 Welcome back to this NPTEL course on game theory. In the previous session we have discussed by matrix games and the Nash equilibrium existence of Nash equilibrium and also a non-linear programming method. Now in this session we would like to point out that there is an intuitive way of solving certain games. So this method is known as solving by dominance which in fact we have seen this already in the context of zero sum games. So we will do this for the by matrix games now and in particular we introduce what is called iterated elimination. Let us look at this. We start with a simple game. We are looking at a simple by matrix game where the payoffs are given by. So let us understand these as the payoffs of the players. This is CD, CD and this is for player 2, player 1. So if we really observe player 1 by playing C he will get either 2 or 0 but if he plays D he is getting 3 or 1. In which case in respect of the player 2 strategy by playing D player 1 is getting higher pay than the corresponding payoffs this thing. So therefore D is here strictly dominating C, strictly dominates C. In fact in a symmetric fashion for player 2 also D strictly dominates C. So therefore because the players are maximizing their utility no one would like to play C because D is a better strategy for them. So therefore both of them will play D and hence DD is going to be the equilibrium in this game. So this is known as the solving the games by dominance. So let me formally define this thing. So let us we consider a game G. So that is basically it has a S1, S2. These are the payoff strategy spaces and then corresponding payoff functions are pi 1, pi 2. So let us take for player i. So let us take Si prime, Si double prime in Si. Let us take 2 strategies. We say that Si prime strictly dominates Si double prime if and only if his utility when he is playing Si prime irrespective of what other player plays. So S minus i is let us say what the other player is playing is this is strictly bigger than Ui of Si double prime S minus i. So when whatever player 2 plays that is given by S minus i basically S minus i is a strategy in S minus i. In fact the notation here is that when I say minus i means except i. So if i is equals to 1 minus i becomes 2. If i is equals to 2 minus i becomes 1 that is the understanding here. So whatever strategy the other player plays S minus i Si prime yields strictly higher payoff than Si double prime. Then I say that Si prime strictly dominates Si double prime. And in fact Si double prime is dominated by Si prime. So this is the definition of a strict dominance. So now let us look at it some simple games. So let us look at the following game. So we take 4, 3, 5, 1, 6, 2, 2, 1, 8, 4, 3, 6, 5, 9, 9, 6, 2, 8. Can we solve this game by dominance? Let us look at it. So if we really look at is the player 1 he is getting 4, 5, 6 and in the strategy when they play the rho U if they play rho M 2, 8, 3 and rho D means 5, 9, 2. We cannot say that for example the rho U is dominated by rho M or rho U strictly dominates rho M none of these things we can say here. So but let us look at the other side. What about the player 2? From a player 2 perspective if you look at it for example here there is 1, 4, 6. So the player 1 is getting 1, 4, 6 when he plays the center column and what about this right most column if he plays 2, 6, 8. So by playing R he is strictly getting a higher payoff than the center column. So therefore in fact for player 2 R strictly dominates. So player 2 will never play C. So let us mark this as a mark this with red here saying that he is never playing this one. Now what we have here is now a player 2 will either play L or R. Now player 1 has all these U, M, D. Now let us look at it. What about the column L and column R 3, 1, 9 and 2, 6, 8. No one dominates the other so we cannot say anything. But let us look at the player 1's perspective now. Now player 1 if he plays U he will get either 4 or 6 in comparison with M where he gets 2 and 3. So 2 and 3 is certainly smaller than what he is getting when he plays U. So therefore he will M is strictly dominated by U. So he will never play M so therefore let me mark this also. Then what else do we have? So player 1 has U and D that is 4 and 6 and compared to 5 and 2. Nothing dominates the other whereas player 2 if you look at it he has the 2 columns L and R where he is getting 3, 9 and 2, 8. 3, 9 is certainly higher than 2 and 8. Therefore player 2 will never play R so I can remove this one also. So when you remove this R also player 1 has is now forced to play only L. So now player 1 has 2 choices U and D when he plays U he will get 4 when he plays D he will get 5. So he will naturally play D. So therefore this is going to be the equilibrium here. So this is known as iterated elimination of strictly dominant strategies. This is iterated elimination of strictly dominated strategies. So when a strategy is strictly dominated by another strategy you do not play with that and if you keep on going on eventually if it leads to a solution then that is going to be a Nash equilibrium. In fact we can state it as a theorem here if iterated elimination of strictly dominated strategies leads to a single pair of strategies then this single pair is Nash equilibrium. So the proof if you really look at it proof is tried forward from the very definition. So the strategy every time when you are removing it it is dominated by something so therefore that strategy if you every time whatever you are removing it both of them they have a better strategy so they are playing with that and so therefore eventually it leads to this equilibrium. The proof is not really difficult so let us try to prove this fact. Suppose the way we prove is we prove by contradiction. Suppose let us say iterated elimination should let us say lead to x star and y star. So let us assume this one and we need to show that x star y star is equilibrium. Suppose x star is not best response to y star. So then let us consider the set x to be all x in S1 such that the pi 1 of x y star greater than pi 1 of x star y this is y star. So this is basically this set because x star is not best response to y star so there must be some x for which pi 1 x y star must be bigger so that is this thing therefore this set is non-empty. So this