.

In reply to Robert Durran:

Why is the textbook answer 1/4 when random presentation of kids leads to 1/2?

Do you mean 1/4 is the answer to the question: 'how many families with three children, at least two of whom are boys, have three boys'? That doesn't seem very much like the question we were asked

In reply to lowersharpnose:

2/3 I think

In reply to John Gillott:




If it had been "I have 2 children. I have a son over 7 feet tall. What is the probability I have 2 sons", the answer would be just less than 1/2 (this is the Boy Born On a Tuesday effect where a rare male characteristic will occur almost twice as often in a two boy family than in a boy/girl family because they get two "shots" at it (first and second born child) - but not quite twice as often because it very occasionally happens twice in the same family)



Are you arguing that the specific information (the name) has the same "Boy Tuesday" effect as the specific information that he is over 7 feet tall? If so, then I'm not sure I am convinced because the name is not a characteristic assigned by chance. If names were drawn out of a hat with John occurring with, say, probability 0.01 then I agree that the answer would be very slightly less than 1/2.

I can't see that being told about a son called John is any different to being told that there is a son and then deciding yourself to refer to him as "John" for convenience (which surely would not change the problem!).

If someone showed you a child and said "This is one of my two children. What is the probability I have two sons?" and you noted that the child was a boy, what would the answer be? I can't see that this can be any different from being shown a boy and told that their name is John. And it is surely no different from (1) above.

In reply to John Gillott:


Because the usual interpretation, in the absence of further information, is the proportion of 3 children families with at least 2 boys which in fact have 3 boys. I do agree that in realistic situations further information will usually be available and there is a strong case for the most useful answer being 1/2 and perhaps that should be the textbook answer!


But, strictly speaking, all we actually know about the family is that there are at least two boys (unless you think knowing names changes it!).

In reply to John Gillott:


I cautiously agree. I am assuming the scenario that one of the children has been chosen at random and then makes a true statement of the form "I am the eldest/youngest boy/girl".

In reply to Robert Durran:



No, I'm not arguing that.


The name is just a marker in this case. As I said, what really matters here is that we have a boy in sight. That is, we have a particular boy and we can register him as such. The crucial difference is between knowing that there is a boy in the house because we know that there are two of them at least (out of three) and the child who walks out being a boy. Once we see a boy or two boys emerge (or run past us, or who are introduced to us) we have new information relevant to our question.


Yes, I agree. If I said naming was crucial I was wrong. In this case the key point is that we see an actual identifiable child we know the sex of. We want to be able to be sure that we are dealing with the same child if they run past us again, that was all I was getting at. So, we want to make sure we know who they are.

In reply to John Gillott:


So does it make a difference whether the information comes from seeing a boy at a window or from the father telling you that he has a son? What if the father had told the boy to stand by the window knowing you were looking? Is this different from him just telling you he has a son? I still think that we cannot answer the question without knowing precisely the mechanism by which you gain knowledge that there is a boy.

In reply to Robert Durran:



I suggest that on an obvious reading of the problem we know more than this. Two of the three children were presented to us, or they ran past us. They were both boys. Of all the three child families this could have happened with, 3/4 of those families contain a girl. But we didn't see a girl, despite there being a decent chance this would have happened had the family been one of the 75% with a girl and two boys. This raised the chance that we were in fact dealing with a family with three boys from 25% to 50%.

In reply to Robert Durran:


If the father is fixing it to be non random and we don't know he is doing it, or don't know how he is doing it, all bets are off. My default is that it is random, so seeing a boy emerge or run past is different from knowing that there are boys in the house. It changes the odds.

In reply to John Gillott:


Ok. I think we agree now! It all comes down to knowing or assuming that the child we see is selected at random. If the father decides specifically to let you know he has a son by any means (introduces you to John/mentions he has a son/tells a boy to stand in the window) then it is different. I don't think there is any reason to assume random selection of two children in the original problem (maybe he just wanted to let you know he had sons of a similar age to yours), so in my opinion, as you say, all bets are off.

In reply to lowersharpnose:

Here's a well-known riddle and arithmetic puzzle:

Diophantus’ childhood lasted one sixth of his life. His beard grew after one-twelfth more. He married after one-seventh more. His son was born five years later. The son lived to half his father’s age. Diophantus died four years after his son.

How old was Diophantus when he died?

In reply to lowersharpnose:

A fruit bowl contains 10 apples and 5 bananas. Through the course of a day people help themselves to a piece of fruit, entirely at random, until only one piece remains.

What is the probability that the last remaining fruit is an apple?

In reply to Paul Robertson:



This is one of my favourite problems because it is so simple when approached the right way (I won't give it away!). I have used it as a very satisfying example of using a sort of symmetry. A more cryptic version is to ask what the probability is that the bananas run out first.

