Tennis - Winning a 5 set match

I read this earlier, at The Atlantic in a story about the possible demise of 5 set matches at grand slams.

A best-of-three format (say, Player A vs. Player B) can produce only four distinct three-set outcomes: ABA, ABB, BAA and BAB. A five-set match, on the other hand, has 12 possible set outcomes, creating a host of ways momentum can twist and turn over the course of several hours.

I hadn’t thought about this before.

The final sentence seemed a little odd, until I realised they were talking about the number of outcomes when then match goes to the full five sets.  They are deliberately ignoring the matches that only last 3 or 4 sets.

Anyway - this is more interesting than the common scenario when we want to know how many arrangements there are if you have two things to choose from.  With five sets in a match (and using Player A and Player B as in the Atlantic article) you would get 2^5 = 32 outcomes.  Some of these, of course, are not real, because if Player A wins all of the first three sets (a straight sets victory) the game ends without the final two sets being played, regardless of whether Player A would have gone on to wins them or whether Player would have done.

Asking secondary pupils to work out how many different ways there are to win a five-set match (when it doesn’t have to go the full distance) seems like a nice idea.  I can think of three ways they might set (!) this out.  Are there more?

One is to use a tree diagram and to cross out the ones that will never crop up because one player has already won.  The bottom row shows the 12 ways mentioned in the original article. 

There are therefore (from this diagram) 20 different possible ways for the match to finish.

 

Another is to make a list of these outcomes and not to count the ones that are the same as other matches that have already finished.  The numbers in the next diagram show the counting, while the shading shows who has won:

The symmetry here is nice and shows that I could have got away with just doing the top half and then doubling the number of ways.  This is just a repeat of the tree diagram, though, so the next way concentrates more on the combinations that work.

Here are the ways that the game can progress if player A wins the first two sets:

If Player A wins the first set and B wins the second then this can happen:

…which gives 10 ways so far.  Then we can exchange A and B to give us the other 10 ways.

Liu Bolin

This link takes you to a wonderful picture.  [I am linking to it because I don’t own the copyright and don’t think I should reproduce it here directly.]  Look at it before you read on.

It is a photograph of a supermarket shelf in China. 

At first glance it is nothing particularly remarkable.  But then, when you look closer, you can see the outline of a person.  The artist, Liu Bolin, has had his clothes, hands, head and face painted to match his background, so he almost blends in, like a chameleon.

He has done this rather a lot in different situations, but this particular picture is my favourite because it can also be used to do some maths.

How I used it in class

As part of the starter activity I projected the picture while the pupils were arriving.  It was good to see how closely they were paying attention (the sudden “oh my goodness - there’s a man there”) was a giveaway!   

We then discussed what questions they would be able to answer.  They came up with:

·         How many bottles are there?

·         How many cans are there?

·         How much liquid is there?

·         How long would all this drink last one person?

·         What percentage of the shelf is covered by the man?

·         What fraction of the picture is taken up by the man?

I gave them a copy of this on paper (I fitted two copies of the image on one sheet and printed them out in black and white).  We decided there were two rows of bottles and assumed that there were also two rows of cans.

The question the pupils then worked on was: “how much weight are the shelves holding?”.

Some pupils wanted to know how much liquid each type of container holds.  Others explained that the large bottles on the bottom shelf were 2 litre bottles, the smaller ones looked like 500ml bottles and the cans hold 330ml. 

Someone pointed out that three cans stacked up could be treated as 1 litre (330ml x 3 = 990ml).  We also knew the answer wouldn’t be exact because we don’t know the weight of an empty bottle/can.

Finally, we needed to know how heavy the liquid is.  Someone knew that a litre of water has a mass of 1kg.  We decided that other drinks were likely to have a very similar density to that of water.

The generally agreed answer was just over half a tonne.

What did the pupils gain from doing this?

They used different conversions (and most learned that a litre of water has mass 1kg), including converting kg to tonnes.

They had to make decisions about what sorts of estimating they could and should do.

They had opportunities to decide on questions to work on (and did work on these after answering the question above).

Unleaded petrol

Here is a real life starting point for a lesson starter.  It was inspired by this advert in my local petrol station:

I plan to present it like this:

Unleaded petrol at my local station costs 133.9 pence per litre.

I filled up and needed 53.27 litres.

I then used the offer in the advert and bought two bottles of drink that cost £1.79 each.

How much did I save?

How much petrol would I need to buy in order to get the drinks for free?

