Prerequisites: Chapters 1, 2, 4, 6, 7, and 8.
Coaches of team sports like soccer supervise the training of their teams, but they also play. They select the players and the system before the game. Of course, this move is extremely important, and even though these decision may not be made simultaneously by both coaches, since it is kept secret and we have imperfect information, it is equivalent to a simultaneous move. Besides trying to motivate their teams, the coaches also have the important task to react to what they see by changing players or changing the system.
In a Champion's League soccer game between Bolton Wanderers and Bayern Munich in November 2007, Munich was leading 2:1 when their coach Ottmar Hitzfeld exchanged Frank Ribery and Bastian Schweinsteiger, maybe the two key players of his team. After that, the Bolton Wanderers were able to score for an end result of 2:2, which caused Rummenigge's remark, indirectly criticising Hitzfeld's move, which in turn prompted Hitzfeld's response.
In this section we will extend the simple static simultaneous soccer model discussed previously into a game with three rounds, where in each round both coaches move simultaneously. We will see that there are too many pure strategies to try the Normal Form approach, and instead we have to try the variant of backward induction on subtrees.
The most obvious feature of the game is its structure of simultaneous moves performed in rounds. But since for DMA soccer we have used the notion of "rounds" already, let's call them "thirds".
Each one of the two coaches Ann and Beth moves three times. Their first moves, their second moves, and their third moves are performed simultaneously.
For the extensive form we must avoid simultaneous moves by assuming that Ann moves always first, but Beth doesn't know Ann's decision after Beth herself has moved. That means, at the beginning of each third, Ann has single-position information sets, but Beth has information sets containing several positions. Each one of Ann's single-position information set is the start of a subgame. So we have subgames starting at Ann's possible positions at the beginning of the last third, larger subgames starting at Ann's positions at the beginning of the second third, and finally the full game starting at Ann's first part start position.
Although such multi-round simultaneous games do not have perfect information, and therefore backward induction can not be performed, we can still use the same idea. First we assign expected payoffs for the Ann's single-position information sets at the beginning of the last third. This analysis also includes move recommendations for both Ann and Beth. Then we assign such values to Ann's single-position information sets at the beginning of the second third. In order to be able to do this, we need the values for the third move positions. If we have done this for all of them, we cann assign expected payoffs for the start position. Different to backward induction, these values do not have to be unique, so it is possible to get different sets of values at some of these positions, which of course complicates matters. But in our game, this problem will not occur.
Again the distributions "DX,MX,AX" for the number of players in defense, midfield, and attack,
are encoded as (DX,MX), since AX then follows from the equation DX+MX+AX=5.
The (undirected) graph to the right visualizes the possible changes in distributions at each move.
Two distributions are connected by an edge if they can be transformed into each other.
Others, like (1,2) and (2,2) would require two such moves.
You would have to move a player from attack to defense, but these fields are not adjacent,
so you would have to move one player from attack
to midfield and one from midfield to defense.
Note that all these allowed moves are symmetric.
If you can transform distribution 1 into distribution 2, then you can also transform
distribution 2 into distribution 1 simply by doing the reverse.
At their first move, both players have only one information set each. At the second and third move, the information sets for the players are in principle a combination of the standing so far and the two distributions so far. But note that in the payoff system used, where all that counts is whether your team is winning, drawing, or losing, a standing of 1:2 would lead to the same move than a standing of 0:1. In each of these two cases your team needs two more goals to win, and one goal for a draw. Therefore all what matters about the standing is the goal difference so far. At the second move, Ann and Beth have each 7·21·21=3087 information sets. Ann may have chosen any of the 21 possible start distributions, Beth may have chosen any of these 21 start distributions, and the goal difference so far (after 3 rounds) might be any of the the 7 possible values 3, 2, 1, 0, -1, -2, -3 (viewed as in favor for Ann). At their third move, Ann and Beth each have 13·21·21=5733 information sets, since the possible goal difference after 6 rounds could be any of the 13 values 6, 5, ... -5, -6. Therefore analyzing, or even formally describing the game takes a huge amount of effort and space. Since at the first move the players have 21 options of choosing the initial distribution, and at the second and third move they have at least 2 options (in most cases they have 5 options: a defense player could move to midfield, or a midfield player could move to either defense or attack, or an attack player could move to midfield, or there is no change), each player has at least 21·23087·25733 pure strategies. This is very huge number with more than 2600 digits! More than the universe has atoms, for all we know. Admittedly the number of reduced pure strategies would be a little smaller, but still, using the normal form to analyze this game is not an option, even for this relatively simple game.
