Thursday, September 17, 2009

For God's Sake Even I Can Do This

Try this little test:

Imagine that you are blindfolded and there is a bag in front of you containing a large number of five different types of sweet. How many must you take out of the bag to be certain of having chosen three of the same type?

Think about it for a few minutes, it's not that hard.

Now if you got the right answer, phone up your local Primary School and offer to teach the staff.

(The answer's in this article)

Apparently only one in five new British Primary School Teachers can solve this problem, which according to their spokesperson; means that 75% can't.

33 comments:

Anonymous said...

If you are taught Maths as a sequence of unrelated processes on numbers (as it is and has been for several years) rather than an analytical problem solving approach (as it used to be) then you won't be able to solve this problem.

First question raised by most modern students would be "We didn't do this. What's the formula ?" ....

Anonymous said...

I checked through a few of the linked articles. In this one:
http://www.telegraph.co.uk/news/uknews/2066676/Economy-under-threat-as-maths-exams-are-dumbed-down-says-think-tank.html
the think tank Reform described sudoko as a 'maths puzzle'.

We're doomed aren't we?

Anonymous said...

Good job it's not a spelling puzzle. I meant Sudoku.

Anonymous said...

I lecture in a teacher training college and I have to admit that there aren't many of the students that I would want teaching my own children.

I know you don't have to be a genius to teach Primary, but the standard of written work from my student teachers really would shock you. Basic mistakes in grammar and spelling (eg 'there' vs 'their') are commonplace and it is depressing to wade through endless pages of work without finding anything that is to do with the question.

Almost all of them pass though...

wonderfulforhisage said...

"Apparently only one in five new British Primary School Teachers can solve this problem, which according to their spokesperson; means that 75% can't."

One in five is 20% leaving 80% that can't.

Physician heal thyself?!

Anonymous said...

I think that was meant as a joke, Wonderfulforhisage.

FrankC said...

I must admit I got it wrong. My answer was "6", but I skimmed the question, assuming "two" where it said "three".

Failure to read the question = FAIL.

Anonymous said...

The point is you could work out how to do it.

It's a classic type of mathematical proof where you (sort of) work out the opposite of what you want (how many balls can I take without having three of the same type) and work the answer out from that.

A lot of modern mathematics students would struggle with it because you have to derive your own solution to the problem and it's a two stage problem where the stages aren't subdivided for them in advance.

Modern "maths" does not work like this. It would (in an exam) give you the answer in two stages, if it asked a question like this at all (which it probably wouldn't), or it would feed you the answer with an easier example first.

It's one of the reasons maths is dumbed down. You don't need to work out *how to do it* any more, you just follow the processes.

This works as long as the problem is not abstract. Almost all advanced maths problems are though !

Anonymous said...

I dont geddit! Seriously, IT geek, good at math, dont get the requirement. I would have said "40% of the total sweets plus one" would be the number need to be required to be absolutely "sure" you got one (assuming even distribution of all 5 sweets in the bag).

Anyone care to correct me? Or is this a poorly worded question?

Anonymous said...

Oops, previous anon here, I said "get one", instead of "get three different". Doh!

Interesting note; it does not actually say how many sweets are there or what ratios, just "large numbers" and "five different". So I assumed even spread and gave a ratio / formula rather than a number. Do I pass?

Anonymous said...

It's not asking about a particular probability of picking out three sweets of the same colour, it's asking for the minimum number needed to make this a certainty. As long as there are at least two of each colour, that number doesn't depend upon the relative proportions at all.

You have totally misunderstood the question.

But of course you've passed.

Anonymous said...

Senior Science PGCE trainee here. Why is the answer '11'? Anonymous' comments at 9:21 are insightful, but no - I don't get it.

Lilyofthefield said...

I said 3 because you could pull out three the same colour, but my husband looked at me with deep pity and said that it doesn't guarantee anything. You pick out the first five and the fewest you can get of each colour is one. Pick out another five and the fewest you can get of each colour is one (if you get none, then you've already got at least two of another colour) so now you've got two of each. The next one you draw will be make three of something.

