I'ma little late on this one but never the less.

On Saturday August 22nd the 4th South African Tertiary Maths Olympiad was held. The final results can be seen here. A large thanks to Stephan Wagner for organizing the contest, and to Margaret Archibald, Steven James, Celeste Johnson and Shunmuga Pillay for helping out with invigilating at Wits.

Also I must thank the Mathematical Science Students Council at Wits for advertising the event far far better than I could ever have done. In the end we had 35 Wits students sat the exam.

I plan to do another post on the statistics of the exam over the last few years fairly soon but for now I'll simply say congrats to the winners and to everyone who took part.

## Sunday, September 6, 2015

## Sunday, May 24, 2015

### Math Bee

I've been asked to set a competition style maths contest for first year students at wits. For some reason or another the organizers decided to call this a (the?) math bee. The goal is 20 multiple choice questions of a level somewhere between the SAMO Seniors 2nd round and the SATMO.

I'd like to avoid the paper having too many of my own personal biases. To that end this post is a call for questions. If anyone has any that seem appropriate then please e-mail me them to me, I will thank you via blog post.

I'd like to avoid the paper having too many of my own personal biases. To that end this post is a call for questions. If anyone has any that seem appropriate then please e-mail me them to me, I will thank you via blog post.

## Sunday, May 17, 2015

### Jury Duty

I've been playing with some problem's from the book "50 Challenginh problems in probability with solutions by Frederick Mosteller (my great grand advisor as it happens).

Anyway one of the first problems compares two different juries.

The first is a one man jury where the juror gets it right with probability p. As there is only one juror his ruling applies.

The second is a three man jury with two jurors getting it right with probability p (independently) and the third coin-flipping juror who gets it right with probability 1/2 (again independently). In this jury majority rules. Which is to sy we need two of the three jurors to get things right.

Now the question asked is "which is better". A little algebra shows that they're both equally good, which is to say that the second jury reaches the right verdict with probability p.

So adding one "regular" (right with probability p) juror and one coin flipper doesn't change anything! What if we ad yet another coin flipper and another regular juror? Now we need three of five correct. Now things

A little more algebra shows that we now get the correct verdict with probability,

ppp+2.25ppq+0.75pqq=

pp(p+q)+1.25ppq+0.75pqq=

pp+0.75pq+0.5ppq=

p-0.25pq+0.5ppq=

p+pq(2p-1)/4

Which means that is p>1/2 that adding two jurors to our three (as we did to our one) now improves the verdict. Similarly if p<1/2 (why we're using jurors who're worse than coin flips I don't know but if we did) adding jurors 4 and 5 makes things worse.

Does anyone have any intuition for why adding the first pair of jurors does nothjign but adding the second pair helps?

Anyway one of the first problems compares two different juries.

The first is a one man jury where the juror gets it right with probability p. As there is only one juror his ruling applies.

The second is a three man jury with two jurors getting it right with probability p (independently) and the third coin-flipping juror who gets it right with probability 1/2 (again independently). In this jury majority rules. Which is to sy we need two of the three jurors to get things right.

Now the question asked is "which is better". A little algebra shows that they're both equally good, which is to say that the second jury reaches the right verdict with probability p.

So adding one "regular" (right with probability p) juror and one coin flipper doesn't change anything! What if we ad yet another coin flipper and another regular juror? Now we need three of five correct. Now things

__do__change.A little more algebra shows that we now get the correct verdict with probability,

ppp+2.25ppq+0.75pqq=

pp(p+q)+1.25ppq+0.75pqq=

pp+0.75pq+0.5ppq=

p-0.25pq+0.5ppq=

p+pq(2p-1)/4

Which means that is p>1/2 that adding two jurors to our three (as we did to our one) now improves the verdict. Similarly if p<1/2 (why we're using jurors who're worse than coin flips I don't know but if we did) adding jurors 4 and 5 makes things worse.

Does anyone have any intuition for why adding the first pair of jurors does nothjign but adding the second pair helps?

## Saturday, April 4, 2015

### Another lightbulb puzzle

So last week was the first weekly Joburg Math meetup. Which basically means a bunch of friends meeting up and playing with math olympiad type problems. If anyone is interested these meetups happen on Sundays at noon in Motherland coffee Rosebank Mall, and of course everyone is welcome. hopefully they'll also give me new ideas for more regular math type blog posts.

Anyway here is a problem from The book Creative Mathematics by Alan Beardon, and soem generalizaions due to those involved in the meetup.

