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# ECON 159: Game Theory

## Lecture 15

## - Backward Induction: Chess, Strategies, and Credible Threats

### Overview

We first discuss Zermelo’s theorem: that games like tic-tac-toe or chess have a solution. That is, either there is a way for player 1 to force a win, or there is a way for player 1 to force a tie, or there is a way for player 2 to force a win. The proof is by induction. Then we formally define and informally discuss both perfect information and strategies in such games. This allows us to find Nash equilibria in sequential games. But we find that some Nash equilibria are inconsistent with backward induction. In particular, we discuss an example that involves a threat that is believed in an equilibrium but does not seem credible.

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html## Game Theory## ECON 159 - Lecture 15 - Backward Induction: Chess, Strategies, and Credible Threats## Chapter 1. First and Second Mover Advantages: Zermelo’s Theorem [00:00:00]
Now today, I want to just draw a slightly grander lesson out of that game. So not only was it the case that the game sometimes has first mover advantages and sometimes has second mover advantages, but moreover, we could tell when it had a first mover advantage and we could tell when it had a second mover advantage. Is that right? When we actually looked at the initial set up of those stones, we knew immediately that’s a game in which Player 1 is going to win, or alternatively, we knew immediately that’s a game which Player 2 is going to win. Now it turns out that that idea is very general and actually has a name attached to it, and that name is Zermelo. So today we’ll start off by talking about a theorem due to a guy called Zermelo, and the idea of this theorem is this. We’re going to look at games more general than just Nim, and we’re going to ask the question, under what circumstances would you know about a game either that Player 1, the person who goes first, can force a win, or that Player 2 can force a win, or will allow a third possibility, which is it’s going to be a tie. So here’s the theorem, suppose there are two players in this game, like the games that we looked at last time, and suppose–I won’t define this formally now–but suppose the game is a game of perfect information. So what do I mean by perfect information? I’ll define this later on in the class, but for now all I mean is, that whenever a player has his turn to move, that player knows exactly what has happened prior in the game. So, for example, all these sequential move games we’ve been looking at are moves of perfect information. When I get to move I know exactly what you did yesterday, I know what I did the day before yesterday and so on. So it’s a game of perfect information. I’m going to assume that the game has a finite number of nodes. So two things here, it can’t go on forever, this game, and also there’s no point in which it branches in an infinite way. So there’s a finite number of nodes, and we’ll assume that the game has three possible outcomes. Actually there’s a more general version of this theorem but this will do for now. The three possible outcomes are either a win for Player 1, so I’ll call it W Either Player 1 can force a win, so either it’s the case that this game is a game that if Player 1 plays as well as they can, they’re going to win the game no matter what Player 2 does. Or 1 can at least force a tie, which means Player 1 can play in such a way that they can assure themselves of a tie regardless of what Player 2 does. Or it could be a game in which 2 can force a loss on 1, so a win for 2. So this theorem, when you first look at it, it doesn’t seem to say very much. You’re staring at this thing–you might think, we already knew that we’re looking at games that only have three possible outcomes win, loss, or tie so it doesn’t seem so surprising if you look at this theorem, it says, well, you’re going to end up with a win, or a loss, or a tie. However, that’s not quite what the theorem says. The theorem says, not only are you going to end up there–we knew that already–but the games divide themselves. Games of these forms divide themselves into those games in which Player 1 has a way of winning regardless of what Player 2 does; or games in which Player 1 has a way of forcing a tie regardless of what Player 2 does; or Player 2 has a way of winning regardless of what Player 1 does, so these games all have a solution. Let’s just go back to Nim to illustrate the point. So in Nim actually there’s no tie so we can forget the middle of these, and in Nim, under certain circumstances it is the case that Player 1 can force a win. Who remembers what the case was, for when Player 1 can force a win? Anybody? The people who played last time. No, yes, Ale, there’s somebody here. Shout it out.
