These are great videos! I have a question for this game specifically though. Is the point to get the most points or just more points than the other player?
If the point is to just beat the other player than player two's strategy should be to get a point advantage and then always play right as she will always get either 2 points or zero. How would a strategy like that be represented?
@FilthParade Players maximize their own payoffs. They are only concerned about their opponents' strategies insofar as it affects which strategies they play.
You are thinking of zero sum games, where the payoffs in every cell sum up to zero, i.e. if my payoff is X, yours is -X. This is a special class of games that I don't really focus on.
I cannot tell you how helpful these videos are. Thank you so much for the time and effort you put into this. I was having a very difficult time grasping the mixed strategy algorithm until I watched your series. Best wishes!
@JimBobJenkins Yeah man you right, i was trying to draw it as vectors but its was really mind boggling... i really like the way you do the 2x2 matrices. Our teacher only showed us the graphical analysis but that becomes quite tedious. The game we got in our assingment has a semi complex payoff structure such that we cant really eliminate parts of it since there are no strictly dominated strategies.
@mrosendal I mean, you can still do it as a function of four equations. Set the three utilities equal to each other and then write out the utilities as a function of \sigma_{up}, \sigma_{middle}, and \sigma_{down}, and then substitute \sigma_{down} = 1 - \sigma_{top} - \sigma_{middle}. But, yeah, it's a mess.
@JimBobJenkins Thanks dude... ill try out some of the things... at least i have until wednesday to hand in my assignment .. :) If i get into some deep doo doo i will email you again if thats alright...
@JimBobJenkins Hey i think i have it. If player 1 chooses Top, Middle or Bottom, player 2 will never choose center hence i can eliminate C as players 2s strategy. If player 2 then chooses either left or right player 1 will never choose bottom hence we can eliminate B as player 1 strategy. We are then left with a new 2x2 matrix L R
T 2,0 4,2
M 3,4 2,3
This i can then solve be equating the utilities and solving for sigma. right ?
@mrosendal Unfortunately, eliminating strategies like that does not preclude the possibility of MSNE. You can only really do that for strictly dominated strategies.
HI William, I love your videos, extremely helpful and extremely interesting. Is there any way (using a similar algorithm) to solve an N-person game with group interatctions (i.e. a public goods game or the volunteers dilemna)?
Quite elegant algorithm in its simplicity, William. Well-done. Based on the solution, Player 1 must play Down 5/6th of the time and Player 2 must play Right 2/3rd of the time. My question, then, is Down-Right the mixed-strategy Nash Equilibrium?
@DubaiGuy08 Not quite. The probability distribution you gave--where 1 goes up 1/6th of the time and down 5/6ths of the time, and 2 goes left 1/3rd of the time and right 2/3rds of the time--is the MSNE. We don't actually know what will happen until the players make their moves, since they are moving randomly.
@JimBobJenkins Great lecture!! But maybe I've got confused: why can't we jump the conclusion that the final trend would end up with Down-Right result, as DubaiGuy08 said below? Or, to more basic, what's the meaning of the calculated result? I mean, for Player 1: Down=5/6 it represents that Player 1 would like to chose more Downs than Ups, doesn't it?
And for Player 2: Right 2/3, that means Player 2 prefer to decide more Rights than Lefts, right? What's my misunderstanding? Thanks.
@CogitoErgoCogitoSum Not entirely sure what you mean here, but in mixed strategy Nash equilibrium, both players are completely indifferent about which strategy they play or any probability distribution between the two.
It sounds like you might want to refer back to pure strategy Nash equilibrium, in which your strategy does depend on what the other player is doing.
Again. Thanks a lot for this. I cant believe I'm learning more from a youtube video than from my ivy league classes with my lecturer from eastern europe..
What will really set one's head spinning is that the definition of a mixed strategy that works in the general case with e.g. a continuous set of pure strategies (e.g. Cournot duopoly) is just "a probability measure on a/the set of pure strategies" (OK, this is slightly imprecise, but close enough for a YT comment).
The analogy in that setting of this algorithm then becomes solving for a meassure when you know the values of integrals. Something that smells like Riesz's representation theorem.
That's effectively 1b) done of my homework, cheers.
jpt591 3 days ago
my..... my BRAIN!!!
Leiland103 5 days ago
God Bless you baba!
daboiisgud 1 week ago
thank you
rhyswaterfield 1 week ago
This is awesome man
TheAnimefever 3 weeks ago
saved my life.... thank you
irondia73 3 months ago
you're a life saver <3
Nahelaa 4 months ago
You are my own personal jesus christ. I missed a week of classes, and was so lost on my homework right now. Thank god for you. Thank you so much man.
SamAt5am 4 months ago
Thanks, pretty much useful! We'll see if it helps me get rid of my algorithmics assignment :)
northfromheree 9 months ago
what if you have 2 players which each have 3 options? such as up middle and down?
swaine55 10 months ago
@swaine55 You have to consider all combinations of mixed strategies, which is quite a headache.
JimBobJenkins 10 months ago
Thank you!
yellowpepsi3 10 months ago
These are great videos! I have a question for this game specifically though. Is the point to get the most points or just more points than the other player?
If the point is to just beat the other player than player two's strategy should be to get a point advantage and then always play right as she will always get either 2 points or zero. How would a strategy like that be represented?
FilthParade 10 months ago
@FilthParade Players maximize their own payoffs. They are only concerned about their opponents' strategies insofar as it affects which strategies they play.
