“The mark of a civilized man is the ability to look at a column of numbers

and weep”Bertrand Russell (1872- 1970)

COPY please

The Hanging paradox!

• A judge tells a condemned prisoner that he will be hanged at noon on one weekday in the following week but that the execution will be a surprise to the prisoner. He will not know the day of the hanging until the executioner knocks on his cell door at noon that day. Having reflected on his sentence, the prisoner draws the conclusion that he will escape from the hanging. His reasoning is in several parts. He begins by concluding that the "surprise hanging" can't be on Friday, as if he hasn't been hanged by Thursday, there is only one day left - and so it won't be a surprise if he's hanged on Friday. Since the judge's sentence stipulated that the hanging would be a surprise to him, he concludes it cannot occur on Friday.

• He then reasons that the surprise hanging cannot be on Thursday either, because Friday has already been eliminated and if he hasn't been hanged by Wednesday night, the hanging must occur on Thursday, making a Thursday hanging not a surprise either. By similar reasoning he concludes that the hanging can also not occur on Wednesday, Tuesday or Monday. Joyfully he retires to his cell confident that the hanging will not occur at all.

The Hanging paradox!

• The next week, the executioner knocks on the prisoner's door at noon on Wednesday — which, despite all the above, was an utter surprise to him. Everything the judge said came true.

Problems

On each table is a mathematical problem. Can you find the answers?

What’s wrong?

x = 0.999999….

10x = 9.99999….

10x – x = 9.99999 – 0.999999….

9x = 9

x = 1

What’s wrong?

x = 0.999999….

10x = 9.99999….

10x – x = 9.99999 – 0.999999….

9x = 9

x = 1

0.99999…. to infinity = 1

Catalogues

• Suppose that every public library has to compile a catalog of all its books. The catalog is itself one of the library's books, but while some librarians include it in the catalog for completeness, others leave it out, as being self-evident.

• Now imagine that all these catalogs are sent to the national library. Some of them include themselves in their listings, others do not. The national librarian compiles two master catalogs – one of all the catalogs that list themselves, and one of all those that don't.

• The question is now, should these master catalogs list themselves?

Catalogues• The 'Catalog of all catalogs that list themselves' is no

problem. If the librarian doesn't include it in its own listing, it is still a true catalog of those catalogs that do include themselves. If he does include it, it remains a true catalog of those that list themselves.

• However, just as the librarian cannot go wrong with the first master catalog, he is doomed to fail with the second. When it comes to the 'Catalog of all catalogs that don't list themselves', the librarian cannot include it in its own listing, because then it would belong in the other catalog, that of catalogs that do include themselves. However, if the librarian leaves it out, the catalog is incomplete. Either way, it can never be a true catalog of catalogs that do not list themselves.

The Monty Hall Problem

• You're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1 [but the door is not opened], and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

The Monty Hall Problem

•  You should always switch to the other door. If the car is initially equally likely to be behind each door, a player who picks Door 1 and doesn't switch has a 1 in 3 chance of winning the car while a player who picks Door 1 and does switch has a 2 in 3 chance. The host has removed an incorrect option from the unchosen doors, so contestants who switch double their chances of winning the car.

Seven Bridges of Königsberg

• The problem is to find a walk through the city that crosses each bridge once and only once. Every bridge must have been crossed completely every time and no swimming!

Seven Bridges of Königsberg

Except at the endpoints of the walk, whenever one enters a vertex by a bridge, one leaves the vertex by a bridge. In other words, during any walk, the number of times one enters a non-terminal vertex equals the number of times one leaves it. Now, if every bridge has been traversed exactly once, it follows that, for each land mass (except possibly for the ones chosen for the start and finish), the number of bridges touching that land mass must be even (half of them, in the particular traversal, will be traversed "toward" the landmass; the other half, "away" from it). However, all four of the land masses are touched by an odd number of bridges. Since two land masses serve as the endpoints of a walk, the proposition of a walk traversing each bridge once leads to a contradiction.

What’s wrong?

Let us begin with an innocent statement, leta = b

Multiply both sides by a to geta2 = ab

Add a2 – 2ab to both sidesa2 + a2 -2ab = ab + a2 -2ab

This can be simplified to2(a2 –ab) = a2 –ab

Divide both sides by a2 – ab2 = 1

?!

What’s wrong?

