6. Calculating Mixed Strategy Nash Equilibrium of a 3 x 3 Game: An Example (Game Theory Playlist 4)

[@mauriciolagosrogers3498]

you have a superpower: the ability of taking difficult things and explain them in an easy way.

[@jakemoore4610]

This video saved me right before a test, thank you!

[@TheDiscountedKnowledge]

Your explanation is absolutely amazing. I took MIT's Game Theory course on Coursera, and unfortunately they cannot explain some of the basic concepts. Knowing something is different from being able to teach what you know.

[@Chris-xq1xp]

why do you switch the inequality sign at 7:51? Please help I have a midterm tomorrow

[@evrenaydn8991]

hocam teşekkür ederiz bu kanal için

[@Mma-vb2io]

this is great but could you please show how to solve a 3x3 game where the NE is both players mixing between all three strategy?

[@kpoper2000]

Thank you!! This was so useful

[@e-k4110]

hocam supersiniz

[@taemi5507]

given that player 1 never chooses the 3rd strategy, that has to mean it is strictly dominated? There is always one strategy better for him than the third one (D), he has no reason to play it?

[@hiddecroes8053]

how are you so quick to remove strategy L, but need to take the time to calculate probabilities for strategy D. Player 2 will never choose L because there is always a better outcome from switching to M, but the same can be said for player 1 and D right. If player 2 chooses M, player 1 can choose C and if player 2 chooses R, player 1 can choose U. why can't you remove D as an option?

[@anthonylewallen9335]

Because the definition of dominance is when one strategy does at least as well as and in some cases strictly better than any other strategy no matter what the opposing player does. Taking this definition into mind then this means if we look at player 2. M dominates L because 4 > 3, 6 > 3, and 5 > 2. However for player 1 U does not dominate D because 10 > 7, 2 < 5, and 6 > 3. Here we see that while U does better than D in 2 cases (when player 2 chooses L and R) U does worse than D when player 2 chooses M. If U were 10 5 and 6 then it would dominate D because it would be 10 > 7, 5 = 5, and 6 > 3 so if that were the case then yes it would dominate D but it doesn't because 2 < 5. So considering that L is never going to be played by player 2 because it is strongly dominated by M, and no strategies for player 1 are dominant, each player is might be better off using a mixed strategy. But you'd have to calculate the expected payoffs. If we are talking about Pure Strategy then the best response is (U, M) payoff of (6,6).

[@PankajKumar-ot3mg]

great lecture

[@nickli7517]

Saved my life :D

[@하이-b7q]

I love it!! Thank you very much😄

[@mahamadykontiliguissomko3229]

Very useful , thanks sir

[@gabrielesacchitella9548]

incredible

[@SamyuktaNeeraj]

If D is strictly dominated for a particular r, does that mean we can eliminate it from the table?

[@amirreza6019]

Of course, this is part of the IESDS which applies for the mixed strategies too. If the mixed strategy has a better payoff for a player than the pure strategy , therefore one can eliminate the dominated strategy.