So this theorem applies to all games of this form. So what games are of this form? Let’s try and think of some other examples. So one example is tic-tac-toe. Everyone know the rules of tic-tac-toe? In England we call it Noughts and Crosses, but you guys call it tic-tac-toe, is that right? Everyone know what tic-tac-toe is? Yeah, which category is tic-tac-toe? Is it a game which Player 1 can force a win, or is it a category in which Player 1 can only force a tie, or is it a category which you’d rather go second and Player 2 can force a win for Player 2 or a loss for Player 1? Which is tic-tac-toe? Let’s have a show of hands here. Who thinks tic-tac-toe is a game in which Player 1 can force a win? Who thinks tic-tac-toe is a game in which Player 1 can only force a tie? Who thinks Player 2’s going to win? Most of you are right. It’s a game in which if people play correctly then it’ll turn out to be a tie. So tic-tac-toe is a game that leads to a tie. Player 1 can still make a mistake, in which case they can lose. Player 2 can make a mistake in which case they would lose, but there is a way of playing that forces a tie. So these are fairly simple games, let’s talk about more complicated games. So what about the game of checkers? So the game of checkers meets these conditions. It’s a two player game. You always know all the moves prior to your move. It’s finite: there’s some rules in checkers that prevent it going on forever. And there are two or three outcomes: I guess there’s a third outcome if you get into a cycle you could tie. So checkers fits all these descriptions and what this theorem tells us is that checkers has a solution. I’m not sure I know what that solution is, or I think actually somebody did compute it quite recently, even in the last few months, I just forgot to Google it this morning to remind myself. But what this theorem tells us even before those people that computed it: checkers has a solution. Let’s be a bit more ambitious. What about chess? So chess meets this description. Chess is a two player game, everybody knows all the moves before them, it’s sequential, it has finite number of moves, it’s a very large number but it is finite, and it has three possible outcomes, a win, a loss, or a tie. So let’s be careful, the reason it’s finite is that if you cycle–I forget what it is–three times then the game is declared a draw, declared a tie. So what’s this theorem telling us? It’s telling us that there is a way to solve chess. Chess has a solution. We don’t know what that solution is. It could be that solution is that Player 1, who’s the player with the white pieces can force a win. It could be that Player 1 can only force a tie, and it could even be that Player 2 can force a win. We don’t know which it is, but there is a solution. There’s a catch to this theorem. What’s the catch? The catch is it doesn’t actually tell us–this theorem is not going to tell us what that solution is. It doesn’t tell us how to play. This theorem, in and of itself, doesn’t tell us how to play chess. It just says there ## Chapter 2. Zermelo’s Theorem: Proof [00:10:17]So we’re going to try and prove this. We don’t often do proofs in class, but the reason I want to prove this particular one is because I think the proof is instructive as part of sort of QR at Yale. So here’s another example here, and some other examples. You could think of chess as being the most dramatic example. So the reason I want to spend actually some time proving this today is because we’re going to prove it by induction and I’m going to sketch the proof. I’m not going to go through every possible step but I want to give people here a feeling for what a proof by induction looks like. The reason for this is, my guess is, well let’s find out, how many of you have seen a proof by induction before? How many have not? So for those of you who haven’t I think it’s a good thing in life, at one point in your life to see a proof by induction, and for those who have, my guess is you saw it in some awful math class in high school and it just went over–it didn’t necessarily go over your head but, the excitement of it doesn’t catch on. This is a context where it might appeal. So proof by induction. We’re going to prove this by induction on the maximum length of the game and we’ll call that N. We’ll call N the maximum length of the game. What do I mean by this? If I write down a tree I can always look at all the paths from the beginning of the tree all the way through to the end of the tree. And I’m going to look at the path in that particular tree that has the largest length, and I’ll call that the length of the game, the maximum length of the game. So we’re going to do induction on the maximum length of the game. So how do we start a proof by induction? Let’s just remind ourselves, those people who have seen them before. We’re going to prove that this theorem this true, the claim in the theorem is true for the easy case when the game is only 1 move long. That’s the first step, and then we’re going to try and show that if it’s true for all games of length < = N, whatever N is, then it must therefore be true for games of length N+1. That’s the way you do a proof by induction. So let’s just see how that works in practice. So let’s start with the easy step and we’ll do it in some detail, more detail than you really need to in a math class. So if N = 1, what do these games look like? Well let’s look at some examples here. So I claim it’s pretty trivial if N = 1 but let’s do it anyway. So the game might look like this. Player 1 is going to move–here’s Player 1 moving–the game’s only length one–so Player 1’s the only player who ever gets to move. So the game might look like this. Let’s put in a fifth branch, it might look like this. And at