I could also ask pupils at what stage it is worth getting the drinks even if I don’t really want them.  Is it sensible when they are free? - definitely.  Is it worth it when they effectively cost me only 8p? - almost certainly.  Is it worth it if I only end up saving 5p?  No.

Also, assuming I do want the drinks, what is the percentage saving on the cost of the petrol if I use the offer?

Should be a brief activity.

Presentation files (SMART and PPT) available on TES.

Bluefin Tuna

A story in the news today has provoked a lesson idea. 

I am taking most of my text from The Daily Telegraph, 5Jan13, but it also appeared elsewhere.

A bluefin tuna was sold at auction in Japan in January 2013 for £1.09 million.

The winning bidder, president of the Sushi-Zanmai restaurant chain, said “the price was a bit high”.

I have got the following on a sheet/PPT for pupils to read:

From The Daily Telegraph:In the year's first auction at Tokyo's sprawling Tsukiji fish market, the 489-pound tuna caught off northeastern Japan sold for the record price of £1.09 million, said Ryoji Yagi, a market official. 

But with a single mouthful-sized piece of sashimi weighing around 1oz, the record breaking tuna is worth around £??? per bite. 

Japanese eat 80 per cent of the bluefin tuna caught worldwide, and much of the global catch is shipped to Japan for consumption. 

The price works out to a stunning 700,000 yen per kilogram. 

In November, the 48 member nations of the International Commission for the Conservation of Atlantic Tunas, or ICCAT, voted to maintain strict catch limits on the species, although some countries argued for higher limits. The quota will be allowed to rise slightly from 12,900 metric tons a year to 13,500. 

From Wikipedia: At maturity it is about 1.5 m (4 ft 11 in) long and weighs about 60 kg (130 lb)

I thought I would then ask the pupils what things we can work out and what information they would need to be able to do this.

Here are the ideas I have had:

•       How much it weighs in kg.

•       The exchange rate between the yen (¥) and the pound sterling (£).

•       How much does a 1oz serving of the tuna cost?

•       How many tonnes of tuna are eaten in Japan each year?

•       How many tuna fish is that?

•       What is the percentage increase in the amount of tuna that can be caught next year?

•       How many times heavier than me is the fish?

•       Any others?

And the information I think the pupils might need?  Conversion between different units:

Imperial:

1 stone = 14 lb (pounds)

1 lb = 16 oz (ounces)

Metric:

1kg = 1000g

1 tonne = 1000kg

Conversion:

1 kg ≈ 2.2 lb

What are the benefits of this activity?

I will use it as a starter with Yr 10.  It is loosely a functional problem, but will involve not only extracting information from the newspaper article but also being creative to decide what we could work out.  It should also allow them to revise metric/imperial units of weight.

It might also happen that some pupils will think of other things they want to work out and will then need to find out more background information (for example, if they want to know how much tuna is consumed per person on average in Japan each year they will need the rough population of Japan).

There are also a few practical issues for us to consider.  Can we just divide the cost of the fish by its weight to work out the price per kg (or per lb or per oz) or do we need to allow for bones, skin, internal organs, etc (or do some of these get eaten as well?).

Resources are on TES.

The jugs problem - but is it maths?

Some of the other websites I have read about the Jugs problem state that it is merely a riddle, or that there is no maths involved in solving it.  So: are the problems in the previous blog posts (Die Hard: With a Vengeance - 3 and 5 gallon jugs and Generalising the jugs problem) actually mathematics or not?

One reason this is worth considering (in my opinion) is that there are lots of other problems that fall into this category (“nice - but is it maths?”).

Clearly this depends on your definition of “mathematics”.  There are, though, lots of different skills that are used (or that can be used) to solve the problem and these are either clearly mathematical or are helpful in solving other mathematical problems.  I think these include:

·         Flexibility.  Working in different ways (practically, using diagrams, with numbers) is helpful.

·         Using trial and improvement.  Very often we don’t know how to get started.  Getting started by doing something is often useful, even if it turns out not to be correct.

·         Getting used to the problem.  Realising that you can pour from one jug to the other, realising that when you pour away water you are emptying a jug, etc, are all helpful.  It might also be useful to realise that the units involved are all the same, so if gallons are off-putting because of their unfamiliarity then working in pints or litres will give the same answer.

·         Tenacity.  I suppose this follows on from the trial and improvement.

·         Mathematising.  While it is possible to answer the question without doing this, the mathematisation makes it easier to generalise.

Even if it isn’t maths in its original form, a problem like this can end up using lots of mathematical skills. 