More challenging extension is to have 10 apples, 5 bananas and 8 oranges (or any other numbers) and ask what the probability is that the apples run out first - a colleague once asked me to solve this because he was seriously trying to work out the probability that one of three isotopes in a mixture is all used up first in a chemical reaction.

I once had an argument about these problems at Higgar Tor which ended up with my climbing partner getting so wound up that he threatened to hit me!

In reply to Robert Durran:

Yes, the fruit bowl one is great.

I would need to get out my old books, but in general can't all these kinds of puzzles be solved in a spoilsport brute force kind of way using the formula for combinations? Or are they still a challenge even if we know the formula - because the difficulty is working out just how to apply the formula (wondering just how you nearly came to blows at Higgar...)

In reply to John Gillott:


It's much easier than that

In reply to Paul Robertson:

I know that the fruit bowl one is (I've seen it before).

In reply to John Gillott:

Yes, often, and it's nice to then see everything cancelling away, suggesting there is a much neater approach!


Yes, sometimes it is not immediately obvious. One I like is "how many ways are there of distributing 20 indistinguishable apples between 5 bowls"

The three fruit problem required some pleasing subtlety which I'd have to spend some time reconstructing......


He refused to accept that "the apples run out first" is precisely equivalent to "the last fruit eaten is a banana" (although there may have been a misunderstanding involved) and that therefore their probabilities are equal and things kind of escalated........... eventually we agreed to divert our aggression to a boulder problem.

In reply to Robert Durran:

the 2 fruit bowl took some working out - i think i came up with 4 different ways but couldn't convince myself they were correct/ not flawed. A sort of 'brute force and cancelling' convinced me (when i realised thats what i was doing) and it had to be simple arithmetic

nice one !

In reply to lithos:

I found it helpful to imagine that each piece of fruit was in an identical box so you couldn't tell which type was inside.
The question then is:

After 14 boxes have been removed, what is the probability that the remaining box contains a banana?

In reply to Paul Robertson:




Surely there is a much more elegant approach (at least to my mind!)

In reply to Robert Durran:

You could turn the box on its head

In reply to Robert Durran:


Go on. Enlighten me.

In reply to Paul Robertson:

The fruits are effectively just laid out at random in a line. By symmetry, the probability of the rightmost fruit being a banana is the same as the leftmost fruit being a banana (ie the probability of the first fruit chosen being a banana, which is easy). The appeal to symmetry appeals to me!

Have you tried the three fruit version?

In reply to lowersharpnose:

I like the tennis tournament one. It's not hard, but it's interesting to see how people solve it.

Numbers can vary, and how it is posed can vary, but this is typical:

In a knockout tennis tournament, how many matches are played if 97 players participate?

In reply to Robert Durran:


Working on it

In reply to Robert Durran:


im sure this is true and im not doubting you - but the phrase 'By symmetry' seems a bit like a magic step
(got a link i can read ?)

In reply to John Gillott:


I like the fact that it provides an indirect proof that 1 + 2 + 2^2 + 2^3 + 2^4 +......+ 2^(n-1) = 2^n - 1

In reply to lithos:



Could look at it like this: shuffle 15 cards, 10 of which are hearts, 5 of which are spades. Take one at random. Chance that it is a heart is 2/3. Or, take one off the top. Chance that it is a heart is 2/3. Or, deal them out. Chance the last one is a heart is 2/3. Symmetry between first and last is the key point.

In reply to lithos:


Subtle, but not magic

Though, when teaching, I admit that I regularly say "What's the magic word?"
Symmetry is a very powerful idea in many areas.

In reply to Robert Durran and John

so wikipedia helps me here - explaining uniform probability distribution as symmetric

http://en.wikipedia.org/wiki/Uniform_distribution_(discrete)

symmetry is usually associated with geometry (i realise its broader - esp. now!) in my mind so thats the issue i had....

cheers

In reply to Robert Durran:

4/27 ?

No, hang on, that's wrong... or if it's right I was lucky... back to the drawing board

In reply to Robert Durran:


44/207 I think / hope!

In reply to John Gillott:

I know the tennis tournament one, really good.

In reply to John Gillott:

w.r.t. to my 'I have one sibling. I am the oldest boy. What is the probability I have a sister?'

I don't agree with your answer. If you consider the case when the speaker is the only boy, he could say 'I am the youngest boy' or 'I am the oldest boy'. I think the question only makes sense if both utterances are equally likely. Hence,...

In reply to Paul Robertson:

I love the fruit bowl one.

In reply to lowersharpnose:



But even if that is true...

Call the boy 'me'

Possibilities for two kids are, in terms of birth order:

me, sister
sister, me
me, brother
brother, me

The last one is ruled out because 'me' is the oldest boy (I'm using the notation that written to the left means born first).

The scenario where 'me' has a sister works whether 'me' considers himself the youngest or oldest boy, so we have 2/3 as the answer.

