1. Algorithmic Applications of Game Theory
Lecture 1

2. Traditional computer science

3. This course
Markets
Routing
Advertisements
Social

4. Game theory
A set of agents (people) who are in a situation of conflict
Each agent has its own goals
Assumption – agents are rational + common knowledge of rationality
What will happen?
Prediction

5. Algorithmic game theory
Using ideas from CS in game theory
Mechanism design and market design
Ad auctions, spectrum auctions, Doctor’s match
Optimization when the players hold the input
Divide the rent of an apartment between friends, when each friend gets a different room
Choose a time for class (see below)
Approximate solutions
Price of Anarchy

6. Let’s start another way…
Everyone choose a number between zero and a hundred, and write it on the piece of paper you have, with your name.
Then pass all the numbers to me.
I will compute the average (hopefully correctly)
The winner is the one who is closest to two thirds of the average.

7. What would be a good strategy?
Well, you never want to put more than 66, right?
But if everyone never puts more than 66, you never want to put more than 44, right? …
So everyone should play zero.
There is a difference between rationality and common knowledge of rationality

8. What do you usually get?
Politiken played this with 19,196 for 5000 krones

9. Administration Lecture once a week, no recitation (TIRGUL) or homework
You need to be responsible and study (not just) before the test
Test in the end of the semester
Textbook: Algorithmic Game Theory by Nisan, Roughgarden, Tardos, and Vazirani
Office hours right after class – do come by
Feedback after every lesson
I will have to skip a couple of classes, and will fill them another day \ Friday according to you. Light refreshments on me whenever I do this.

10. Ups and downs of this course
New lecturer – will work hard
Material can be relevant to your career
Game theory changes the way you see life
New lecturer – no experience (and maybe no talent as well)
New course – no tests to study from
Will be non trivial – you will have to study

11. Hopefully some of you are still here So we can start with some definitions

12. Games Each player selects a strategy
Given the vector of strategies, each player gets a payoff
A game is summarized by the payoff matrix:
Same idea for more than two players…
Columns / Rows R1 R2 R3 C1
1 / 4
-1 / 6
2 / 7 C2
4 / 3
3 / -2
3 / 4

13. Notation
Vector (profile) of strategies: s, or . That is s = (s1,..., sn)
Player i’s utility when s is played is denoted Ui(s)
Suppose we want to state player’s i utility when all players play s, but instead of playing si he plays . This is denoted as Ui(s-i, )

14. Practicing notation on the example
Rows/ Columns C1 C2 C3 R1
1 / 4
-1 / 6
2 / 7 R2
4 / 3
3 / -2
3 / 4
Denote s = (R1, C1)
URows(s) = 1
URows(s-Rows,R2) = 4
UColumns(R2,C2) = -2

15. But what will the players do?
I don’t know
We have a semester to talk about this
In some cases it’s obvious
No matter what Rows does, Columns is better off with C3
Rows/ Columns C1 C2 C3 R1
1 / 4
-1 / 6
2 / 7 R2
4 / 3
3 / -2
3 / 4

16. Analysis continued Suppose player Columns plays C3. What will Rows do?
Rows/ Columns C1 C2 C3 R1
1 / 4
-1 / 6
2 / 7 R2
4 / 3
3 / -2
Suppose player Columns plays C3. What will Rows do?
Play R1
So the outcome will be 2 / 7

17. Dominant strategies
The last game was easy to analyze: no matter what Rows did, Columns played C3
In this case we say that C3 is a Dominant strategy.
Formally: consider player i. If for any strategy profile s we have Ui(s-i,i) ≥ Ui(s) We say i is a dominant strategy for player i

18. Domination
A dominant strategy is the optimal action for a player i, no matter what the other players do.
Can we say that some strategy i is “better” than i even when i is not a dominant strategy?
We say that i dominates i if for every profile s Ui(s-i, i) ≥ Ui(s-i, i)

19. Dominated strategies
We already know that if i is a dominant strategy we expect it will always be played.
Suppose i dominates i
Then we expect i will never be played, since player i is always better off playing i
If for every other strategy i, we have that i dominates i then i is a dominant strategy