It just didn't sound right and to his bitter contempt I have been through all the "what ifs" - 2 of one colour + 2 of another +1, and none of colours 4 & 5 etc, and eleven works every time.

I started school in 1960, did old-school Maths and have a measured IQ of 148 (or did when I was young) and am still crap at stuff like this. And I'm a teacher hahahahaha.

Lilyofthefield said...

My son's just come in. I asked him the question. It took him about eight seconds to come up with "11".
Good job he didn't get my mathematical gene.

MarkUK said...

This is the simplest mathematical proof of all, for small numbers. It's called "proof by exhaustion" (and other terms).

You simply try every scenario, each time looking for the one that will not lead to the result you want.

A simpler version is where you have a drawer of socks, half black and half grey (n>2). the light is out. How many socks do you have to pull out of the drawer to ensure you have a pair? Obviously the answer is three.

If you pull out a black one first, you may pull out a black or grey one second. Taking a third sock, it must be either black or grey so if you haven't already got a pair (i.e. the first two chosen were one black and one grey) you are bound to match one of them with your third choice.

You can use a simple tree diagram to see this.

Now, such a problem is trivial - and it was the kind of puzzle my teacher set in what is now Y5 & Y6.

If new teachers can't do this, we have failed not only to teach them basic mathematics but also how to work things out logially.

Anonymous said...

Anon9:21 again.

Senior Science PGCE, no you can't. It's not your fault, you aren't stupid, but you haven't been taught Mathematics properly and you haven't been taught to think Mathematically.

Modern Mathematics teaching is about processes. Knowing how to solve a quadratic equation, or whatever. But not about how to do things with it, how to take Mathematics and make it do things, how to manipulate algebra to learn something you didn't know previously.

Now, to some extent O-Level always has been a bit process orientated and that allowed a bright non mathematician to do okay. You can learn how to do trigonometry, algebra, geometry, whatever. A-Level, though, wasn't.

The real difference between old and new A-Level papers is that new ones walk you through the answers.

There was a great example of this (from a GCSE paper) where a more complicated multiplication (say 27 x 62) was broken down into factors to make the sums easier (3 x 3 x 3 x 62). To someone like me, I'd do it automatically.

The point was, though, the question didn't set you the sum and ask you to evaluate it. It went something like:

1) show that 3 x 62 = 186
2) using this answer, show that 9 x 62 is 558
3) use this answer to show the value of 27 x 62

so you are effectively hand held through the question. Same with A-Level maths, even though it's more complicated, obviously.

The problem with this is you never have to really think, WTF do I do with this to get the answer - and that's where you really learn Mathematics.

Thicko said...

"a large number of five different types of sweet"

So, for argument's sake, let us say that the bag contains:
100 of sweet type 1
100 of sweet type 2
100 of sweet type 3
100 of sweet type 4
100 of sweet type 5

I pull out 11 sweets all of which are sweet type 1. How can 11 possibly be the correct answer?

Mark H said...

I tagged this on to the end of a maths test for my top set year 10 (ie 14 and 15 year olds) last Friday.

23% of them got the correct answer of eleven.

This is rather worrying-my group isn't that brilliant and yet their results apparently beat those of new Primary Teachers!

Anonymous said...

Hmmm, IT geek here again. Would you say "11" if there were only 2 of each type of sweet? (2 x 5 = 10)
"Thicko" has defined the limit problem quite nicely there.

I agree they are asking for a number but unfortunately not giving sufficient data. The "11" works if there are 5 of each type of sweet (draw out 5 jelly babies, then 5 toffees, and then a trebor mint in the worst case). If there are 6 of each type, you could draw 6 jelly babies, 5 toffees, giving 11 drawn and still only have 2 types of sweet.

I think I see the flaw, in that drawing 11 would be "likely" to give 3 different sets (assuming an even distribution), but not "certain".

Just goes to show, its not just the figures, its how you communicate the problem. (And remember, No-one expects the Spanish Inquisition!)

Anonymous said...

My wife was a teacher and simply said "why would I want to do this?" meaning take sweets out of a bag without looking until she got the 3 she wanted.