In reply to John Gillott:
But, when he is the only boy, he will only say 'I am the oldest boy' half of the time.

I think you are answering a question like:

Consider all families that have two sibling and don't have two girls. If there is an oldest boy, what is the probability that the sibling is a girl. This is not the question posed.

In reply to lowersharpnose:


Reading your latest post, I'm going to say Zero.

In reply to John Gillott:

Don't you need an even number of players to start a knockout tournament?

In reply to Andy Hardy:

No you can have a bye into the next round.

In reply to John Gillott:


Correct!

In reply to lowersharpnose:

OK then 1 winner = 96 losers

In reply to lowersharpnose:


No need for the "if" (all such families have an oldest boy).
I took the question to be equivalent to this; the oldest boy HAS spoken (and he said he was the oldest boy) and in two out of three such families, his sibling is a sister.

In reply to lowersharpnose:



But we are TOLD that he said "I am the oldest boy".

If I toss a coin and tell you it came up heads, the probability that it says tails is no longer 1/2!

In reply to Robert Durran:

LSN's suggestion is, I think, that a boy with a younger brother would have said that for sure, but some boys with a sister might not have said that because they see themselves as being a younger brother / youngest boy. So, the relevant population is boys who will make that statement and it is to be expected that half the boys with a sister would not make that statement, meaning that the answer is 1/2 rather than 2/3.

It might be a matter of whether they were asked the question or just made the statement. If asked, they probably ought to answer yes to the original statement if posed as a question. But if we looked at who would make the statement un-prompted, I can imagine that somewhere between 0 and 50% of boys with a sister wouldn't make the statement.

In reply to lowersharpnose:

I have smiley stickers which I put on pupils' good work. They come in 5 different colours and I dispense the colours at random. Pupils like to collect the full set and so will sometimes ask for a particular colour. I refuse. On average, how many stickers will a pupil need to get in order to collect the full set of colours?

In reply to Robert Durran:

If the rule is that if he is the only boy then he ALWAYS says I am the oldest boy, then yes, I'll agree to 2/3.

In reply to Robert Durran:

11 and 5/12 (not that they can get 5/12).

So between 11 and 12?

In reply to John Gillott:


I agree. What approach did you use? I didn't find this problem very easy!

In reply to Robert Durran:

Yes, I also found it tricky. The breakthrough for me was the realisation / recollection that if the probability of an event happening is P then the average number of goes before it happens is 1/P.

First off they get a sticker, which takes one go.

Then there is a 4/5 chance they get a different one, so this takes on average 5/4 goes. Then it is a 3/5 chance of a different one, so this takes 5/3 goes etc.

The answer therefore is 1 + 5/4 + 5/3 + 5/2 + 5

In reply to John Gillott:



I really like that! And it generalises easily to any number of stickers. The connection to a harmonic series is nice. Of course you have used two standard results without proof (the 1/p one and the fact that the expectation of a sum is the sum of the expectations).........

I set up a recurrence relation between the expected waiting times to get r out of 5 stickers a r-1 out of 5 stickers and used that.

In reply to Robert Durran:

Sounds like a close call as to whether or not I get my sticker!

How did you formulate the recurrence relation?

N(r) = N(r-1) + 5/(6-r), where N(r) is the number of goes to get r stickers for r between 2 and 5 and N(1) is 1

Would do it. But then that's pretty much the same as mine, so has the same problems....?

In reply to John Gillott:

eg N(4) = 1+4/5 N(3) +1/5 N(4) (1 sticker, then need 3 more with probability 4/5 or still 4 more stickers with probability 1/5 )

This gives N(4) =N(3) + 5/4 leading to the same sum you got.

In reply to lowersharpnose:

suprised not to see this posted yet. I think the people posting here would make short work of it


"A group of people with assorted eye colors live on an island. They are all perfect logicians -- if a conclusion can be logically deduced, they will do it instantly. No one knows the color of their eyes. Every night at midnight, a ferry stops at the island. Any islanders who have figured out the color of their own eyes then leave the island, and the rest stay. Everyone can see everyone else at all times and keeps a count of the number of people they see with each eye color (excluding themselves), but they cannot otherwise communicate. Everyone on the island knows all the rules in this paragraph.

On this island there are 100 blue-eyed people, 100 brown-eyed people, and the Guru (she happens to have green eyes). So any given blue-eyed person can see 100 people with brown eyes and 99 people with blue eyes (and one with green), but that does not tell him his own eye color; as far as he knows the totals could be 101 brown and 99 blue. Or 100 brown, 99 blue, and he could have red eyes.

The Guru is allowed to speak once (let's say at noon), on one day in all their endless years on the island. Standing before the islanders, she says the following:

"I can see someone who has blue eyes."

Who leaves the island, and on what night?


