What's my score?

I hadn't played Yahtzee in years, but the other night I settled down to possibly the best couple of games I've ever played.

For those not in the know, Yahtzee is a game from Hasbro (My copy's from MB Games... that's how long it's been!) that's been around since 1956. Put simply, it's a game in which players roll five dice and are given a score based on what is rolled.

We played two games, and they were probably my best ever scores (though that doesn't say much; I'm not exactly a pro). In my first game, I scored exactly 80% of the maximum possible mark (assuming no bonuses for extra "yahtzee*"s), and in the second I scored 89% of the maximum, rounded to the nearest whole

1. What was my score in the first game?
2. In the second game, what's the highest score I could have got? What's the lowest?

Have a go at the questions before looking at the answers below!

The solutions

I'll remind you here as well that there are usually many, many ways of solving problems, so if you haven't done it the same way as me that doesn't mean that you've done it wrong. In fact, if you do it a different way I'd like to see an explanation in the comments- it might even be a better** way of doing it!

Part 1)

Look at the information I've given you: I got 80% of the total possible score. Assuming you know something about percentages*** then all you need to know is the total possible score. If you don't know the game, here's a scorecard to look at:

You just need to think of the best possible score for each roll, and record it, then add anything up, taking into account any bonuses. The overall total should come to 375. Comment and ask if you're not sure how to get that.

Now all you need to do is work out 80% of 375, then you've got my score for game 1:

One way of doing this is to find 10% (divide by 10) and then multiply by 8 (as 8 lots of 10% is 80%). So:

375 / 10 = 37.5

37.5 * 8 = 300

• My score for game 1 was 300 points exactly!

Part 2)

This one's trickier. The question asks for my highest and lowest possible scores because I didn't get exactly 89% - I rounded the answer. This means that finding 89% of the maximum possible score won't give the score I got, as it could have been a bit smaller or a bit bigger than that. In fact, if you work out 89% of 375...

375 / 100 * 89 = 333.75

... you don't even get a whole number, so it's actually impossible to get exactly 89% of the maximum score in a game of Yahtzee. So we'd have to make a guess. Luckily we can use the information given to make that guess as accurate as possible- we can find a sensible low- end for our guess, and a sensible high- end, and go somewhere in between. That's what the question means when it asks for the highest and lowest scores that I "could have" got.

To do this we need to look at that percentage first: I said I rounded it to the nearest whole percent, which was 89. So what are the lowest and highest answers I could have got before I rounded it? Any more than 89.5%, and I'd have rounded it up to 90%. Any less than 88.5%, and it'd have gone down to 88%. If we work out the two percentages in bold, we'll have our answers:

375 / 100 * 88.5 = 331.875

375 / 100 * 89.5 = 335.625

They're the highest and lowest scores, but now our common sense should kick in: there's no way of getting a decimal score in Yahtzee! That means the for the lowest score, we've got to pick 331 or 332, and for the highest, we must choose 335 or 336. But which one? Well, the lowest score that would end up being rounded to 89% would be 331.875. Any lower than that and we won't get 89%, so it must be higher than that: we pick 332. Using similar reasoning, but at the other end of the scale, we must pick 335 as the highest score that I could have got.

• My score for game 2 was somewhere between 332 and 335 inclusive.

Is it possible to be any more accurate than this?





* In the game Yahtzee, a 'yahtzee' is rolling five of the same number in one turn, i.e. after three rolls you end up with 5 twos, or 5 sixes, etc.
** What makes one answer 'better' than another, assuming that both are correct? That might be a subject for another post...
** If you don't, ask me for a post ;-)

If you fold a piece of paper 50 times, will it reach the Moon?

This post was prompted by @mikemcsharry 's tweet of a similar nature the other day. But is it true? For the pedants out there, I'm assuming that we're folding our imaginary sheet of paper exactly in half each time.

Stuff we need to know:

• How far away is the Moon?