Problem 1. I have a line of n lightbulbs. Each lightbulb has a switch associated to it. The switch toggles (if on changes to off, if off changes to on) all bulbs except the one it's associated to. If all the bulbs start out as off, for whcih n can they be changed to all on?

Problem 2. What if I put the bulbs in an n1 by n2 grid. Again each bulb has a switch associated to it. The switch toggles all bulbs sharign either a row or coloumn with it's bulb. Another way to say that is that it toggles bulbs agreeing in with it's position in exactly one coordinate.

Problem 3. What if I have an n1 by n2 by... by nk grid. Now each switch toggles bulbs which agree in exactly k-1 co-ordinates.

Problem 4. We have the same grid as problem 3 but now the switches toggle bulbs agreeing with our bulb in exactly l coordinates for general l.

I won't say how far we got with these at the meetup because it would spoil the fun somewhat. I will say that only problem 1 actually occurs in Beardon's excellent book, so it's not to be taken as a given that there are nice solutions beyond that.

Anyway here is a problem from The book Creative Mathematics by Alan Beardon, and soem generalizaions due to those involved in the meetup.

Problem 1. I have a line of n lightbulbs. Each lightbulb has a switch associated to it. The switch toggles (if on changes to off, if off changes to on) all bulbs except the one it's associated to. If all the bulbs start out as off, for whcih n can they be changed to all on?

Problem 2. What if I put the bulbs in an n1 by n2 grid. Again each bulb has a switch associated to it. The switch toggles all bulbs sharign either a row or coloumn with it's bulb. Another way to say that is that it toggles bulbs agreeing in with it's position in exactly one coordinate.

Problem 3. What if I have an n1 by n2 by... by nk grid. Now each switch toggles bulbs which agree in exactly k-1 co-ordinates.

Problem 4. We have the same grid as problem 3 but now the switches toggle bulbs agreeing with our bulb in exactly l coordinates for general l.

I won't say how far we got with these at the meetup because it would spoil the fun somewhat. I will say that only problem 1 actually occurs in Beardon's excellent book, so it's not to be taken as a given that there are nice solutions beyond that.

## Monday, January 12, 2015

### The condom problem

Today I'm going to share a problem which I learned from Prof Vishnu Jejjela . As usual I won't spoil the solution for a few days so that people can mull over it. In the mean time I'll lay out the problem and a few special cases which I think are worth thinking about.

n men and m women want to have sex with each other. In particular they want all mn pairings to occur. The problem is that they all have a different STD and they all want to avoid getting anyone else's STD. For this they have condoms, these condoms have some fairly well defined properties.

1. They can be nested.

2. They can be used in either direction.

3. They never break.

4. Diseases never get through them.

5. On the other hand if a side of condom A is in contact with a side of condom B with disease x then disease x jumps to condom A.

6. If a person with disease y uses condom C then disease y appears on the side of condom C actually used.

The problem of course is to minimize the number of condoms used. The naive solution of course involves using mn condoms (1 for each encounter).

Two classically special interesting cases are

1. m=n=2.

2. n=1, m an arbitrary odd integer.

n men and m women want to have sex with each other. In particular they want all mn pairings to occur. The problem is that they all have a different STD and they all want to avoid getting anyone else's STD. For this they have condoms, these condoms have some fairly well defined properties.

1. They can be nested.

2. They can be used in either direction.

3. They never break.

4. Diseases never get through them.

5. On the other hand if a side of condom A is in contact with a side of condom B with disease x then disease x jumps to condom A.

6. If a person with disease y uses condom C then disease y appears on the side of condom C actually used.

The problem of course is to minimize the number of condoms used. The naive solution of course involves using mn condoms (1 for each encounter).

Two classically special interesting cases are

1. m=n=2.

2. n=1, m an arbitrary odd integer.

## Sunday, January 4, 2015

### Happy New Year!!!

Happy new year guys (yes both of you). Anyway it's 2015 and I'm going to try to be a more active blogger this year (yes, yes easy to say at the start of the year when everything else going on is comparatively quite).

Anyway what better way to kick of the year than with the 2015 problem of the year . I've played with it a bit but I think I'll leave the ones I have out for a bit. There are a few nice tricks which I don't really want to spoil for anyone reading. I'll leave it for a commenter to spoil. :-)

Anyway what better way to kick of the year than with the 2015 problem of the year . I've played with it a bit but I think I'll leave the ones I have out for a bit. There are a few nice tricks which I don't really want to spoil for anyone reading. I'll leave it for a commenter to spoil. :-)

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