There are no mirrors or reflecting surfaces, nothing dumb. It is not a trick question, and the answer is logical. It doesn't depend on tricky wording or anyone lying or guessing, and it doesn't involve people doing something silly like creating a sign language or doing genetics. The Guru is not making eye contact with anyone in particular; she's simply saying "I count at least one blue-eyed person on this island who isn't me."

And lastly, the answer is not "no one leaves.""



similar to the people wearing hats puzzle. Try with smaller numbers and build up if you can't work it out.

In reply to Sink41:

I know this one. It is a top class problem.

In reply to Sink41:

Is it as simple as: there are 4 people left on the island, the guru and two people with brown eyes, and one with blue eyes, but none of the latter three knows the colour of their own eyes. So when the guru says this, the blue eyed person knows he/she has blue eyes and so leaves that night. That's result of five minutes thought, because I'm working ... I've probably made some huge logical howler ...

In reply to Gordon Stainforth:

Not a logical error as such Gordon - you'll just need to tell us what happened to the other 197 people!

In reply to Sink41:

All the Blue eyed people leave on the 100th day.

With only one blue eyed person, they will know there are 0 or 1 (if they are blue eyed) blue eyed people and the initial statement resolves it on day 1 - they must be blue eyed and leave on day 1.
With two blue eyed people, they will know there are 1 or 2 blue eyed people, if no-one leaves on day one there must be two. So they would both leave on day 2.
With three blue eyed people, they would know there were 2 or 3 blue eyed people and would resolve it on day 3 when no-one left on day 2.

With 100 blue eyed people they know there are either 99 or 100 blue eyed people and it is resolved on day 100.

In reply to lowersharpnose:

Anyone up for a game of Modified Russian Roulette?

Two bullets in adjacent chambers (six in total), you have to pull the trigger twice.

If you survive the first attempt, should you spin again or pull the trigger immediately?

CB

In reply to cb294:

I'd pull the trigger immediately.

Probability of dying (Pd) on the first pull = 1/3, so if you spin again Pd = 1/3. If you've survived the first pull, there is only 1 position the chamber can be in which causes you to die on the second, so Pd = 1/6.

In reply to tom_in_edinburgh:

Some great discussions on stats / probability sites about this one. Why does the guru saying they can see a blue pair start it off is a good follow up question, given that everyone can see at least 99 blue pairs and everyone knows that everyone can see at least 98 blue pairs.

In reply to cb294:

Pull immediately (no spin). Good puzzle - my immediate reaction was that it would make no difference

In reply to cb294:


Pull immediately (Probability of dying 1/4)
Spin (probability of dying 2/5)

So pull immediately

In reply to Robert Durran:




2/6 not 2/5, surely. He can land on the same empty chamber he first landed on.

Or, looking at the two events together:

Chance of surviving the game using the spin strategy = 4/6 x 4/6 = 4/9

Chance of surviving using the no spin strategy equals chance of avoiding the two loaded chambers and the one before them on the first spin, so 3/6 = 1/2

In reply to tom_in_edinburgh:

All the blue eyed people leave together 100 days after the announcement . Nobody else leaves?

In reply to gazhbo:

That's another good follow up question discussed on the various sites. Nice twist to it!

In reply to John Gillott:


I noticed that. The guru doesn't actually provide any information, he only serves to start the clock. My take is the problem is internally inconsistent because nobody would stay on the island for more than 100 days.

In reply to lowersharpnose:

Good thread, though increasingly difficult to tie answers to original questions!
Here is one my 9 year old came home from school with.

4 4 4 4 = 1
4 4 4 4 = 2
4 4 4 4 = 3
4 4 4 4 = 4
4 4 4 4 = 5
4 4 4 4 = 6
4 4 4 4 = 7
4 4 4 4 = 8
4 4 4 4 = 9
4 4 4 4 = 10

use maths symbols, including brackets, to make each sum correct. e.g.
4 + 4 + 4 - 4 = 8

(Then see if you can keep going...)

In reply to tom_in_edinburgh:


Re whether starting the clock is the issue, see here:

http://math.stackexchange.com/questions/489308/blue-eyes-a-logic-puzzle/490...

As to everyone leaving after 100 days, why? All the blue eyed people leave on day 100. How can the remaining people leave?

In reply to John Gillott:

Pull immediately is correct. The chance of dying on the second pull is then 1/4 vs. 1/3 for spinning again.

You are in your boat carrying a ballast of rocks. Will the level of the lake you are on rise, fall, or remain unchanged when you throw these rocks overboard?

CB

In reply to John Gillott:


Yes. My error.

In reply to cb294:


It falls. When in the boat, the rocks displace a volume of water with the same weight as the rocks (Archimedes principle). When sitting at the bottom of the lake, they displace their own volume which is less because they are denser than the water. It is pretty intuitive if you imagine small rocks of enormous density.