20. Relations between strategies
Suppose i dominates i. Can it be that i dominates i ?
Yes, but then player i is indifferent between them. Proof:
For every profile s we have Ui(s-i, i) ≥ Ui(s-i, i) and Ui(s-i, i) ≥ Ui(s-i, i) gives Ui(s-i, i) = Ui(s-i, i)
Note that other players may get different utility if i plays i or I
In particular, player i can have multiple dominant strategies

21. Are dominant strategies an optimal predictor?
Well, only in theory
Think about chess
A strategy is what I will do in every board situation
Given white’s strategy and black’s strategy, the result is either white wins, black wins or tie
So in theory (and also in game theory), the game is “not interesting” and white will play a strategy which will let him always win or tie.
In practice (and taking a CS perspective) there is a computational question of finding the strategy…

22. Example – prisoner’s dilemma
Board?

23. Prisoner’s Dilemma is a theoretical concept with no real life interpretation
Show of hands: Please raise your hand if you did a preparation course for the psychometric exam ובעברית – מי עשה קורס הכנה לפסיכומטרי?
This is just a (multiplayer) prisoner’s dilemma

24. פסיכומטרי Suppose there are n students A1…An ranked A1>A2>…An
If no one takes the course, the ranking is correct, and only the good students get to study CS.
No matter what the other students do, it’s dominant for Ai to take the course, and increase his chances of studying CS.
If all take the course, we get the same ranking again, but everyone wasted three months and a ton of money.

25. Here is another game No dominant strategy for any player…
Rows/ Columns C1 C2 C3 R1
0 / 7
2 / 4
7 / 0 R2
4 / 2
5 / 5 R3
No dominant strategy for any player…

26. So how can we predict something?
Rows/ Columns C1 C2 C3 R1
0 / 7
2 / 4
7 / 0 R2
4 / 2
5 / 5 R3
Imagine that the game is played many times.
Imagine that at some point, the profile (R2, C2) is being played
Then no player has incentive to move

27. Nash equilibrium The strategy profile is called a Nash equilibrium.
Named after John Nash who proved existance in ‘51 (Nobel in ‘94). Original concept due to Von Neumann
Formally: A strategy profile s = (s1, …sn) is a Nash equilibrium, if for every i and i we have Ui(s) ≥ Ui(s-i, i)
So given that everyone else sticks, no player wants to move

28. Examples of Nash Equilibrium
Suppose each player i has a dominant strategy di. Then (d1… dn) is a Nash equilibrium
Proof: volunteers? Should I do it?

29. Multiple Nash equilibria
The battle of sexes
See multiple equilibria on the board
Note different equilibria are better for some players

30. No (Pure) Nash equilibrium
0,0
1,-1
-1,1
We will get back to this – players can randomize

31. Back to the Prisoner’s Dilemma
Both players confessing is a Nash equilibrium
But it sux…

32. How much can a Nash equilibrium Suck?
Or in a more clean language:
Consider a game G. The social welfare of a profile s is defined as Welfare (s) = iUi(s)
Let O be the profile with maximal social welfare
Let N be the Nash equilibrium profile with minimal social welfare (worst Nash solution)
The Price of Anarchy of G is defined to be Welfare(O) / Welfare(N)

33. Price of Anarchy
Suppose a 100 people want to get from BIU to Jerusalem
The train takes two hours
Driving the car takes 1 hour+1 minute for every other driver
How many people will drive to Jerusalem?

34. The train game Each player has two strategies – Car and Train.
The utility of each player is 0 for taking the train (regardless of the number of passengers)
The utility of taking the car is 60 – the number of car drivers.
The Nash equilibrium is that 60 drivers take the car and 40 take the train. Social welfare = 0
The optimal solution is that 30 people take the car, for a social welfare of 30*30 = 900
Think about the new entrance to Tel Aviv
Rigorous treatment of Price of Anarchy later

35. Questions?
Feedback
Office hours

36. Extra Slides

37. Chicken

38. Road example 50 people want to get from A to B
1 hour
N minutes
50 people want to get from A to B
There are two roads, each one has two segments. One takes an hour, and the other one takes the number of people on it

39. Nash in road example A B
1 hour
N minutes
In the Nash equilibrium, 25 people would take each route, for a travel time of 85 minutes

40. Braess’ paradox
1 hour
N minutes A
Free B
1 hour
N minutes
Now suppose someone adds an extra road which takes no time at all. Travel time goes to 100 minutes