non-emptiness this set is essentially the best response of y star it is not really best response better response I can say what are all the strategies x which give you better value than x star against y star. So this x is non-empty. So the thing is that all the strategies in x must have been eliminated. So all these strategies must have been certainly eliminated from this x because iterated elimination led to x star and y star that is the this thing. Let us look at the last stage look at last stage where a strategy x in x is eliminated. So look at what we are saying that you are taking some x in x and then when this x is eliminated that last place where it is eliminated that is what we are looking at it. So for this for x to be eliminated there must be some x prime let me put it S1 such that x prime dominates x. So this should have happened therefore what we have is that pi 1 of this x prime y star is bigger than pi 1 of x y star but this is bigger than pi 1 of x star y star that means what we are saying is that even x star is dominated by x prime. So this contradicts the fact that x star is the eventual strategy that remained after the iterated elimination. So this particular thing is saying that x prime is giving higher payoff than corresponding to x star for the player 1 but that cannot happen that means it is dominating even this thing. So this contradicts this argument and which proves that the strategy pair x star y star is Nash equilibrium. Now there are some interesting facts here. So when you are looking at iterated elimination it always leads to a unique solution it cannot lead to multiple solutions. So these are all some facts which we can easily verify in fact if we go through the proof that it will be evident from there as well. So but there is one more aspect of iterated elimination. So what we have seen so far is about a strictly dominant solution. There is another thing called weakly dominant solution, weakly dominant strategies let me introduce them. So what we say that for a player s prime is weakly dominated by s double prime for player i if what we need is pi i s prime s minus i is less than or equals to pi i s double prime s minus i. This should be true for all s minus i and there must exist at least 1 s minus i where this inequalities is strict. So what I am saying is that if it is never strict that means you are getting the same payoffs in this thing that means the corresponding rows are columns they are identical. So we are just avoiding that. So for example if you take from the player one s perspective the one row all the entries of that row are less than or equals to another row but one of the entry must be bigger strictly bigger. So then we are calling weakly. Now what about what about iterated elimination of weakly dominant strategies. So the interesting fact here is that when you are looking at the iterated elimination of a weakly dominant solutions we may not end up getting a Nash equilibrium. So this need not lead to Nash equilibrium. So let us look one simple example. So look at this 1 1 0 0 3 2 2 0 0 1 1. So let us look at this example what is going to happen here. So let me call this as L r this is t this is m b. So m dominates t in this case m dominates t here 1 you are getting 3 0 and 2. So therefore this one is dominated by this thing. So therefore this we do not need to consider here. Therefore now what we have is that 3 2 0 0 1 1 this is going to be the this thing. Now for example we can always say so for example here is an interesting situation where now what remains here is that 3 2 2 2 0 0 1 1. Now this is the game that remind now. Now in this case what we have this is m b L r. So for a player he is getting let us say this is a L if he removes L because he R is weakly dominates L. So he can say that he does not want to play this. Then player in the next round player 1 will choose m. So therefore m r comes is m r Nash equilibrium. So of course I look at this one if for example player 2 fixes r the player 1 is best to play m and when player plays m what is the best to player he can play either of them. So is m r is a Nash equilibrium here m r is going to be a Nash equilibrium here. Now but there is another interesting point here to see is that even m L is also Nash equilibrium. So what is really happening here? So if you observe this procedure the order in which you are eliminating the strategies matters a lot which equilibrium you are arriving at. So this is not really an example to show that weak iterated elimination of weak dominant strategies does not lead to Nash equilibrium but this is an example which shows that the order in which you are eliminating the strategies is a very very important which equilibrium you are getting and other things. So this is a in that sense it is a very interesting example and in fact here I would like everyone to ask to construct an example where the iterated elimination need not lead to a Nash equilibrium. So this I would like people to try this one. Before concluding this session would like to point out a few points about this iterated elimination. So this iterated elimination when you are looking at a strict dominant strategies it always leads to a Nash equilibrium the order in which you are using you are eliminating that may give different equilibrium. Now even though we have seen it only for a figure pure strategies the same argument can be applied to mixed strategies which I will leave it to you to figure out you can study the same argument and if at all iterated elimination leads to some solution that will give you a Nash equilibrium but not in the weak dominance for which I ask everyone of you to come up with an example and this method does not apply to all sort of games there are most games cannot be solved by this dominant so for that we require a other kind of algorithms. The one algorithm that we have seen already in the previous class is non-linear programming and the other algorithm that we are going to see it in the next set of lectures which is known as a Lemke-hausen which is more of a combinatorial argument. In fact this Lemke-hausen algorithm also proves the existence of a Nash equilibrium using fairly simple arguments no fixed point argument required. Okay with this I will stop this session we will continue in the next session.