It's about 384,400 kilometres away, and counting.

• How thick is a sheet of paper?

That all depends on the quality of paper and the manufacturer. Typical office paper (80 "gsm"*) is about 0.1mm thick, I think. As it makes sense to have everything in the same units, 0.1 mm is 0.0000001 kilometres.

Answering the question, part 1: How thick is a piece of paper folded 50 times?
I've seen this question tackled before, and the biggest mistake is to rush straight in and say "if we fold the paper 50 times, the paper will be 50 times thicker".

Think about it like this (feel free to grab a sheet of paper and try it):

• Fold a piece of paper in half once. It's twice as thick as it was, right?
• Fold it in half again. It's now twice as thick as it was last time- if you've got the paper in front of you, you can count the layers: it's four times as thick as the original piece of paper.
• Fold it in half again. if you count the layers, you'll see that it's now eight times thicker than the original.

By now, it's fairly easy to see what's happening: each time you fold, you're doubling the thickness of your lump of paper. That means it's fairly easy to see the long, slow way of calculating the thickness of the paper after 50 folds:

Start off with the thickness of one sheet, and then double it. Double it again. Double once more. Double again. And again. And again. Keep going until you've done it 50 times. If you're into button-mashing** on a calculator, you could type this in:

• 0.0000001 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 112,589,991

Because the original number (0.0000001) was in kilometres, we know that the answer must be. So if we fold a piece of paper 50 times, it would end up being 112,589,991 kilometres thick!

Answering the question, part 2: Would this reach the Moon?

This is the easy bit: we just have to compare the two numbers:

• Distance to Moon: 384,400 km
• Height of paper folded 50 times: 112,589,991 km

The second number is bigger (and, importantly, given in the same units), so we have our answer: yes, it would!

But it'd be boring to stop there...

• 112,589,991 / 384,400 = 292.897999...

That means that our folded paper would reach nearly 300 times further than the Moon! So what's that far out?

• Travelling inwards from Earth to the Sun, we'd cross the orbit of Venus just after climbing a bit more than a third of our stack of paper.
• About 80% of the way up our stack, we'd cross Mercury's orbit.
• We'd be about 3/4 of the way to the Sun when we were standing on top of our paper tower.
• Travelling outwards from the Earth away from the Sun, we'd cross Mars's orbit a bit more than a tenth of the way up the stack.
• We wouldn't quite reach the main asteroid belt, and we'd need more than 400 more of these paper towers to reach the next planet, Jupiter.

Why don't we do it, then?!

Give it a try! If you get past 8 folds (remember, you're folding the paper exactly in half each time!), I'll happily send you a Mars bar. Any more than 10 and I could probably be persuaded to give you my car.****

* Why they can't use standard notation, I don't know. "gsm" stands for 'grams per square metre', which would properly be written as "g/m²"
** I'm not into button mashing. There has to be an easier way - we can use powers:

• 2 x 2 means the same as 2².
• 2 x 2 x 2 means that same as 2³.
• 2 x 2 x 2 x 2 means that same as 2⁴.

See the pattern? So,

• 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 means the same as 2⁵⁰.

This means that we can write our calculation much more easily and accurately*** as:

• 0.0000001 x 2⁵⁰ 

*** We're typing less, so there's less chance of' 'user error'.
**** As a disclaimer, you're not getting my car.

Your birth year + your age = 111. How?

There's a meme flying round the internet at the moment that tells you to...

... add the last two digits of your birth year to the age you will turn on your birthday this year (2011, for any time travellers). It tells you the answer will be 111.

For me, born in 1982 and turning 29 in October, this would be 82 + 29 = 111. Try it with your own details...

It works!

But how?
To complete this task, you need to know two pieces of information: the last two digits of your birth year, and the age you will turn this year.

Your birth year is easy- just take the two digits off the end: for me, it's 82.