the end of this game, rather than putting payoffs let me just put the outcomes. So the outcome must either be a win, or a tie, or a loss, so suppose it looks like this. Suppose it’s a win, or here we could have a tie, or here we could have a win again, or here we could have a loss, and here we could have a tie. So in this particular game I claim it’s pretty clear this game has a solution. It’s pretty clear what 1 should do. What should 1 do? 1 should pick one of her choices that leads to a win. To be careful I’ll put 1 in here just to distinguish who it is who’s actually winning and losing. So in this game I claim that this game has a pretty obvious solution, either Player 1 is going to choose this branch that leads to a win or this branch that leads to a win, and either way Player 1 is going to win. Is that obvious? It’s so obvious, it’s kind of painful. So I claim this game, we could actually replace this first node with what’s going to happen which is Player 1’s going to win. Is that right? That was easy. Let’s look at a different example, we’ll do three. So here’s another possible example, and this again, Player 1 is going to move here and this time the possible outcomes are tie, or a loss, or a loss. Once again it’s a one player game, this time he has three choices and the three choices lead to a tie or he loses, so I claim once again this is simple. What Player 1 should do is what? Choose the tie since there’s no way of winning. But in that case he could actually force it. He could actually choose the tie in which case he’s going to tie. So this game has a solution called choose the tie. And again, we could replace the first node of the game with what’s actually going to happen which is a tie. Everyone happy? There’s one other possibility I guess. The other possibility is that here Player 1 has a lot of choices, maybe four choices in this case, once again Player 1 is going to move but in each case–unfortunately for Player 1–in each case the outcome is that Player 1 loses. So this is a game in which Player 1 is going to lose no matter what they do. Once again it has a solution: the solution is Player 1 is toast and is going to lose. So I claim this has really captured all the possibilities of games of length 1. I mean you could imagine that there could be more branches on them, but basically there are these three possibilities. Either it’s the case that one of those branches leads to a win, in which case the solution is Player 1 wins; or it’s the case that none of the branches have wins in them, but one of them has a tie, in which Player 1 is going to tie; or it’s the case that all the branches have loss in them, in which Player 1’s going to lose. So I claim I’m for done for games of length one. Everyone okay? So that step is deceptively easy but there it is. All right, now we do the inductive step, the big step. What we’re going to do is we’re going to assume that the statement of the theorem, which I’ve now hidden unfortunately–let me unhide it if I can, it’s not going to be easy. This now visible? ## Chapter 3. Zermelo’s Theorem: Generalization [00:17:06]So now let’s assume that the statement of this theorem is true for all games length < = N, suppose that’s the case. Suppose the claim is true for all games of this type, of length <= some N. What we need to show is–what we’re going to try and show is therefore it will be true for all games of length N + 1. What we want to show is–we claim, therefore, it will be true for games of length N + 1, that’s the key step in inductive proofs. All right, so how are we going to do that? Well let’s have a look at a game of length N + 1, and obviously I can’t use thousands here so let me choose from relatively small numbers, but the idea will go through. So let’s have a look at a game. So here’s a game in which Player 1 moves first, Player 2 moves second. Let’s suppose that if Player 2 does this, then 1 only has little choice here, up here then one has a more complicated choice, and perhaps 2 could move again. So this is quite a complicated little game up here, and down here let’s make it a little bit less complicated. Here the game looks like this and then comes to an end, so the tree looks like this. I haven’t put the end, I haven’t put the outcomes on it but the tree looks like this. So how long is this tree first of all? Well I claim that the longest path in this tree, from the beginning to the end has four steps. Let’s just check. So one, two, three, four that’s what I claim is the longest path. Any path going down this way only has three moves in it, so this is a game of length four. So we can apply it to our claim, let’s assume that the theorem holds for all trees length three or fewer, so that N + 1 is 4 for the purpose of our example. So in this example–I don’t want to put it here, let’s put it here–in this example N = 3, so that N + 1 = 4. Everyone happy with this in the example? Now what I want to observe about this is the following: contained in this N + 1, or the game of length 4 are some smaller games. You could think of them as sub-games. They’re games within the game, is that right? So let’s have a look. I claim, in particular–draw around them in pink–there’s a game here that follows Player 1 choosing Up and there’s a game here that follows Player 1 choosing Down, is that right? Okay, so these are both games, and what do we have here? So this little game in here–the game in the top circle, the game that follows Player 1 moving up–that’s a game and it has a length. So this little thing is a game and I’m going to call it a sub-game, some jargon that we’ll be getting used to later on in the semester. It’s a game within the game, but it’s basically just a game: it’s a sub-game and this is a sub-game following 1 choosing Up–let’s