Therefore we have to stick to the extensive form, which is also huge, with thousands of vertices, impossible to attach. And the methods used for normal form are out of reach, therefore the recursive method discussed above for multi-round simultaneous games is our only option. Still, we will not solve the game completely, but we will see that given more time, it could be done.
Assume that six rounds have been played, and the two coaches are about to make their final third move, meaning we are the beginning of a last third subgame. The distribution that can be achieved next obviosuly depend on the team's present position. The decision to which position to move will depend on two factors: First of course it depends on the position the other team has presently. Secondly it depends on the standing so far, or to be more precise, the goal difference so far: A coach whose team is behind would act differently (probably strengthening midfield or attack) than a coach whose team is leading. If a team is behind, a higher probability for goals is preferable to a low probability for goals, even if higher probability for goals of the own team might imply also a higher probability for goals one gets.
Each of the two coaches has up to five options for the third move---keeping the distribution, or changing it to one of the up to four possible distributions as displayed in the graph. For each pair of third moves chosen, the probabilities for each of the ten possible goal patterns achieved in the last part --- 3:0, 2:0, 1:0, 0:0, 2:1, 1:1, 0:1, 1:2, 0:2, 0:3--- can be computed in the same way as in DMA soccer I. The ten possible outcomes of the whole game are obtained from the present standing by adding these ten goal patterns. The convention nowadays is to give a payoff of 3 points for a win, 1 point for a draw, and 0 points for a loss. Therefore the expected number of points, when proceeding with a certain combination of distributions for team A and team B, can be computed rather easily.
Assume team A played distribution (2,2) during the middle third, whereas team B played distribution (3,1). Then Ann has five options: She can move to one of the four distributions (1,3) or (2,1) or (2,3), or (3,1), or stay at distribution (2,2) In the same way, Beth has five options of moving to (2,2) or (3,0) or (3,2) or (4,0), or keeping the distribution (3,1). Since the cases where one team leads already by four or more goals are easy to analyze---there is nothing the team behind can do---we only look at the other seven goal differences where the outcome is still not determined.
| (2,2) | (3,0) | (3,1) | (3,2) | (4,0) | |
| (1,3) | 0.48, 2.28 | 0.89, 1.69 | 0.52, 2.21 | 0.51, 2.17 | 0.70, 1.92 |
| (2,1) | 0.37, 2.42 | 1.49, 1.08 | 0.51, 2.23 | 0.45, 2.25 | 1.22, 1.33 |
| (2,2) | 0.42, 2.35 | 0.89, 1.69 | 0.47, 2.27 | 0.42, 2.30 | 0.70, 1.92 |
| (2,3) | 0, 3 | 0, 3 | 0, 3 | 0, 3 | 0, 3 |
| (3,1) | 0.26, 2,56 | 0.89, 1.69 | 0.33, 2.46 | 0.27, 2.52 | 0.70, 1.92 |
Let me explain how this entries are obtained for one cell, the upper middle cell where Ann moves to distribution (1,3) and Beth keeps her distribution (3,1). As explained in DMA Soccer I, in each of the remaining three rounds the probability for a goal for Ann's team is pA=(3/4)·(1/4)=3/16 (remember that Ann has a midfield of 3 and Beth a midfield of 1, wheras Ann has an attack of 1 and Beth a defense of 3), the probability for a goal for Beth's team is pB=(1/4)·(1/2)=1/8, but most likely is no goal in a round with a probability of p0=1-3/16-1/8=11/16. For the last three rounds combined, 10 goal patterns are possible, having the following probabilities:
| Goals for Ann/Beth | 0 | 1 | 2 | 3 |
| 0 | p03=0.325 | 3·pB·p02=0.177 | 3·pB2·p0=0.032 | pB3=0.002 |
| 1 | 3·pA·p02=0.266 | 6·pA·pB·p0=0.097 | 3·pA·pB2=0.009 | |
| 2 | 3·pA2·p0=0.073 | 3·pA2·pB=0.013 | ||
| 3 | pA3=0.007 |
Of course, all these values add up to 1.