You probably know your own age, but if we're going to figure out how this works, we need to think about it in a different way. To find out your age knowing only the year you were born in and what year it is now, you could subtract your birth year from this year. Using my info, that'd be...

 2011 - 1982 = 29

... which I can confirm is correct!

The thing is, we're only supposed to be using the last two digits of the year, so lets try...

 11 - 82

That gives us a strange value: -71. But consider that the leading two digits of the years we're using are different: 20 and 19 respectively, and remember that this is because we're in different centuries: a century is 100 years, so add this on and what do we get? 29!

So to find out the age this year of anyone who was born during the previous century using only the last two digits of the year, we could do the following:

 Age = 11 - the last two digits of their birth year + 100

This looks a bit clumsy, so I'm going to use the letter 'x' to represent 'the last two digits of their birth year':

 Age = 11 - x + 100

Lets clean it up further: We start off with 11, take something from it, and then add on 100. Why not deal with those two numbers at the same time? If I start with 11 and then later add 100, I may as well start off with 111 in the first place:

 Age = 111 - x

Right, that's the slightly complicated bit sorted. Lets put it all together:

The info we need:

• Last two digits of birth year: we're calling this x.
• Age this year: we're saying this is 111 - x

What we have to do:

• birth year + age;
• Using the notation we've defined above, that's: x + 111 - x

But wait... we're starting off with whatever x is, then adding on 111, then taking x off again. Whatever x turns out to be, it doesn't really matter because we're just getting rid of it, leaving just...

111

... all by itself!

An important point...
... just pointed out to me by @justfin is that, due to the fact that two-digit years go in 100-year cycles, anyone who's over 100 years old this year will find that they get an answer of 211 instead of 111!

How does binary work?

Like the decimal system that we all use on a daily basis, binary notation is just another way of writing down numbers. To get the idea of how it works across, I'll start off by briefly explaining how the decimal number system works: 

How to read decimal numbers

31

The number above is 'thirty-one'. You know that because you've been brought up to count using the decimal number system. We, as a developed society, use it to a large extent not only because there are some nice, easy to remember patterns that make themselves present with a decimal number system, but because that's what our parents used, and their parents, and their parents...

How does it work?
The number system we're familiar with uses ten 'bases' - the symbols 1, 2, 3, 4, 5, 6, 7, 8, 9 and 0 - displayed in a place value system. That means that where we put them in relation to each other is important. We know that '31' means 'thirty-one' because written like that the 3 represents 'three tens' and the 1 represents 'one unit'. Put together, three tens and one unit make thirty one (of whatever we're counting) altogether.

We can represent ever bigger numbers by adding in more columns to the left- the next one, for example, allows us to say how many hundreds we want (0 - 9), the one after that describes how many thousands there are, and so on for ever and ever and ever, if we wanted to.

That's a crash course in the decimal "base ten" number system- remember that's the one you use every day. On to binary.

How to read binary numbers

11111

The number above is also 'thirty-one', or rather it represents the same amount as the number 31 does in the decimal number system.

How does it work?
The binary system works in a similar way to the decimal one, except it uses only two 'bases': 0 and 1 (i.e. there are no 2s, 3s, 4s, 5s, etc). Instead of each digit telling us how many 'ones', 'tens', 'hundreds', etc are in the number, the columns are labelled in a different way. The first column (starting from the right) is still 'units' (or 'ones'), but the next one to the left now represents how many twos we want in our number. The one after that is how many fours, and then it's how many eights. Can you see the pattern? Each column (as we move to the left) represents double the value of the one before it, so:

11111 means:

One 'sixteen', one 'eight', one 'four', one 'two' and one 'one'. Added together this makes thirty-one.

Another example, on binary day:
Today's date is the tenth of October, 2010, or 10 10 10, and is being called 'binary day' for fairly obvious reasons.

so as a binary number, 101010 means:

One 'thirty-two', no 'sixteen', one 'eight', no 'four', one 'two' and no 'one'. Add the thirty-two, eight and two together and you get:

42

Maths tutoring now available!

