GAMINGWITHMATHEMATICS

E = mc²

AGENDA

Game Theory Game 1: Monty Hall Game 2: Pick Up Sticks Fibonacci Series Game 3: Brick Wall Game 4: Centipede Game Nash Equilibrium

GAME THEORY

A branch of applied mathematics used in social sciences, economics, biology, computer science and philosophy

Attempts to mathematically capture behavior in strategic situations

Traditional applications attempt to find sets of strategies where individuals are unlikely to change their behavior

Also known as equilibria

These equilibrium concepts are motivated differently depending on the field of application

APPLICATIONS OF GAME THEORY

Political Science Economics and Business Biology Computer Science and Logic Philosophy

GAME 1: MONTY HALL

Player is given a choice of three doors where only one contains a prize

The player first chooses an initial door

The host will then reveal an empty door among the other two un-picked doors

Player is given a second chance if he or she would like to switch to the other un-picked door

The host will open up the final chosen door and reveal the result

The player wins if the door he or she picked in the end contains the prize

http://www.shodor.org/interactivate/activities/SimpleMontyHall/?version=1.6.0_03&browser=MSIE&vendor=Sun_Microsystems_Inc.

You think the chance of winning is 50-50?

Think again…

The probability of winning by using the switching technique is in fact 2/3

While the odds for winning by not switching is 1/3

Let’s take a closer look…

The probability of picking the wrong door in the initial stage of the game is 2/3

If the player picks the wrong door initially, the host must reveal the remaining empty door in the second stage of the game, then the last door must be the prized one

Thus, if the player switches over after picking the wrong door initially, he or she will win the prize

GAME 2: PICK UP STICKS

Suppose that there are 10 sticks on the board

Player 1 and 2 will take alternate turns to pick up the sticks

In each turn, the player may either pick up 1 or 2 sticks only

The player who picks the last stick wins.

LET’S HAVE A TRIAL PLAY!

Any 2 volunteers?

STRATEGY TO WIN

At any stage, after player 1's move, the number of sticks left on board must be a multiple of 3

Such that at the beginning, player 1 will only picks up 1 stick (so 9 sticks are left)

Subsequently, if player 2 picks up 1 stick, player 1 will pick up 2

If player 2 picks up 2 sticks, then player 1 picks up 1

WHAT DOES SUCH A STRATEGY ENSURE?

Consider the second last round, player 1 would have played so that 3 sticks are left

Hence, no matter how many stick/s player 2 picks, player 1 will always has the winning move

GAME TREE

IN GENERAL…

In a two-player Pick Up Stick game, each of the two players can pick 1 through K sticks, with M sticks initially

A winning strategy corresponds to picking sticks such that the number of sticks left is a multiple of (K+1)

The player who can ensure this strategy will surely win the game.

In the game of Pick up Stick mentioned earlier on, we have M=10 and K=2.

FIBONACCI SERIES

a sequence of numbers named after Leonardo of Pisa, known as Fibonacci in 1202

The sequence is defined by the following recurrence relation:

i.e. 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987 …

)2()1(

1

0

)(

nFnF

nF

1,

1,

0,

n

n

n

APPLICATIONS OF FIBONACCI SERIES

Appear in biological settings

Numerous poorly substantiated claims of Fibonacci numbers are also found in popular sources

Let’s see one example…

FIBONACCI RABBIT

Suppose a newly-born pair of rabbits, one male, one female, are put in a field

They are able to mate at the age of one month so that at the end of its second month a female can produce another pair of rabbits

Suppose that our rabbits never die and that the female always produces one new pair (one male, one female) every month from the second month onwards

At the end of the first month, they mate, but there is still one only 1 pair.

At the end of the second month the female produces a new pair, so now there are 2 pairs of rabbits in the field.

At the end of the third month, the original female produces a second pair, making 3 pairs in all in the field.

At the end of the fourth month, the original female has produced yet another new pair, the female born two months ago produces her first pair also, making 5 pairs.

GAME 3: BRICK WALL

Build a brick wall, using bricks with a length twice as long as its height

The wall is to be two units tall

We can build the wall in a number of patterns, depending on how long we want it

Let f(n) be the number of patterns for walls of height 2 and length n

The solutions we have at present are

f(1) = 1

f(2) = 2

f(n)=f(n-1)+f(n-2)

.

.

.

The number of patterns of length n is the sum of the number of patterns of length (n-1) and the number of patterns of length (n-2)

In terms of the "f" notation in mathematics, we have f(n) = f(n-1) + f(n-2)

Since f(1) = 1 and f(2) = 2, f(n) numbers are indeed the Fibonacci numbers!

GAME 4: CENTIPEDE GAME

One of the games used in Game Theory

The game was developed by Rosenthal in 1981

RULES Two players, A and B

Players do not cooperate

Initially A has two piles of coins: one pile contains 4 coins and the other 1 coin

Each player has two options of moves: either take the larger pile and give the smaller pile or pass both piles across

Each time the pile of coins pass across the table, the number of coins in it doubles

The game continues for a fixed number of rounds or until a player decides to end the game by keeping the larger pile and giving the other away

APPLICATIONS

Aimed to understand legislative bargaining concepts in political science

Laboratory tested theory would be more accurate and can be precise in predicting actions in an empirical way

Used to predict behaviors of political balances

LET’S PLAY!

Any two volunteers?

Condition: Maximum number of moves is 10

STRATEGY

Decision to move

right is known as

a "pass“

Moving down is described as a "take“

The numbers at the end of the down arrow are the payoffs to A and B respectively

MATHEMATICAL REASONING

If B passes, he or she will get 16 while A gets 64. By "taking“, B gets 32 while A only gets 8

At the prior vertex, A knowing what B will do, will chose to take first

Therefore, it is suggested that the first player take right at the start

Such reasoning is known as Nash Equilibrium (backward induction)

As the stakes increase, the play tends to approach Nash equilibrium

But it will never reach an equilibrium

By continuing playing, the winnings increases

But the risk is that your opponent might get more than you at the end of the game

NASH EQUILIBRIUM

If there is a set of strategies with the property that no player can benefit by changing her strategy while the other players keep their strategies unchanged, then that set of strategies and the corresponding payoffs constitute the Nash Equilibrium

EXAMPLE

Two suspected persons, A and B, are arrested by the police. Both of them are interrogated separately.

If one betrays the other and the other remains silent, the betrayer is released and the silent suspect gets penalized with the full 10 year imprisonment

If both remain silent, both suspects will be sentenced to 6 months in jail

If they betray each other, both will receive a 5-year arrest

Each prisoner must make the choice of whether to betray the other or to remain silent

However, neither of them knows what choice the other prisoner will make

So how should the prisoners act?

Payoff matrix of prisoner dilemma

Each prisoner wants to minimize the time of imprisonment

Betraying is the dominant strategy

If you knew the other prisoner would stay silent, your best move is to betray

If you knew the other prisoner would betray, your best move is still to betray

Thus, they will get a lower payoff than what they would get by staying silent

BACK TO NASH EQUILIBRIUM

Condition 1: The player who did not change has no better strategy in the new circumstance

Condition 2: The player who changes is now playing with a strictly worse strategy

If both cases are met, the equilibrium is said to be stable

If condition 1 does not hold, then the equilibrium is unstable

If only condition 1 holds, then there are likely to be an infinite number of optimal strategies for the player who changed

That’s the end of our presentation.

THANK YOU