So: it is mathematics?  I would say "yes", but if you disagree I would say "who cares?".  There are so many useful things that arise from doing it that it doesn’t really matter!

Generalising the jugs problem

This problem appeared in the previous blog: do think about it before reading on.

“With any two jugs, holding a and b gallons when full (where a and b are integers), can you make all of the values from 1 to a+b?”

This clearly cannot be done if a and b are both multiples of the same thing.  For example, if they are both multiples of 3 then the only answers that can be made must also be multiples of 3 (since we will be adding or subtracting multiples of 3 each time).  I think the easiest one to consider is where a and b are both even.  The only numbers you can make are also even.

If they are not multiples of the same thing (ie they are co-prime) then I think we can make every number up to a+b and I reckon the addition/subtraction method used in the previous blog helps to show how it works. 

Here is an example for 7 and 4:

1 = 4+4-7

2 = 7+7-4-4-4

3 = 7-4

4 = 4

5 = 4+4+4-7

6 = 7+7-4-4

7 = 7

We could of course just use the values for 1 each time and repeat as many times as is necessary.  This would mean that 3 would be made by doing three lots of (4+4-7), but 4+4-7+4+4-7+4+4-7 is significantly less neat than 7-4.

For the rest of the numbers up to 11 we can put the numbers already obtained into the small jug and fill the bigger one to the brim (so 8 is 7+1, etc).

This doesn’t prove it is possible for all pairs of co-prime jugs, nor does it explain how to do it, but it does give a method that can be applied to work it out.

Die Hard: With a Vengeance - 3 and 5 gallon jugs

This old problem is particularly well-known from the film “Die Hard: With a Vengeance”. 

You have a 5 gallon jug and a 3 gallon jug and need to measure out exactly 4 gallons.  How can you do it?

In the film they have a time limit and a reason for getting it right (a bomb will go off if they aren’t accurate).

If you haven’t seen the problem before then please do work on it before looking for the answer. 

There are lots of webpages that give the solution to the problem (Google is your friend), and the answer is also given in the film itself.  It also appears near the bottom of the trivia page about the film.  

There are a couple of interesting things about this problem that I haven’t seen noted elsewhere.

1]  It is great that a Hollywood movie uses problems like this and bring them to a wider audience.

2]  The problem itself needs to be stated more carefully than it appears above: for example the jugs need not be uniform in cross-section (so you can’t just fill both to half-way and then combine) and there are no markings on the jugs (you can’t just read off the “4 gallon line”).

3]  Real life issues need to be ignored (the 3 gallon jug does hold exactly 3 gallons, and these jugs can be poured from without any water spilling - I can’t get a measuring jug to do this!).

Mathematising the problem

Many webpages give the answer as a list of things to do.  Sometimes it is noted that there are two solutions to the problem (although there are clearly lots of solutions - you could just pour water from one jug to another for as long as you want before getting down to solving it) and sometimes the most efficient solution is sought (using the fewest number of moves).  This problem is significantly easier if it is mathematised. 

Using the first solution from the IMDB page:

1. Fill the 5 gallon jug and decant the water into the 3 gallon jug. This leaves two gallons in the big jug. 2. Empty the 3 gallon jug and pour in the two gallons from the 5 gallon jug, leaving space for one gallon in the small jug.  - this is equivalent to 5 - 3.

3. Refill the 5 gallon jug - this is equivalent to adding 5.

and pour water from it into the 3 gallon jug until the small jug's full. - Because the small jug is not used the water is essentially poured away, so this is equivalent to subtracting 3.

So: we did 5 - 3 + 5 - 3, which equals 4.

The beauty of this is that it can be generalised.  As long as we use 5s and 3s to make the answer 4, we can generate alternative answers.

3+3+3-5 = 4.  Now we just need to finish it off.  Start with 3 gallons - pour it into the big jug.  Fill the small one again and pour into the big jug.  Empty the big jug (subtract 5) pour the 1 gallon left in the small jug into the big jug and then pour a full small jug into the big jug.  In fact we are doing 3+3-5+3=4

Generalising the problem

It is also rare to see a generalisation of the problem. 

“With any two jugs, holding a and b gallons when full (where a and b are integers), can you make all of the values from 1 to a+b?”

My thoughts on this will appear in the next blog (and that is a resolution!).

New Year’s Resolution?

I haven’t decided whether blogging regularly really is a new year’s resolution or not, but I do hope to post fairly often.

We’ll see how I get on …