Just a quick note to say that I'm now offering private mathematics tuition, primarily to people in the Kettering area of Northamptonshire but if you're further away and still want to hire me get in touch and I'll see what I can do.

I'm offering sessions on anything up to GCSE level at the moment, whether you're at school or a more mature student. Get in touch and we can discuss your needs, whatever they are. My prices start at £25 per hour (but may be more if you want me to travel further).

I have profiles on the following tutoring websites:
Flying Colours Maths | Tutors4me | TutorNet | TutorHunt | Tutor in UK
Feel free to get in touch through any of those, or use the contact me page on this site.

Why me?
I've got experience and understanding: when I first started out in the teaching business I was a teaching assistant in a local primary school. I know my stuff and I got on really well with the kids so I started working with individuals and small groups, sometimes helping those who were struggling to catch up, sometimes taking the more confident kids and pushing them further. I earned myself a good reference from the head teacher and started a PGCE* course, training to teach mathematics at secondary level. I've been teaching maths in a local secondary school ever since.
I've also worked in industry, fast food, a mobile telephones shop and the job centre, so I do have some experience of the real world outside of a classroom!
In terms of understanding, I earned myself a BSc in Mathematics with Astronomy from the University of Leicester. Maths teachers who actually have a maths degree are not as common as you might think.

I know my stuff, and if you're the kind of person who wants to learn and is willing to put the time and effort in, then I am as well.

• Maybe you've had some time off school and want to catch up with what your class has been doing in the mean time: I can help.
• Maybe you're headed for a B in your GCSE but you really want that grade A: I can help.
• Maybe you're not all that keen on maths but you know the value of getting that all important C and want to put that little extra in to get it, if only you can find someone to guide you: I can help.

I don't care if you're eight or eighty: if you want to improve your game in maths: I can help.

Who am I?
Why would you want to hire anybody without knowing whether you were going to get on with them? If I tell you a bit about me then hopefully you'll see that we can get on and have a bit of fun as well as learning some maths.

First and foremost, I'm a geek (but don't change channel just yet...). I'm into most things science related which is where my love of maths comes from: a good understanding of maths can help you to understand and explain pretty much anything else in a bit more depth. I love learning new things and almost equally love passing on the things I've learnt to anyone who wants to learn too. I'm especially interested in astronomy**, which is an incredibly mathematical science, and run a blog in which I answer questions on related issues. Head over to Blogstronomy and have a look!

I don't just do maths and science, though. I also play guitar, sometimes in a band that plays classic rock covers (and the odd slightly more modern number). I have long hair and am fond of leather jackets and a decent Status Quo riff, but I went to see - and thoroughly enjoyed - the Hairspray musical recently, and I'm looking forward to the 25^th anniversary showing of Les Miserables: my tastes are varied.

I'm young (I'll hit 28 next week, but you don't have to get me a card), enthusiastic, flexible and committed to helping you hit the targets that you set for yourself.

If you have any questions about me, maths or what I offer, or would like to enlist my services, please read the new tutoring page on this site, and then get in touch.

I've been given a lot of advice and guidance by Colin over at Flying Colours Maths who has also listed me as part of his expanding empire.




* PGCE stands for Post-Graduate Certificate of Education - it's the most popular route for gaining Qualified Teacher Status in the UK.
** Not to be confused with astrology..,

How many connections?

This post was inspired by @squiggle7's Challenge That 'let's get connected' Post.

In it, @squiggle7 wonders about the number of connections able to be made by people in a 'room', be that virtually or physically.

Let's keep this up to date and use connecting with like-minded individuals on twitter as an example:

Imagine there's only one person on twitter who shares a particular interest of yours- how many connections can be made? By 'connections' I'm meaning mutual followships (to keep it simple), so in this case it's obvious: only one connection can be made.