In reply to cb294:

That's a nice one - bit physicsy though!

I've seen it before so won't give the answer, though to hint at it I remember imagining tying an inflated balloon to the bottom of the lake to confirm my answer

In reply to Robert Durran:

Here's one that jcm gave me years ago.

Boy in a circular pool

A boy is treading water at the centre of a circular pool. At the edge of the pool a friend challenges him. ‘If’, he says to the boy in the water, ‘you can get to the edge before I reach the point you touch, you win, if not I win.’

If the friend on the edge can run at a steady maximum speed of 10 mph, what is the minimum speed the boy in the water has to be able to swim steadily at in order to win the challenge?

In reply to Robert Durran:

What if they are pumice stones?

In reply to John Gillott:

10 / pi mph plus a smidge (or is that too obvious / wrong!)

In reply to lithos:

I'm guessing you did this:

If we use a compass notation, and imagine the friend at West, then swimming in an easterly direction gives a benchmark minimum speed for the boy of 10/Pi (for a dead heat).

However, it is possible to improve on this, i.e. swim more slowly using a better strategy.

In reply to John Gillott:


Done in a hurry without really checking my argument........

10/(pi +1)

In reply to John Gillott:

Boy in pool...

Swim in some kind of spiral to make the way around the pool longer? Do they have boys have to start swimming/running at the same moment?

CB

In reply to John Gillott:

yeah i assumed (doh) that he swimmer wuld/had to swim in a straight line ( but it doesnt mention that at all !)

runner could swim until the runner is half way (north) then turn 90 degrees (head south) and repeat when
runner gets to E turn and head back to W point (or in a spiral as suggested)

In reply to cb294:

It's up to the boy in the middle to make the first move (let's assume he can only tread water for so long).

In reply to cb294:

It is a spiral, yes (though I tend to think of it making the dash to the edge shorter)

In reply to Robert Durran:

That's my answer as well.

The following strategy gives a minimum speed of 10/(Pi + 1).

If the boy swims at a speed that is less than 10/Pi he can’t dash for the side in a straight line and expect to win. But, no matter what speed he swims at, if he chooses a circle small enough he can circle at a faster rate than his friend can circle around the edge of the pool. So, at a speed less than 10/Pi, he gradually spirals out until he approaches the radius of a circle at which their circling rates are equal. Since he is approaching this radius from points less than this distance from the centre of the circle, he is spiralling faster than his friend can circle the edge of the pool. Therefore he can arrange it such that when he gets to the radius at which their circling speeds match he is as far apart from his friend as it is possible to be while satisfying the restriction that they are both on circles centred on the centre of the pool (for example, if his friend is at West again, the boy would be at the easterly most point of the smaller circle). Then, he dashes directly east.

In reply to Robert Durran:

I have other strategies that give a speed in between 10/(pi + 1) and 10/pi, but none better (slower) than 10/(pi + 1). And intuitively, the spiralling approach feels like it can't be improved upon.

But can it be proven that there is no better strategy?

In reply to John Gillott:


Good! Pleased to get it quickly, but been pondering it ever since..........

>............ Then, he dashes directly east.

This is precisely my solution and effectively my argument. The bit I've been worrying about is the dashing east. I think it's ok: if the swimmer now spirals at all, then the pursuer can run counterclockwise to him, closing the angular gap faster, while the swimmer's radial speed has been compromised so he is also further from the edge - surely a worse position to be in...........maybe not an entirely rigorous argument!

In reply to Robert Durran:

At any point in time the boy can change direction and head for the point diametrically opposite the pursuer's position at that moment. I calculated that this was a poorer strategy if the moment chosen was such that the boy headed due south (so made a right angled turn; the pursuer running round from West via North)). 'By inspection' I reckoned no other change direction strategy was any better. It's all a bit hand-wavy though.

In reply to John Gillott:


I think that the our solution comes down to the assumption that, at any given radius, the best position for the swimmer to be in is directly opposite the runner. Seems obvious!

One of my favourite mathematical curiosities:

Q1) (A warm up) How many squares are there on a chess board (Hint: more than 64)

Q2) How many cubes are there in a 3-dimensional chess board (an 8x8x8 cube)

Q3) How many rectangles are there on a chess board.

In reply to Robert Durran:

What I had in mind is this: boy spirals out, swimming at 10/(pi + 1). Gets to the point where he is at the eastern side of his smaller circle, pursuer sitting tight at West. Boy then cuts loose and heads directly east, so heading for East. Pursuer sets off via North. This will result in a dead heat - they'll get to East at the same time. Can the boy, still swimming at 10/(pi + 1), get to the edge before his pursuer if, say, he heads for SE (thinking in terms of a compass) when the pursuer is at NW, so after the pursuer has run a quarter of the way from West to East round the edge? As far as I can tell, it is better that the boy carries on heading for East, because although this means the pursuer has only 135 degrees to run around rather than the 180 he would have if the boy headed for SE, this is more than offset by the fact that the distance for the boy is smaller heading East (so straight out on a radius still) rather than the greater distance involved in getting to SE.