Therefore, the probability that in the last three rounds the goal difference will
For the same distributions (1,3) versus (3,1), if Ann's team is one goal behind, she will have won at the end with probability 0.007+0.073=0.079, there will be a draw with probability 0.279, and Beth's team will have won with probability 0.422+0.186+0.032+0.002=0.642. The expected value of points for Ann's team in that situation equals 3·0.079+1·0.279=0.516 ∼ 0.52. The expected value of points for Beth's team is 3·0.642+1·0.279=2.205 ∼ 2.21.
This analysis is done automatically in the Excel sheet DMA3.xls. Go to the "after 2 thirds" sheet. In the black cells, fill in the distributions for Ann and Beth used so far, and the goal difference so far. Also you can change the number of points for win, draw and loss in cells L4, M4, and N4. After having changed the values, the expected points bimatrix is displayed. Please confirm the bimatrix above.
Let us now look at the payoff bimatrix above. Not too surprisingly, Ann's options (2,3) and (3,1) are strictly dominated by option (2,1). Moving one player from midfield to attack is better for Ann than moving one player from attack to midfield, or moving one player from midfield to defense. If we delete these last two rows, then Beth's move (2,2), moving one player from defense to midfield, strictly dominates all her other moves. When deleting these options, Ann's move (1,3) strictly dominates the other moves, therefore the process IESD results in the only Nash equilibrium (1,3) versus (2,2). Instead of strengthening her attack, Ann will move one defense player to the midfield, and Beth will do the same, moving one defense player to midfield, even though her team is leading. This could be interpreted as Beth anticipating Ann's strengthening of the midfield and trying to counter it.
Obviously the expected numbers of points change if the goal difference after six rounds is different, and the recommendations for the third move might also change. For the calculations of the bimatrix however, a lot of work already done for Ann one goal behind can be reused. We get the same set of moves, and for each pair of moves, the probabilities pA, pB, and p0 are the same as above. The same for the probabilities of change of the goal difference by 3, 2, 1, 0, -1, -2, or -3. In the example of Ann's move of (1,3) versus Beth's move of (3,1), these numbers are again 0.007, 0.073, 0.279, 0.422, 0.186, 0.032, and 0.002. What differs is that these seven probabilities are added differently. If Ann is two goals behind so far, and they continue with moves (1,3) versus (3,1), then Ann will win the game with probability 0.007, draw with probability 0.073, and lose with probability 0.279 + 0.422 + 0.186 + 0.032 + 0.002, and Ann's expected number of points is therefore 3 · 0.007 + 1 · 0.073, whereas Beth's expected payoff is 3 · (0.279 + 0.422 + 0.186 + 0.032 + 0.002) + 1 · 0.073.
Of course, we can also calculate the 5 × 5 bimatrices for all these 7 cases with the Excel sheet. Analyzing these bimatrices, we obtain the following move recommendations and expected number of points for Ann and Beth:
| goal difference | Ann's move and expectation | Beth's move and expectation |
| -3 | (1,3): 0.003 | (3,2): 2.993 |
| -2 | (1,3): 0.068 | (3,2): 2.875 |
| -1 | (1,3): 0.480 | (2,2): 2.280 |
| 0 | (1,3): 1.320 | (2,2): 1.320 |
| 1 | (2,2): 2.347 | (2,2): 0.417 |
| 2 | (2,3): 2.900 | (2,2): 0.053 |
| 3 | (2,3): 2.995 | (2,2): 0.002 |
Here are the results for different goal differences: Ann should keep her distribution in case the present standing is a draw. If she is behind, she would move one player from midfield to attack, with a little mixing between this or not changing the distribution if she is just one goal behind. If she leads, she moves one player from midfield to defense. Beth on the other hand almost never (with the exception of a little mixing if she leads by one goal) should keep her distribution. If Beth leads by two goals or more, she should move one player from attack to midfield, even leaving her without attack. Otherwise she would move one player from defense to midfield.