put Up in here–following up. And I claim that this game has a length–this sub-game has a length. So we knew that we started from a game of length 4, we’ve taken one of the moves, and let’s just make sure this is actually a game of length 3. So the game started here, this would be a game of length 3 because you could go one, two, three moves, is that right? So this is a game of length 3. So this is a sub-game following 1 choosing Up and it, the sub-game, has length 3. Down here, this is also a sub-game, it’s a game that follows Player 1 choosing Down and this little sub-game has length–now here we have to be a little bit careful–you might think that since we started from a game of length 4, after the first move we must be in a game of length 3. But actually that’s not true, and if we look carefully, we’ll notice that this game down here, the game starting here actually isn’t of length 3. It’s of length 2. Is that right? So even though started from a game of length 4, by going this way we put ourselves into a game of length 2. Is that right? So 1,2 or 1,2–whatever–it has length 2. Okay, all right so in this example N is 3 and N + 1 is 4, and our assumption, our induction assumption is what? We’ve assumed that the claim of the theorem holds for all games <= N, which in this case means <= 3. So what does that tell us? That tells us, by our assumption, this game, which I’ve put a pink circle around on the top, this is a game of length 3, this game has a solution. It must have a solution because it’s of length 3 or less. So by the induction hypothesis as it’s called–by the induction hypothesis–this is a technical term, but what’s the induction hypothesis? It’s this thing. By the assumption that we made this game has a solution. So it’s a game of length 3, 3 <= N in this case. So this game must have a solution. Now I don’t immediately know by staring at it what it is, but let’s suppose it was W, say it was W. And the game down below, this is also a game, and it’s a game of length 2 but 2 is less than 3 as well, 2 <= 3. So this game also has a solution. This game also, by the same assumption, has a solution, so its solution is say–I don’t know–maybe its L. So what does that mean to say that these games have solutions, in this case we’re going to assume W is this one, and L is this one, what does it mean? It means that, just as we did with the games up there, we could put the solution at the beginning of the game. We know that if we get into this game, we’re going to get the solution of this game; and we know if we get into this game, we’re going to get the solution of this game. So we can replace this one with its solution which by assumption was W, and this one by its solution which by assumption was L. And here I want to be really careful, I need to keep track of which person it is it’s a win or loss for. So this was a win for Player 1, hence it was a loss for Player 2, and this was a loss for Player 1, hence it was a win for Player 2. So these games, each of them have some solution, and this case I’ve written down the solutions as W or L. So now what could we do? Let me just bring down this board again. I can translate this game into a very simple game. I’m going to translate it up here, so this game can be translated as follows. Player 1 moves, if he goes Up then he hits a W and if he goes Down he hits a L. So in this particular example, Player 1 is effectively choosing between going Up and finding himself in a game which has a solution, and the solution is he wins it, or going Down and finding himself in a game which has a solution, and the solution is he’s toast. But that’s what? That’s a one move game. So we know this has a solution, and in particular, he’s going to choose Up. This one has a solution. This has a solution, it is a game of length 1. So what do we do? Somewhat schematically, we took a game of length N + 1, in this case that was 4; and we pointed out that once Player 1 has made her first move in this game, we are in a game, or in a sub-game if you like, that has length less than 4, it could be 3, it could be 2, whatever. Whatever sub-game we’re in, by assumption, that game has a solution. So it’s really effectively, Player 1 is choosing between going into a game with solution W or going into a game with solution L. And if there were 15 other sub-games here, each one would have a solution, and each one Player 1 would be able to associate that solution with what he’s going to get–what she’s going to get. So I claim that, if it in fact is true that each of these sub-games of length 3 or less had a solution, then the game of length 4 must have a solution, which is what? It’s the solution which is: Player 1 should pick the best sub-game. Are people convinced by that step? People are looking slightly “deer in the headlamps” now, so let me just say it again. We started by assuming that all games of length 3 or less, or N or less, have a solution. We pointed out that any game that has length N + 1 can be thought of as an initial move by Player 1 followed by a game of length N or less, N or fewer I should say. Each of those games of N or fewer steps has a solution, so Player 1 is just going to choose the game that has the best solution for her and we’re done. In this particular example, Player 1 is going to choose Up and therefore the solution to this parent game is Player 1 wins. Now the hard step I think in proofs by induction is realizing that you’re done. So I claim we’re now done, why are we done? Well we know that it was pretty easy that all games of length 1 have a solution. That was pretty trivial. And we’ve shown that if any game of length of N or fewer has a solution, then games of N + 1 have a solution. So now let’s see how we can proceed. We know that games of length 