Trembling Hand Perfect Nash Equilibrium

[@jschoete3430]

Helpful explanation :)

[@rheak.5470]

thxx! you really explained it well! I seldom understand stuff from YT but i understood this rly well. My assignment was on trembling hand perfection without it being actually taught in class, so this is rly helpful!

[@happyfrogkhmel]

Thanks for explanation. You look great btw!)

[@user-hg3ui7gt6p]

Really helpful!

[@andresramasco5644]

gracias pa

[@gutyhuy3817]

I think it would be a better example if (B,C) and (M,R) were (0,3) and (3,0) respectively. Then the probabilities would be more important.

[@Priyal_Doshi]

Hey, how can I get in touch with you for further questions?

[@lowkeybeams5663]

I really don't get it.

[@AUwe-yo2ox]

Let me do a little explanation. For Player1, Player 1 Plays T, and Player 2 palys L. but Player1 knows Player 2 may play C or R some of the time. Hence if Player 2 plays C or R, Player 1 has a profitable deviation Playing M or B. Next, we find out what are Player1's payoffs playing M or B. The largest Payoff for Player 1 knowing that Player 2 plays L with probability q1, Player2 Plays C with probability q2 and Player2 Plays R with probability 1 - q1- q2 ( Note that the sum of the probabilities must equal 1 i.e q1+q2+1-q1-q2=1) will be his THP NE. Solving the payoffs for Player 1 playing M and for playing B we have U(M) = 0(q1) + 1(q2) +2(1-q1-q2) = 2 - 2q1 - q2 U(B) = 0(q1) + 0(q2) +2(1 - q1 - q2) = 2 - 2q1 - 2q2 Since q1 is a positive number between 0 and 1, and q2 is a positive number between 0 and 1. It implies the payoff for player 1 playing M is greater than her payoff playing B. Hence M is the THP NE for Player 1. i.e U(M)>U(B) (because subtracting q2 from 2 -2q1 is greater than subtracting 2q2 from 2 -2q1) Similar argument can be made for Player 2 by calculating the payoff for player 2