Let us begin with an innocent statement, leta = b

Multiply both sides by a to geta2 = ab

Add a2 – 2ab to both sidesa2 + a2 -2ab = ab + a2 -2ab

This can be simplified to2(a2 –ab) = a2 –ab

Divide both sides by a2 – ab2 = 1

?!

If a = b

a2 –ab = zero

You can’t divide by zero!

Birthday coincidence

What is the chance/probability that two people in a group of 23 randomly selected people have the same birthday?

Birthday coincidence

We need to look at how many possible pairs there are.

Person 1 and person 2Person 1 and person 3Person 1 and person 4 etc.

There are 22 possible pairs with person 1

Birthday coincidence

There are 22 possible pairs with person 1

Then person 2 with person 3

person 2 with person 4

person 2 with person 5 etc.

There are 21 possible pairs with person 2

Birthday coincidence

There are 22 possible pairs with person 1

There are 21 possible pairs with person 2

It follows there are 20 possible pairs with person 3, 19 with person 4, 18 with person 5 etc.etc.

Birthday coincidence

The total number of different pairs of people is therefore;

22 + 21 + 20………….+ 3 + 2 + 1 = 253

Since there are 365 possible birthdays (we’ll ignore 29th February for simplicity!), the chance of two people having the same birthday is 253/365 = 0.69 (69%)

Birthday coincidence

What is the chance/probability that two people in a group of 23 randomly selected people have the same birthday?

69%

If there are over 30 people the chance is even higher, worth a bet at a party!

Class size

There are three classes in year 3. One class contains 12 students, one class contains 25 students, and one class contains 23 students. What is the average class size from

• the teachers’ point of view?

• the students’ point of view?

The teacher’s point of view

The teacher has to teach 3 classes, one of 12 students, one of 25 students, and one of 23 students.

Average class size = (12 + 25 + 23)/3 = 20

The students point of view

There are 12 students in a class of 12. When asked the size of their class they will all say 12

12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12.

25 students will say 25!

25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25.

23 students will say 23!

23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23.

The students point of view

Average class size for a student =

((12 x 12) + (25 x 25) + (23 x 23))/(12 + 25 + 23)

= 21.6

(Remember the average from the teacher’s point of view was 20!

A TruelA truel is similar to a duel, except there are three participants rather than two.

One morning Mr Black, Mr Grey and Mr White decide to resolve a conflict by truelling with pistols until only one of them survives.

Mr Black is the worst shot, he hits the target only 33% of the time.Mr Grey is a better shot, hitting the target 66% of the time.Mr White is the best shot, he hits the target every time (100%)

To make the truel fairer, Mr Black shoots first, followed by Mr Grey (if he is still alive!), then Mr White (if he is still alive). They go round again until only one is left alive.

Where should Mr Black aim his first shot to give him the best chance to survive?

Option 1

If Mr Black shoots at Mr Grey and hits him Mr White will then shoot Mr Black and Mr Black will die because Mr White never misses.

I

Option 2

If Mr Black shoots at Mr White and hits him the next shot will be taken by Mr Grey. Mr Grey is only 66% accurate so there is a chance Mr Black may survive to fire back at Mr Grey and win the Truel.

This would appear to be a better option then option 1.

Another option?

Is there a third option?

Option 3

Mr Black could aim in the air. Mr Grey has the next shot and will aim at Mr White because he is more dangerous. If Mr White survives he will aim at Mr Grey who is a more dangerous opponent.

By aiming in the air, Mr Black is allowing Mr Grey to eliminate Mr White or vica versa.

The best strategy

Option 3 is the best strategy. Eventually Mr Grey or Mr White will eliminate each other, allowing Mr Black the first shot in a duel instead of a truel.

Oslo to Kristiansand

A student drives from Oslo to Kristiansand (300 km) and back again. Her average speed for the first half of the journey is 50 km/h (Oslo to Kristiansand) and her average speed back is 20 km/h. What is her average speed for the entire journey?

Oslo to Kristiansand

For Olso to Kristiansand, time = distance ÷ speed = 300 ÷ 50 = 6 hours

For Kristiansand/Oslo, time = distance ÷ speed = 300 ÷ 20 = 15 hours

Average speed = distance ÷ time

= (300 + 300) ÷ (6 + 15)

= 28.6 km/h.

(not (50 + 20)/2 = 35 km/h)

Not always obvious is it?

Mathematics

“Math – that most logical of sciences – shows us that the truth can be highly counterintuitive and that sense is hardly common”

K.C. Cole