1 have a solution, so let’s consider games of length 2. Games of length 2, do games of length 2 have a solution? Well let’s set N = 1. We know that if games of length 1 have a solution, then any game of length 2 could be thought of as an initial step followed by a game of length 1, but that has a solution, so therefore games of length 2 have a solution. Now let’s think of games with length 3, we’ve shown that games with length 1 have a solution and games with length 2 have a solution. Any game of length 3 can be thought of as a game in which there’s an initial step followed by either a game of length 1 or a game of length 2, each of which have a solution, so once again a game of length 3 has a solution and so on. ## Chapter 4. Zermelo’s Theorem: Games of Induction [00:31:20]So games of induction, they work by building up on the length of the game, and we’re ending up knowing that we’re done. For those people who have never seen a proof by induction don’t worry I’m not going to test you on a proof with induction on the exam, I just wanted you to see one once. Let’s all take a deep breath and try and digest this a little bit by playing a game. I’ll leave it up there so you can stare at it. We’ll want it down again later. Let’s try and actually play a game and see if we can actually throw any light on this. So I’m going to pick out volunteers like I did last time, but first of all, let me tell you what the game is. So once again I’m going to have rocks in this game, and once again I’m going to approximate those rocks with marks on the blackboard. So here’s an example. The example is this. The game involves an array of rocks, so here the array has five rows and three columns. So the rocks are not arranged as they were before in piles, but rather in a sort of in a pretty array. I should have made it more even but this is: one, two, three, four, five rows; and one, two, three columns, at least in this example. But in general there’s an array of rocks, N x M. We’re going to play sequentially with the players. And the way the game is going to work is this–is when it’s your turn to move, you have to point to one of these rocks and whichever rock you point to I will remove. I will delete that rock and all the rocks that lie above or to the right of it: all the rocks that lie to the northeast of it. So for example, if you pointed to this rock, then I will remove this rock and also this one, this one, and this one. If the next person comes along and chooses this one, then I will remove this one, and also that one. Okay, everyone understand? All right and the game is lost–this is important–the game is lost by the person who ends up taking the last rock. The loser is the person who ends up removing the last rock. Could I have some volunteers? Can I volunteer people then? How about the guy with the white t-shirt with the yellow on there? Okay, come on up front, and who else can I volunteer. Everyone’s looking away from me, desperately looking away from me. How about the guy with the Yale sweatshirt on, the Yale football sweatshirt on, okay. So come on up. Your name is? I should get a mike, let me get a mike up here, thank you. Great, your name is?
But here’s the challenge, the challenge is, I know from Zermelo’s theorem that no matter what N is, and no matter what M is, this game has a solution. So Zermelo’s theorem tells us this game has a solution. Now, notice it could have a different solution depending on the N and the M, depending on how many rows and how many columns, is that right?. So depending on N and M, each such game has a solution–is that right–which could depend on N and on M, on the number of rows and columns. So just as in Nim, it depended on how those piles started out. So what’s going to be the homework assignment? Find the solution. Homework- what is the solution? So I claim, let me give you a hint, I claim it is useful to remember that there is a solution. It turns out to be useful to remember that. That wasn’t much of a hint. You knew that already, but let me just emphasize that. ## Chapter 5. Games of Perfect Information: Definition [00:40:27]Okay, so I want to do a bit of a transition now away from these games like chess and like checkers, and like this game with the rocks or Nim, and I want to be a little bit formal for a while. So one thing that we haven’t done for a while, actually since the mid-term, is write down some formal definitions. So I’m going to do that now. I want at least one of these boards back. So as I warned very early in the class, there are some points of the day when we have to stop and do some work and the next twenty minutes or so is such a point. So what is this? This is formal stuff. The first thing I want to do is I want to give you a formal definition of something I’ve already mentioned today and that’s the idea of perfect information. What we’ve been looking at, or at least since the mid-term, are games of perfect information. So a game of perfect information is one in which at every node, or at each node in the game the player whose turn it is to move at that node knows which node she is at. So it’s a very simple idea. Every game we’ve seen since the mid-term has this property. When you’re playing this game you know where you are. So you know which node you’re at, you know what your choices are So the whole history of the game is observed by everybody. When I get to move I know what you did yesterday, I know what I did the day before yesterday, and if I’m playing with a third person I know what they did the day before that. So a very simple idea but it turns out to be an idea that actually allows us to use things like backward induction very simply. Now so far all we’ve been doing is thinking about such games, games like perfect competition, sorry