| goal difference | Ann's move and expectation | Beth's move and expectation |
| -3 | (1,2): 0.008 | (2,3): 2.984 |
| -2 | (1,2): 0.120 | (2,3): 2.784 |
| -1 | 70% of (1,2) and 30% of (1,3): 0.551 | 85% of (1,3) and 15% of (2,2): 2.227 |
| 0 | (1,3). 1.344 | (1,3): 1.344 |
| 1 | (2,2): 2.280 | (1,3): 0.480 |
| 2 | (2,2): 2.832 | (1,3): 0.096 |
| 3 | (2,2): 2.984 | (1,3): 0.008 |
The expected values are shown to the right, for the different goal differences.
If we compare corresponding values, it is clear that Ann has a slight advantage.
Here are the results for different goal differences:
If we compare corresponding values, it is clear that Ann has a slight advantage.
In this case, mixed strategies occur, see the table below:
| goal difference | Ann's mixed strategies and expectations | Beth's mixed strategies and expectations |
| -3 | 50% of (0,4), 50% of (1,2): 0.013 | 50% of (1,4), 50% of (2,2): 2.974 |
| -2 | (1,2): 0.141 | (2,2): 2.766 |
| -1 | 69% of (1,2), 31% of (1,3): 0.551 | 85% of (1,3), 15% of (2,2): 2.227 |
| 0 | (1,3): 1.344 | (1,3): 1.344 |
| 1 | 85% of (1,3), 15% of (2,2): 2.227 | 69% of (1,2), 31% of (1,3): 0.551 |
| 2 | (2,2): 2.766 | (1,2): 0.141 |
| 3 | 50% of (1,4), 50% of (2,2): 2.974 | 50% of (0,4), 50% of (1,2): 0.013 |
Assume Ann's team started with distribution (1,3) and Beth's team with distribution (2,2). How should both teams proceed, depending on the goal difference after the first third, which may be any number between -3 and 3: Should they change their distributions for the second third, and if, in which way? And how many points can Ann and Beth expect now?
These positions can be analyzed just like the positions in the previous section provided we know the expectations of the players in a handful of third move positions. To be more precise, we should know these expectations for the 5·5·13 = 325 such positions where Ann's team plays any of the 5 distributions (0,4), (1,2), (1,3), (1,4), (2,2) that can be achieved from (1,3), Beth's team plays any of the 5 positions (1,3), (2,1), (2,2), (2,3), (3,1) that can be achieved from (2,2), and the goal difference is any of the 13 integers from -6 to 6. Fortunately, this is only a fraction of the total number of 21·21·13=5733 of third move positions. But of course, all third move positions with goal difference -6, -5, -4, 4, 5, or 6 are easy to analyze--- they are lost respectively won no matter what you do. So there only remain 5·5·7 = 175 such positions to analyze. Above we analyzed already 2·7=21 of them, so the additional task is doable, using the Excel sheet again. The data is collected on the tab with name "after 1 third" on the Excel sheet.
We also need calculate the probabilities pA = MA/(MA+MB)·AA/(AA+DB), pB = MB/(MA+MB)·AB/(AB+DA), and p0=1-pA-pB for all 25 combinations of distributions in the second third. This implies 25 different tables for the probabilities for number of goals in 3 rounds, similar to the table above, just plugging in the corresponding values of pA, pB, and p0. Therefore, during the second third, the goal difference will
So if a certain pair of distributions is selected in the second third, seven positions are possible at move 3, and their probabilities are as described above, where pA, pB, and p0 depend on the chosen distributions as explained above. Then we get the expected value for this choice by multiplying the expected values for these seven positions by the seven probabilities, and adding these seven products both for both teams This is also done in the "after 1 third" tab in DMA3.xls. But note that only the calculation of the probabilities and the reference to the seven positions is done automatically there--- if one would try a different case, one would have to manually insert the data in the 13 tables below the main table, which is a tremendous amount of work---solving 5·5·7 third move cases!