You are thinking of zero sum games, where the payoffs in every cell sum up to zero, i.e. if my payoff is X, yours is -X. This is a special class of games that I don't really focus on.
JimBobJenkins 10 months ago
@FilthParade Matching pennies is a zero-sum game, btw.
JimBobJenkins 10 months ago
oh god my head will explode if i watch any of your vids today
RealDeathstrikeable 11 months ago
I cannot tell you how helpful these videos are. Thank you so much for the time and effort you put into this. I was having a very difficult time grasping the mixed strategy algorithm until I watched your series. Best wishes!
ToppyT 1 year ago
What do you do if its a 3x3 payoff matrix ?
mrosendal 1 year ago
@mrosendal You learn matrix algebra or your head will explode. More on that in future videos.
JimBobJenkins 1 year ago
@JimBobJenkins Yeah man you right, i was trying to draw it as vectors but its was really mind boggling... i really like the way you do the 2x2 matrices. Our teacher only showed us the graphical analysis but that becomes quite tedious. The game we got in our assingment has a semi complex payoff structure such that we cant really eliminate parts of it since there are no strictly dominated strategies.
The matrix we got is as follows.
T 2,0 0,1 4,2
M 3,4 3,2 2,3
B 1,3 1,2 3,0
mrosendal 1 year ago
@mrosendal I mean, you can still do it as a function of four equations. Set the three utilities equal to each other and then write out the utilities as a function of \sigma_{up}, \sigma_{middle}, and \sigma_{down}, and then substitute \sigma_{down} = 1 - \sigma_{top} - \sigma_{middle}. But, yeah, it's a mess.
JimBobJenkins 1 year ago
@JimBobJenkins Thanks dude... ill try out some of the things... at least i have until wednesday to hand in my assignment .. :) If i get into some deep doo doo i will email you again if thats alright...
mrosendal 1 year ago
@JimBobJenkins Hey i think i have it. If player 1 chooses Top, Middle or Bottom, player 2 will never choose center hence i can eliminate C as players 2s strategy. If player 2 then chooses either left or right player 1 will never choose bottom hence we can eliminate B as player 1 strategy. We are then left with a new 2x2 matrix L R
T 2,0 4,2
M 3,4 2,3
This i can then solve be equating the utilities and solving for sigma. right ?
mrosendal 1 year ago
@mrosendal Unfortunately, eliminating strategies like that does not preclude the possibility of MSNE. You can only really do that for strictly dominated strategies.
JimBobJenkins 1 year ago
@mrosendal Actually, there IS a strictly dominated strategy. But it is in mixed strategies. I will PM you a link.
JimBobJenkins 1 year ago
HI William, I love your videos, extremely helpful and extremely interesting. Is there any way (using a similar algorithm) to solve an N-person game with group interatctions (i.e. a public goods game or the volunteers dilemna)?
Icecoldpoker 1 year ago
Quite elegant algorithm in its simplicity, William. Well-done. Based on the solution, Player 1 must play Down 5/6th of the time and Player 2 must play Right 2/3rd of the time. My question, then, is Down-Right the mixed-strategy Nash Equilibrium?
DubaiGuy08 1 year ago
@DubaiGuy08 Not quite. The probability distribution you gave--where 1 goes up 1/6th of the time and down 5/6ths of the time, and 2 goes left 1/3rd of the time and right 2/3rds of the time--is the MSNE. We don't actually know what will happen until the players make their moves, since they are moving randomly.
JimBobJenkins 1 year ago
@JimBobJenkins Great lecture!! But maybe I've got confused: why can't we jump the conclusion that the final trend would end up with Down-Right result, as DubaiGuy08 said below? Or, to more basic, what's the meaning of the calculated result? I mean, for Player 1: Down=5/6 it represents that Player 1 would like to chose more Downs than Ups, doesn't it?
And for Player 2: Right 2/3, that means Player 2 prefer to decide more Rights than Lefts, right? What's my misunderstanding? Thanks.
denilson12001 9 months ago
What if the probability that player 1 plays up is a function of which strategy player 2 plays?
CogitoErgoCogitoSum 1 year ago
@CogitoErgoCogitoSum Not entirely sure what you mean here, but in mixed strategy Nash equilibrium, both players are completely indifferent about which strategy they play or any probability distribution between the two.
It sounds like you might want to refer back to pure strategy Nash equilibrium, in which your strategy does depend on what the other player is doing.
JimBobJenkins 1 year ago
very interesting.
TeppTennis 1 year ago
Again. Thanks a lot for this. I cant believe I'm learning more from a youtube video than from my ivy league classes with my lecturer from eastern europe..
Cheers mate!! Please keep it up.
corporalcharlie 1 year ago 10
@corporalcharlie I'm at one of your fellow institutions with the same problem. Cheers indeed
bigbrun13 9 months ago
MSNE:{(1/6,5/6),(1/3,2/3)}
Persian5life 1 year ago
I just want to thank you for posting this amazing video! You have no idea how much you've helped me! Thanks again!
blanc11 1 year ago 5
What will really set one's head spinning is that the definition of a mixed strategy that works in the general case with e.g. a continuous set of pure strategies (e.g. Cournot duopoly) is just "a probability measure on a/the set of pure strategies" (OK, this is slightly imprecise, but close enough for a YT comment).
The analogy in that setting of this algorithm then becomes solving for a meassure when you know the values of integrals. Something that smells like Riesz's representation theorem.
inf0phreak 1 year ago