In reply to Robert Durran:

204 for the first one?

In reply to John Gillott:


8^3+7^3+6^3+5^3+4^3+3^3+2^3+1 = 1296 for the second

In reply to Paul Robertson:


Yes. The third one is the interesting one (remember that squares count as rectangles!)

A fourth question (should be easy if you have done the third) is the number of cuboids in the 3-d chessboard, but the answer is not so interesting.

In reply to Robert Durran:


Sum of x*y for x=1..8 and y=1..8, but only counting the square rectangles once

i.e. (1+2+3+4+5+6+7+8)^2 - 204 = 1296 - 204 = 1092

The 'coincidental' occurrence of 1296 in both results is an interesting result.

1^3+...+N^3 = (1+...+N)^2

It's easy to prove with algebra, but also interesting to illustrate with childrens' building blocks

In reply to Paul Robertson:




I've never seen a satisfactory way of making the result obvious, only rather unsatisfactory illustrations (how would you do it with building blocks?). When I stumbled on the fact that the number of rectangles on the 2-d board is the same as the number of cubes in the 3-d board, I wondered whether there was natural one-to-one correspondence between the rectangles and cubes (some sort of projection perhaps) but I have never been able to establish one.

I've always thought that the result was really just a nice coincidence.

In reply to John Gillott:

Re: Boy in lake

I don´t know whether this is the optimal strategy, but at least it is a strategy for which the minimum ratio of swimming speed to running speed should be calculatable:

Swimmer moves away from the center, then swims in a circle until directly opposite runner, then makes a straight dash for the edge. For this strategy to work, the swimmer first has to be fast enough that for a given radius of his initial circle his angular velocity is greater than that of the runner. Thus, the ratio of speeds must exceed the ratio of the radii.
Second, the swimmer has to be fast enough that he can cover the difference of the radii faster than the runner can cover the semicircle.

It should therefore be possible to express the ratio of speeds that results in simultaneous arrival as a function of the initial radius chosen by the swimmer, and find a minimum for this function.

Unfortunately there is also work to be done,

CB

In reply to cb294:



My son was asked this question at his interview for Imperial College last November. He was quite chuffed as the teacher had gone through it the day before. The physics answer (fall) is, of course, quite different to what most lay folk like myself imagine!

In reply to cb294:

That's exactly right, and leads to the answer 10/(pi + 1) when the runner runs at 10mph, or x/(pi + 1) when the runner runs at x, to generalise.

It's the best strategy I can come up with, and intuition tells me it can't be improved upon.

In reply to John Gillott:



I started looking at your suggestion that the swimmer changes direction and heads for a different point part way through the dash (presumably optimally in a straight line). It looked like a general case would soon get bogged down in some messy trigonometry and algebra. I might return to it when I have time. It certainly seemed unsatisfactory!

In reply to Robert Durran:


I can't remember all the details. You 'slice' the cubes up into 2D squares, and try to rearrange the pieces into a single square.
It doesn't quite work - there are some gaps, and squares with even sides (2x2, 4x4, etc) overlap.
But you can cut out the overlapping pieces and use them to fill in the gaps.

In reply to lowersharpnose:
Here's one to test your mathematical intuition - Which of the following infinite series have a finite sum?

1) 1+1/2+1/4+1/8+1/16+...
2) 1+1/2+1/3+1/4+1/5+...
3) same as 2) but raise each term to the power of 1.00000001
4) same as 2) but omit any term where the denominator has a 9 in it
5) sum of reciprocals of primes: 1/2+1/3+1/5+1/7+1/11+...

Don't need to see proofs. Just interested.

In reply to Paul Robertson:




I think I've seen that. The cutting up needed to fill the gaps seemed rather unsatisfactory and I've always felt that a much more transparent demonstration ought to be possible for it to be more than a nice coincidence.

In reply to Paul Robertson:

Re: infinite series?

1) converges to 2
2) does not converge (as it includes 5)
3) should not converge
4) should not converge
5) does not converge if my memories of school calculus are correct...

CB

In reply to Robert Durran:



We can go from squares to cubes, and and also from squares to rectangles, regarding the latter transformation as as another kind of dimensional transformation (stretching).

Consider the 1x1 square. There are 64 of these on the chess board. Add a dimension to get a cube and consider the 8x8x8 cube. There are 8x64 positions for the 1x1x1 cube to occupy

Now consider all the rectangles with one side equal to 1 and regard them as stretched versions of the 1x1 square - so only stretched in one dimension as it were.