games like quantity competition between firms or games like the games where we were setting up incentives or games like Nim, and we’ve basically been solving these games by backward induction, rather informally. What I want to add in now is the notion of a strategy in these games. So when we had simultaneous move games, strategies were really unproblematic. It was obvious what strategies were. But when we have games which are sequential, that go on over a period of time, and information is emerging over a period of time, we need to be a little careful what we mean by a strategy. So what I want to do is I want to define a pure strategy at least as follows. So a pure strategy for Player i in a game of perfect information–so in the games we’ve been talking about, games which we can represent by trees is what? It’s a complete plan of action. So in other words, what it does is it specifies which action I should take–let’s not make it should take–let’s say What I want to do is I want to look at an example, and this example I’m hoping is going to illustrate some subtleties about this definition. So here’s an example. In this example Player 1 moves first and they can choose in or out, or if you like let’s call it up or down since that’s the way we’ve drawn it. So they can choose down or up–and I’ll use capital letters U and D. If Player chooses U then Player 2 gets to move and Player 2 can choose L or R. If Player 1 moves U followed by Player 2 choosing L, then Player 1 gets to move, in which case Player 1 could move up or down and I’ll use little letters this time, u and d. Let’s put some payoffs in this game. So the payoffs are 1,0 here, 0,2 here, 3,1 here and 2,4 here. So on paper–it’s on the board–when you first look at it, this is a perfectly simple game. It’s just like many of the games we’ve been looking at since the mid-term. It has a tree. We could analyze it by backward induction, and in a moment we will analyze it by backward induction. But what I want to do first is I want to say what are the strategies in this game? So let’s start with Player 2 since it’s easier. Player 2’s strategies here are what? Pretty simple: player 2 only has one decision node–here’s Player 2’s decision node–and the strategy has to tell Player 2 what he’s going to do at that decision node. So there’s only two choices, L or R. So Player 2’s strategies are either L or R. Notice, however, already one slight subtlety here. Player 2 may never get to make this choice. So Player 2 is choosing L or R, but Player 2 may not ever get to make this choice. In particular, if 1 chose D it’s really irrelevant whether Player 2 has chosen L or R. Nevertheless, the strategy has to specify what Player 2 would do were he called upon to make that move. Now let’s look at Player 1’s strategies. So what are Player 1’s strategies in this game? Any takers? Anyone want to guess? So I claim it’s tempting, but it turns out to be wrong, to say that Player 1 has three strategies here. It’s tempting to say either Player 1 moves D, in which case we’re done, or Player 1 moves U, in which case at some later date she may be called upon to choose u or d again. So it’s very tempting to think that Player 1 has just three strategies here, D in which case we’re done, or U followed by either u or d, if she’s reached that point. But actually, if we follow this definition carefully, we notice that Player 1 actually has four strategies here. So let me say what they are and you’ll see why it’s a little odd. So here’s one of the strategies we talked about, U followed by u, and here’s another one we talked about, U followed by d, but I claim there are two others, D followed by u and D followed by d. So what’s a little weird about this? What’s weird is we know perfectly well that if Player 1 chooses D she’s not going to get to make the choice, the second choice u or d. But nevertheless the strategy has to tell us what she would do at every node, regardless of whether that node actually is reached by that strategy. Let me say it again, the strategy has to tell you what Player 1 would do at that node, regardless of whether that node is actually reached by playing that strategy. So a little weird. There’s a bit of redundancy here. It’s a bit redundant. Now why? Well we’ll see why in a second. Let’s first of all consider how this game will be played according to backward induction. So how do we play this game according to backward induction? Where do we start? Shouldn’t be a trick question, where do we start the game according to backward induction? At the end okay. So the end here is Player 1 and Player 1 will choose here to go d, is that right? Because 3 is bigger than 2. So if we roll back to Player 2’s move, Player 2 if she chooses L she knows that she’ll be followed by Player 1 going d, in which case she’ll get 1, but if she chooses R she’ll get 2. So Player 2 will choose R. So Player 1 here, if she chooses U then Player 2 will choose R and Player 1 will get 0, and if she chooses D she’ll get 1, so Player 1 will choose D. So, backward induction suggests that Player 1 will choose D; and followed by Player 2–if Player 2 did get to move–she’d choose R; followed by Player 1–if she did get to move again–choosing d again. Notice that backward induction had to tell us what Player 2 would have done had she got to move. Backward induction had to consider what Player 2 would think that Player 1 would have done were Player 1 to get to move again. So to do backward induction, Player 1 has to ask herself what Player 2 is going to do; and Player 2 has to therefore ask herself what Player 1 would have done; which means Player 1 has to ask herself what Player 2 thinks Player 1 would have done; which means we actually needed to say what Player 1 did over here. Let me