Still, in our concrete case the matrices for the seven second move positions can be created on this tab, by changing the "goal advantage for Ann" entry---in fact, the only cell value you can change on this tab. Analyzing these seven games, we see that all but one have a pure Nash equilibrium. The mixed Nash equilibrium for the remaining case is done using Brown's fictitious play. We get the following best distributions and expected values:
| goal difference | Ann's move and expectation | Beth's move and expectation |
| -3 | (1,2): 0.117 | (2,2): 2.840 |
| -2 | (1,2): 0.338 | (1,3): 2.559 |
| -1 | 40% of (1,2) and 60% of (1,3): 0.759 | 60% of (1,3) and 40% of (2,2): 2.057 |
| 0 | (1,3): 1.390 | (1,3): 1.390 |
| 1 | (2,2): 2.066 | (1,3): 0.737 |
| 2 | (2,2): 2.593 | (1,3): 0.292 |
| 3 | (2,2): 2.873 | (1,3): 0.082 |
The expected values are shown to the right, for the different goal differences.
Since we are earlier in the game, the divide is not as extreme for goal difference 3, 2, or 1
as in the third move case.
Most of the move recommendations are as expected. Moreover, they are almost identical to those the analysis of the third move above, for the same distributions in section 3.2, provided, with two notable exceptions: While at the end of the second third, the recommendation was to order the only attack player into midfield when Beth's team was leading with at least two goals, this now is not appropriate. Maybe it is too early for that. Instead, Beth now keeps the distribution if she is leading with 3 goals, and interestingly, gets more offensive and orders one player from defense to midfield if she is leading with 2 goals, and mixes between both if she leads with 1 goal.
Yes, but it is tedious. Above we analyzed 7 of the 21·21·7= 3087 second move positions. For this we had to analyze 325 of the 21·21·13=5733 third move positions. To analyze the start position, we would have to analyze all these other positions. This task might be a little to big for Excel sheets. Any programmer among you?
Of course we know one entry of the 21 × 21 matrix for the first move---the entry where Ann plays (1,3) and Beth (2,2). Using the same procedure as discussed in the previous section, we can conclude that Ann's expected value then is 1.414 and Beth's 1.400. Remember that (1,3) and (2,2) formed a Nash equilibrium in part I for goal difference, but that the only Nash equilibrium in the static 3 rounds DMA game was (1,3) versus (1,3) (provided a win yields 3 points and a draw 1).
Models never represent reality one-to-one. Mostly rules of the game have to be simplified, like in our example: Soccer itself is far too complicated to be modeled, and even if we could model it, we would not have the data. Of course, the outcome of a real soccer game does not only depend on the distribution of the forces of the team on defense, midfield, and attack, and exchanges can be done at any time. Still, it is the hope for a model to get recommendations that carry over to real situations. At least qualitative recommendations, like "When behind, shift forces from defense to midfield" should be derivable from a valid model.
In our case, any recommendation to real coaches would be, of course, ridiculous!
| goal difference | Ann | Beth |
| -3 | (0,2): 0.016 | (3,2): 2.969 |
| -2 | 31% (0,2) and 69% (1,2): 0.137 | 79% (2,2) and 21% (3,2): 2.78 |
| -1 | (1,2): 0.563 | (2,2): 2.203 |
| 0 | (1,2): 1.344 | (2,2): 1.344 |
| 1 | (1,2): 2.203 | (2,2): 0.563 |
| 2 | (1,2): 2.766 | (2,2): 0.141 |
| 3 | (1,2): 2.969 | (2,2): 0.016 |