Don't stretch at all (a square is still a rectangle) and we can place this 8x8 ways on the board
stretch to form a 1x2 rectangle and we can place this 2x7x8 ways on the board.
stretch to form a 1x3 rectangle and we can place this in 2x6x8 ways on the board
stretch to form a 1x4 rectangle and we can place this in 2x5x8 ways on the board
......
......
stretch to form a 1x8 rectangle and we can place this in 2x1x8 ways on the board.

Add them up and we have(8+14+12+10+8+6+4+2)x8 ways
ie 64x8, the same number as we had for the 1x1x1 cube inside the 8x8x8 cube

The next set of relations is between 2x2x2 cubes and rectangles having one edge equal to 2.

etc etc

Can we make a visualisation out of this, and a satisfying set of 1-1 relations?

In reply to cb294:


You score 3 out of 5

In reply to Paul Robertson:

1 and 3 do; the rest don't?

In reply to John Gillott:

I suspect all except 2 do, but I am not certain about 4 and 5.

Edit: I've just googled 5, but won't give it away!

In reply to John Gillott:


You score 4 out of 5

In reply to Paul Robertson:

Just 1)

In reply to John Gillott:


You score 2 out of 5

In reply to Robert Durran:

I've finally done the three fruit problem.
Given x apples, y bananas, and z oranges, the probability that apples run out first is:

(y*z)/(x+y+z)*(1/(x+z)+1/(x+y))

So for 10 apples, 5 bananas and 8 oranges the probability that apples run out first is 44/207

In reply to Paul Robertson:

I've changed one answer and lost two marks... Another puzzle!

In reply to John Gillott:


Sorry 'just 1)' scores 3 out of 5

In reply to Paul Robertson:

Ah yes, I see it now (well, I've had enough hints!)

4) will converge. Very good!

In reply to John Gillott:

How do you prove 4?

CB

In reply to Paul Robertson:





Yes. I took some time convincing myself that you could just multiply the probabilities together and not worry about the fact that the others of the last type of fruit eaten was mixed in with the other two.

In reply to cb294:

Qualitatively, as the number of digits in the denominator increases the proportion of those containing a 9 approaches 100%.

I don't know about a proper proof though.

In reply to John Gillott:


This seems a promising approach from my day's ponderings............shall have to get pencil and paper out.........good for a wet looking weekend

In reply to lowersharpnose:

Interesting point. I was looking for a suitable series know to converge to match in a comparison test (the inverse approach I used to deduce 2), but didn´t come up with a good example).

Anyway, I find it slightly odd that I could spontaneously remember that the sum of the inverses of all primes diverges, as we applied this in school to somehow deduce that there is an infinite number of primes, but failed to remember the much simpler proof (or even the fact) that the harmonic series is divergent.

20 years after my last formal lesson my maths is really only bits and pieces.

CB

In reply to lowersharpnose: No hard maths required, but this one catches a surprising number of people.


You are driving a bus. It has space for 30 sitting and 20 standing passengers. There are 6 stops before you return to the bus station, and the bus starts with 13 passengers.

At the first stop 5 people get on and 7 get off.

At the second stop 3 people get on.

At the third stop 8 people get on and 12 get off.

At the fourth stop one person gets off.

At the fifth stop 9 people get on and 2 get off.

At the sixth stop 4 people get on.

What is the name of the bus driver?

In reply to davidbeynon:

Yes, I like this one.

In reply to Robert Durran:


Set theory solution: Let X be Apples run out before Bananas; Y be Apples run out before Oranges.

Then use P(XintersectionY) = P(X) + P(Y) - P(XunionY)

= 1/3 + 4/9 - 13/23

= 44/207

In reply to cb294:


I'd like to know that too!
I have it on good authority that 4) converges but I don't know the proof.
Given that 5) diverges I find it astonishing that 4) converges

In reply to John Gillott:





Very nice. Again, you have come up with a more sophisticated approach than me!

In reply to all:

Brilliant thread, this has really given me food for thought, thanks all.

In reply to Paul Robertson:

I was amazed by the convergence of 4 as well. I don't have a proof, but the intuition is that for numbers of n digits, the density of the primes is about ~ 1/ln(10^n) ~ 1/(2.3n) (prime number theorem), which decays inverse linearly; the numbers that contain no 9s have density exactly (9/10)^n, which decays exponentially.

Put n points on the circumference of a circle. Join every point to every other point with a straight line. What is the maximum number of regions into which the circle can be divided (ie the number of regions when no three lines are concurrent)?

In reply to Robert Durran:

A surprising answer -- that one's in David Acheson's book "1089 and all that", along with loads of other great stuff, well worth reading.

In reply to Robert Durran:

Cheers. I had recently skimmed an old book on set theory, which probably helped. I was mulling over the puzzle, 'before bananas and oranges', then it hit me: ah, that's set theory!