say that again. So backward induction here tells us that Player 1 chooses D in which case the game is over. But to do backward induction, to think through backward induction, we really needed to consider not just what Player 2 was going to do were he to get to move, but also what Player 2 would think Player 1 would do were Player 2 to get to move and were to do choose L. So all of these redundant moves were actually part of our backward induction. We need them there so we can think about what people are thinking. So backward induction tells us that the outcome of the game is D, but it tells in terms of strategies, Player 1 chooses D and, were she to get to move again, would choose D again. And Player 2 chooses R. Now let’s analyze this game a totally different way. We now know what the strategies are in the game: Player 2 has two strategies, L and R; and Player 1 has four strategies: U/u, U/d, D/u and D/d. So let’s write up the matrix for this game, so here it is. It must be a 4 x 2 matrix, Player 2 is choosing between L and R and Player 1 is choosing between her four strategies Uu, Ud, Du and Dd and we can put all the payoffs in. So (Uu,L) gets us here which is (2,4). (Uu,R) gets us here which is (0,2). (Ud,L) gets us here which is (3,1). (Ud,R) gets us here again which is (0,2). And Du gets us, going right out of the game immediately, so we’re going to go to (1,0). And in fact that’s also true for (Du,R), and it’s also true for (Dd,L) and it’s also true for (D/d,R). So all of these four strategies at the bottom started off by Player 1 choosing D, in which case the game is over. So once again in this matrix you can kind of see the redundancy I was talking about. In this matrix, the third row, the Du row looks the same as the Dd row. Everyone happy with that? So when we’ve had matrices in the past how have we solved the game? Well we’ve been doing this backward induction stuff for a couple of weeks, but, prior to the mid-term–if I had given you this on the mid-term what would you have done? You’d look for a Nash Equilibrium right? So let’s do that: let’s look for Nash Equilibrium. So if Player 2 chooses L then Player 1’s best response is Uu; and if Player 2 chooses R, then Player 1’s best response is either Du or Dd. Conversely, if Player 1 chooses Uu then Player 2’s best response is L. If Player 1 chooses Ud then Player 2’s best response is R. If Player 1 chooses Du then Player 2 doesn’t care because they’re getting 0 anyway; and if Player 1 chooses Dd then again Player 2 doesn’t care because they’re getting 0 anyway. So the Nash Equilibria in this game are what? So one of them is here, so that’s one Nash Equilibrium. One of them is Dd followed by R, and that’s a Nash Equilibrium we actually found by backward induction. The Nash Equilibrium (Dd,R) corresponds to the one we found by backward induction. But there’s another Nash Equilibrium in this game. The other Nash Equilibrium in this game is this one. That’s also a Nash Equilibrium. What does it involve? It involves Du and R–sorry Du and R. So what happened in that equilibrium? Player 1 went down which made everything else kind of irrelevant. Player 2 played R and Player 1 in their plan of action was actually going to choose u. Say it again, Player 1 in fact went D, Player 2 had they got to move would have chosen R and Player 1 had they got to move at this second time would have chosen u. So how can that be a Nash Equilibrium? It doesn’t correspond to backward induction. In particular, Player 1 up here is choosing a strategy that seems silly. They’re choosing u rather than d which gets them 2 rather than 3. So how could it possibly be that that’s an equilibrium strategy? The reason is–the reason it can be an equilibrium strategy is it doesn’t really matter what Player 1 chooses up here from Player 1’s point of view because it’s never going to be reached anyway. As long as Player 2 is going to choose R here, it really doesn’t matter what Player 1 decides to do up there. From Player 1’s point of view, it doesn’t make a difference. It isn’t reached anyway. So we’re highlighting here a danger, and the danger is this. If you just mechanically find the Nash Equilibria in a game, you’re going to find people choosing actions that, if they ever were called upon to make them, are silly. Say it again, if you just mechanically find the Nash Equilibria in the game, just as we did here, you’re going to select some actions, in this case an action by Player 1, that were she called upon to take that action would be a silly action. The reason it’s surviving our analysis is because in fact she isn’t called upon to make it. Now to make that more concrete, let’s take this to an economic example, this same idea. ## Chapter 6. Games of Perfect Information: Economic Example [01:01:56]So what I want you to imagine is a market, and in this market there is a monopolist who controls the market, but there’s an entrant who is thinking of entering into this market. So right now this market has a monopoly in it, and the monopolist is an incumbent–let’s call him Inc–and an entrant who is trying to decide whether or not to enter the market or whether to stay out. If they stay out then the entrant gets nothing and the incumbent continues to get monopoly profits. So think of this as 3 million a year. If the entrant enters then the incumbent can do one of two things. The incumbent could choose to fight the entrant, by which I mean charge low prices, advertise a lot, try and drive him out of the market. He could play very competitively, in which case the entrant will actually make losses, and the incumbent will drive her profits down to zero. Conversely, the incumbent could choose not to fight. If they don’t fight then they’ll simply share the market and they’ll both