Spotting the pattern
How about two other like- minded people? That's three altogether. I'll call them A, B & C (because they're cool names, right?). Because we're looking at mutual followships only, we can assume that if A is linked with B, then B is linked with A, so we can have the following links:

• A with B
• A with C
• B with C
• 3 connections altogether.

O.k, now how about four people? A, B, C & D this time:

• A with B
• A with C
• A with D
• B with C
• B with D
• C with D
• 6 connections available to be made.

[I'm being systematic when listing these possibilities- can you see the system I'm using?]

And five people (A, B, C, D & E):

• A with B
• A with C
• A with D
• A with E
• B with C
• B with D
• B with E
• C with D
• C with E
• D with E
• 10 connections possible.

We could carry on, but it's fairly clear by now that it's going to get a lot more complicated a lot more quickly, so can we do anything to simplify things? Can we find a pattern to follow?

People  2 3 4 5
Connections  1 3 6 10

Look at the numbers in that table: as the number of people goes up, the number of connections increases too. But by how much? It's fairly easy to see that as the number of people increases by 1, the number of connections increases by 2, then 3, then 4... so we can be confident that the number of connections for 6 people would be 10 +5= 15, and the number of connections for 7 people would be 15+6= 21 and so on. If we wanted to find out how many connections would be possible for, say, 20 people, we could just carry on the sequence. But what about 100 people? 150 people? 2000? 6000? 132,947 people?

Using algebra
Time to think about it a different way... Imagine there's a large number of people all with the same interest who all connect with each other. How many connections will there be? How many people exactly doesn't matter, so we'll call it "n" (so it can stand for any number).

• Person A can make a connection with everybody (except him/herself), so that's n-1 connections.
• Everyone else can make the same number of connections, so that's n-1 connections, n times: n(n-1)
• But we're forgetting that once a person makes a connection, they're connected both ways, so we've actually counted each connection twice. That's easy to fix, though, we just need to halve what we've got so far: 

And that's that. In simple(ish) terms, this means that to find the number of connections that can be made with any number of people in this situation you have to multiply the number of people by one less than the number of people, and then halve the answer.

So, for 132,947 people, there are (132947 x 132946) / 2 = 8,837,385,931 connections that can be made. That's well over eight billion!

12 balls: The solution.

At least, I think it's a solution. Please have a good, long look at it, figure out what it's talking about and then tell me if I've got anything wrong.

If you haven't had a go yourself, don't look at the image, just go here for the original problem, and here for a clue. It wouldn't be any fun if you went straight for the solution, would it!


The solution
Warning! This is the solution and contains spoilers!

O.k, that's not very easy to read, is it? Might be a good idea to click on it and view it full-size.

Just in case you're not sure what's going on, here's a walk-through of just one of the possibilities to give you an idea:

1. First up, you need to name your balls. I've gone for the rather unimaginative (but not too confusing) 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 & 12. You could go for A-L if you wanted to, or you could think of 12 different names, but then you might get too attached.
2. Weigh* any four balls against any other four. I've gone for 1, 2, 3 & 4 against 5, 6, 7 & 8. Three things can happen here:
   i) The two sets of balls balance;
   ii) The two sets don't balance, with the first set rising and the second falling;
   iii) The two sets don't balance, with the first set falling and the second rising.
   Let's pretend they're balanced:
3. If the two sets are balanced, you know that balls 1 - 8 are fine, and the odd one out is either 9, 10, 11 or 12. To find out which one, first you need to weigh** two of the suspects against one suspect and one fine ball. In my example, I've gone with 9 & 10 against 11 & 1. Again, three things can happen:
   i) The two sets balance;
   ii) The two sets don't balance, with the first set rising;
   iii) The two sets don't balance, with the first set rising.
   Let's pretend they're unbalanced, and that the first set (9 & 10) rise.
4. If 9 & 10 rise, then one of two things is true:
   i) Either 9 or 10 is too light;
   ii) 11 is too heavy.
   To work out which it is, weigh*** the balls in the first set against each other (e.g. 9 against 10). Once more, three things can happen, and the solution depends on which:
   i) The two balls are balanced - in which case, it must be that 11 was the odd one out, and it was too heavy;
   ii) The balls are unbalanced, with 9 rising - in which case, it must be that 9 is odd, and it's too light (as 10 has already been in a rising set, so can't be too heavy);
   iii)The balls are unbalanced, with 10 rising - in which case, it must be that 10 is odd, and it's too light (for similar reasons as in the previous case).