Guarantee of 3 the same = (2 x possible colours) + 1

The one that kids tended to relate to was the "painting doors" question. If I can paint a door in 20 minutes and it takes 20 minutes to dry how long will it take me to paint 20 doors with 2 coats of paint.

The best set of answers I ever heard was for a chemistry quiz where the clue was nude element.
a. Barium
b. Titanium
c. Arsenic

Guess what the little buggers chose.

Lilyofthefield said...

Thicko, if you pull out 11 sweets of the same type first go, then you've already got three of the same type!

Harry said...

We need to be clear on this:

there's nothing wrong with the wording of the question, nor is there anything missing.

It is just a simple case of looking at the worst scenario, ie on your first ten goes you pick two of each type of sweet. Then on the 11th go whichever sweet you pick must guarantee you three of a kind.

There's no need to know how many sweets in the bag, what their ratios are or anything else at all.

Harry said...

Sorry I forgot to add that the question clearly says "large number of sweets" in the bag, ie obviously not less than 11.

Thicko said...

Lilyofthefield, the object is to be certain of having 3 different types of sweet not 3 of the same type.

Harry, the number of sweets in the bag IS important. In your example you say the worst scenario would be to pick 2 types of sweet in the first 10 goes. But, if there are 10 or more of one type of sweet, the worst scenario would be to pick 10 of the same type. Even if the 11th pick revealed a sweet of a different type, you are still only left with 2 different types.

Most people seem to be assuming that because there are 5 TYPES of sweet there must also be 5 of each type. This is not so, as, according to the question there is merely a "large number". As IT Geek says, there is insufficient data to answer the question.

Thicko said...

To labour a point (and because I'm supposed to be cleaning the oven):

If there are 1 of each type of sweet, the answer is 3

If there are 2, the answer is 5
If there are 3, the answer is 7
If there are 4, the answer is 9
If there are 5, the answer is 11
If there are 6, the answer is 13

etc, etc.

Which of these is the "large number"?

Thicko said...

Or, I could read the question properly and realise that Lilyofthefield and Harry are correct. If the object is to be sure of having 3 of the SAME TYPE, then the answer is 11 regardless of quantity and distribution.
I am shamefaced and will place myself on the naughty step. Probably do some spanking too.

Anonymous said...

Thicko,
You only did what most of the uk population do, not read the question properly. As a long suffering science teacher trying to get kids through multiple choice papers,I found that setting problem solving questions like this forced them to admit they weren't absorbing all the information provided in the questions correctly.
Like everything else practice makes perfect.
Now get that oven finished!

Anonymous said...

As an unemployed Mathematics and IT teacher, I would explain it by turning the problem on its head. What is the largest number of sweets I can remove before I MUST have three the same.
There are five types of sweet, therefore I can remove two of each, giving ten. The next sweet MUST complete a set of three, the specific sweet is not mentioned and is therefore irrelevant. Algebraicly, I think, that gives us 2n+1= number of sweets, where n is the number of classes (or types) of sweet.
TTFN

Anonymous said...

1) show that 3 x 62 = 186
2) using this answer, show that 9 x 62 is 558
3) use this answer to show the value of 27 x 62

I may be being dense, but I don't see how this line of calculation leads to the answer.

1) Show that 2x62=124
2) Show that 7x62=434
3) use this answer to show the value of 27 x 62

would do it, as you just add 1240 to 434

Anonymous said...

You are all wrong. There is no answer to this question.

On the other hand if the question said at leastthree then the answer would be eleven.

Anonymous said...

and another thing. The UK is a lot better at this kind of question than, say, France or Eastern Europe.

Sample response from a Russian PhD applicant "give me an equation then I can solve it ..."

Rick said...

Anonymous @ 19:20 - pedantry is so much more effective when right, don't you think? "Three" in English does not mean "exactly three" by default.

dicky-bird said...

Of course knowing the answer, and being able to explain it, are quite different.

In the best case scenario, you pick out three sweets of all the same kind.

But to guarantee that you pick out three, consider having 2 of each kind of the five types of sweet, (that's ten sweets in total).

The 11th pick must be one of the five types, making three of the same kind.

So three sweets if you are lucky, 11 sweets will guarantee it, and you can do this with your eyes closed!