make, let’s say Cournot profits or duopoly profits of a million each. So in this game, let’s just say again, there’s a market there. The monopolist is in the market and the monopolist is making 3 million a year which seems pretty nice for the monopolist. The entrant is trying to decide whether to invade this market. If she invades this market, if she’s fought she’s in trouble, but if the monopolist accommodates her, then they just go to duopoly profits and the entrant does very well, gets a million dollars in profit. Let’s have a look at this game. Analyze–first of all this time, let’s analyze it by Nash Equilibrium. So I claim in this game, this is pretty simple: the entrant has two strategies. I’ll put it straight into the matrix. The entrant has two strategies, they’re either to go in or to stay out; and the incumbent has two strategies, they are either to fight the entrant or not to fight. The payoffs of this game are as follows–let’s put them in. In and fight was (-1,0), in and not fight was (1,1) and out led to the incumbents maintaining the monopoly profits. Let’s have a look at the Nash Equilibria of this game. So the first thing to do is to look at the best responses for the entrant. So if the incumbent chooses fight, then the entrant’s best response is to stay out. If the incumbent chooses not fight, then the entrant’s best response is to enter. How about for the incumbent? If the entrant chooses to come in then the incumbent’s best response is not to fight because 1 is bigger than 0. But if the entrant stays out it really doesn’t matter what the incumbent does, they’ll get 3 either way. So the Nash Equilibria here are either in followed by not fighting, you end up with a duopoly, or out followed by fighting. Of course the fight doesn’t take place: out followed by “we would have fought had you entered.” So these are the Nash Equilibria. Let’s analyze this game by backward induction. From the incumbent’s point of view, if the incumbent gets to move then is she going to choose fight or not fight? She’s going to choose not fight, so the entrant should do what? Somebody? The entrant should enter, because if she enters she gets 1, if she stays out she gets 0. So backward induction just gives us this equilibrium. Now the question is what do we think is going on with this other equilibrium? Let’s talk it through. So here you are, you’re about to enter the market, you leave Yale, you set up your business, and your business is challenging some monopolist or quasi-monopolist like Microsoft say., And you go out there and you’re about to put out your new operating system. And you know that your new operating system will make plenty of money provided Microsoft doesn’t retaliate and drop its prices by half and advertise you out of the market. So what does Bill Gates do? Bill Gates is the head of Microsoft. He says, wait a second before you graduate from Yale and go set up this company, let me just tell you I’m going to fight. If you enter I’m going to fight. And if you believe him, if you believe that he’s going to fight, what should you do? You’re not going to bother to enter his market. And if you believe that you’re going to get fought by Bill Gates then you believe that if you enter you’re going to make losses, so you choose to stay out. And this threat that Bill Gates makes here it doesn’t cost him anything, provided you go out, he doesn’t actually have to fight at all. That’s why it is in fact an equilibrium to imagine you staying out believing that you’re going to get fought if you enter, and it is a best response for the guy to fight knowing that you’re going to stay out. But this doesn’t sound right. Why doesn’t this sound right? It doesn’t sound right because you really shouldn’t believe the guy who says he’s going to fight you when you know that if you did enter fighting would cost the incumbent money. If you did enter, if he fights you he gets 0, if he doesn’t fight he gets a million dollars. So this threat that the guy is going to fight you, this threat is not credible. This is an equilibrium but it relies on believing an incredible threat. It’s true that if you think about entering the market that Microsoft is in, you’re very likely to get a little email from Bill Gates. (But it won’t be an email because that can be taken to Court, but some little threatening remark in your ear from Bill Gates.) It’s true if you entered into the market that the people who build aircraft in, the head of Boeing might pay you a call one day or send someone round, but these threats are not credible threats. Is that right? They’re not credible threats because we know that if you did enter, it isn’t in their interest to fight. They’re making a lot of noise about it but you know that if you did enter, backward induction tells us they’re not going to fight. But now we’re in slightly an odd situation, why are we in an odd situation? Two things. One, we seem to be finding that there are Nash Equilibria in games–we found one here and one up here–there are Nash Equilibria that are not supported by backward induction. That’s the first reason we’re a little worried here, and the second reason we’re worried is even the economics of this doesn’t quite smell right. For example, if you in fact did enter–sorry, if you did in fact announce you were going to operate, you were going to build a new operating system and got a threatening phone call from Bill Gates–there might be a reason why you might actually believe that call. Why might you believe that call? Can we get a mike out, I’ll do it. So why might you believe a call from Bill Gates saying he’s going to beat you up–not beat you up but he’s going to charge low prices if you produce an operating system?
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