I hope that makes some kind of sense. If not, or if you'd like any parts of my flow chart explaining in greater detail, feel free to get in touch (leave a comment, or send me a message on Twitter (@TeaKayB), or something).

* This is the first weigh.

** This is the second weigh.

*** This is the third weigh.

12 balls: A clue

Carrying on from the 12 pool balls problem posted the other day, here's a clue for anyone who's not too sure where to get started.

The clue comes in the form of a photograph of my whiteboard after a colleague and I had spent part of a free period muddling the problem through. It's not complete and it's not explicit, but then it wouldn't be a clue if it was, would it?

If you want to have a go without seeing the clue first, then CLICK HERE and close your eyes for a few seconds!



The clue
Warning! This is a clue and contains spoilers!

Click to view full-size image!

I'll post the full solution in another few days... subscribe to make sure you catch it!

I've you've got a maths-related question of your own, you could ask it in the comments or use the form at my main blog.

The 12 pool balls problem

Not my pic... click for originator

O.k, here's one for ya...

You have twelve pool balls and visually they're all exactly identical. One of them, however, has a different mass/weight to the other eleven, but you don't know whether it's heavier or lighter.

You have at your disposal a set of weighing scales so you can weigh the balls against each other, but no weights to go with it. You can only use these scales three times*.

Your job is to find out two things:

1. Which ball is the odd one out?
2. Is it heavier or lighter?

The rules

1. You're only allowed to use the instrumentation and apparatus as described above. No magicking weights or digital scales out of thin air, or inventing science prep rooms nearby.
2. No cheating. There are solutions available online, but if you look them up you're only cheating yourself. If you look it up and post it here, then you degrade yourself to a point a little beneath most traffic wardens.
3. There is a solution.
4. Feel free to comment and ask for clues. My HOD and I have worked out what we think is a pretty nifty solution, so I might post a clue-post in a few days or so if anybody asks for one.
5. I will post my solution at a future date as yet undecided, then edit this post to link to it. If you're reading this after that's happened, please give it a shot on your own (or with friends/colleagues) first. It's more fun that way!
6. Feel free to post your own solutions on your own blogs or websites and then let me know so I can link to them.
7. Subscribe to MathsQS for updates and new posts!

* No, I don't know why. Maybe they're set to explode, or something. Think Die Hard II with the jugs problem.

Movies and maths

Another maths blog that I follow, Mathspig, has made a series of posts debunking ten popular movie clichés. I won't go into too much detail here, as I'll just be repeating the posts themselves, so here they are:

10 Movie Clichés Debunked with Maths (introductory post)

1. You can duck bullets
2. You can out run a fireball
3. You can out run machine gun fire
4. If you are involved in a car chase, hijacking, explosion, volcanic eruption or alien invasion, you will not go into shock
5. You can jump out of an airplane [sic] that is about to crash and catch the baddie who took the last parachute
6. If the pilot dies the control tower can talk you down
7. You can safely jump from a burning skyscraper / bridge / aircraft into water
8. You can survive any amount of beating in a fight
9. A single match or candle will light a room the size of a football stadium
10. You can out run a lava flow

Number 5 reminded me of this (if you get bored you might want to skip to around 6 minutes to get what I'm talking about...):

Hare Lift - Watch more Videos at Vodpod.