In reply to Robert Durran:


Do you mean as a function of n, or for any n? The latter seems limitless....

In reply to lowersharpnose:

Another one from jcm:

Five men in a boat puzzle

Five men are in a lifeboat after the ship they were on sank to the bottom of the ocean. They have with them 100 bars of gold and a loaded gun. It will take many days to get to land. They decide that instead of fighting for the gold, or sharing it out evenly, they will vote on it, in a peculiar way.

Call them A, B, C, D, E.

A will make a proposal as to how it will be shared. This will be voted on (A gets to vote as well). If it is accepted by a majority decision, then that is it. If it is rejected, by majority decision, then A is shot. If the latter happens, then it is B’s turn. And so it goes until a proposal is accepted.

A tied vote leads to the proposer being shot.

All things being equal, the men enjoy seeing the others shot. So, if they calculate they will get the same amount of gold with or without someone dying, they will vote down a proposal.

How will the gold end up being divided among A, B, C, D, E?

In reply to John Gillott:

As a function of n.

Once you've done that you might like to work out the maximum number of 4-dimensional hyperchunks a 4-d hypersphere can be cut up into with n-1 3-d hyperplanes. Actually, before doing this you would probably be best to do the 1-d (line cut by n-1 points), 2-d (circle cut by n-1 lines) and 3-d (sphere cut by n-1 planes) versions and then extend the idea naturally to 4 dimensions. Once you get the key idea, generating the sequences is quite straightforward even if getting the actual formulae is a bit tedious algebraically. I used n-1 rather than n because this makes the connection with the original problem more direct!

I stumbled across this connection while, like with the chessboard one, playing around with investigations for school pupils, but I have no idea why the connection is there (maybe someone will enlighten me!). The original points on a circle one is a classic trap for pupils - draw some, look for a pattern, then make and check predictions! The 2-d version above also makes a nice investigation. The formula for the 3-d version was once, a long time ago, in the hardest Cambridge maths entrance paper.

In reply to John Gillott:
Re men in boat:

I assume that people prefer surviving over getting another person shot.

If A and B were voted down and shot, C could propose to grab all the money.

In this case D would have to accept even though he could get C shot as well, as he cannot risk E voting against him in the next round, when only two voters are left.

Thus, D and E should accept an earlier vote that gives them anything at all. If A were killed, B would therefore survive if he proposed a B:98% C: 0%/ D: 1%/ E: 1% split.

A will therefore survive with the votes of A, C and one of D/E if he proposes 97% for himself, 0% for B, 1% for C, and 2% for one out of D and E.

In reply to cb294:

Perfect answer. Did you have an intuition on reading it that was roughly the opposite of this, or was it obvious to you straight away?

In reply to Robert Durran:



Er, thanks (the circle one is proving tricky enough!)

In reply to John Gillott:

Seen the question for the first time, but it seemed obvious to work backwards from only two persons left.

CB

In reply to lowersharpnose:

An insect starts at one corner of a wire-frame cube and crawls along the edges. Each time it reaches a vertex it makes a random choice about which edge to crawl along next (including doubling back). When it reaches the opposite vertex from where it started, it stops.
What is the average number of edges which the insect will have traversed?

In reply to Paul Robertson:



10

In reply to Robert Durran:


I'd have thought it would be some kind of limit and the result would be a nasty decimal since the bug could go on forever but when I do the math I get 10 too.

Let M(n) be the average number of moves starting n edges away from the goal. M(3) is the case of interest.

M(3) = 1 + M(2) --- the first move towards the opposite vertex always takes you closer
M(3) = 1 + ( 1 + 2/3(M(1)) + 1/3(M(3)) -- when you are two away you have a 2/3 chance of moving to 1 away
M(3) = 2 + 2/3(M(1)) + 1/3(M(3))
2/3(M(3)) = 2 + 2/3(M(1)
M(3) = 6/2 + 6/6 M(1)
M(3) = 3 + M(1)

M(3) = 3 + (1 + 2/3(M(2) + 1/3(0)) -- when you are one away you have a 2/3 chance of moving back to 2 away and 1/3 chance of reaching the goal

M(3) = 4 + 2/3(M(2))

Already know M(3) = 1 + M(2) so solve the simultaneous equations:
M(3) = 4 + 2/3(M(3) - 1)
M(3) - 2/3(M(3) = 4 - 2/3
1/3(M(3) = 10/3
M(3) = 10

In reply to tom_in_edinburgh:

Yes, that's basically how I did it:

M(3)=1+M(2)
M(2)=1+1/3M(3)+2/3M(1)
M(1)=1+2/3M(2)

Then just solve for M(3)

I was surprised how big the answer was.

If you repeatedly toss a coin, how many tosses does it take on average to get a run of HTH ?

If you repeatedly toss a coin, what is the probability that a run of HTH occurs before a run of TTT ?