# Game Theory – Trees in Games

Many games can be solved using the minimax algorithm for exploring a tree and determining the best move from each position.

## Basic Tree

Take a look at the game tree below. The circular nodes represent player positions and the lines represent possible actions. The “root” of the tree is the initial state at the top. We have P1 acting first, P2 acting second, and the payoffs at the leaf nodes in the standard P1, P2 format.

The standard way to solve a tree like this is using backward induction, whereby we start with the leaves (i.e. the payoff nodes at the bottom) of the tree and see which decisions the last player, Player 2 in this case, will make at her decision nodes.

Player 2’s goal is to minimize the maximum payoff of Player 1, which in the zero-sum setting is equivalent to minimizing her own maximum loss or maximizing her own minimum payoff. This is equivalent to a Nash equilibrium in the zero-sum setting.

She picks the right node on the left side (payoff -1 instead of -5) and the left node on the right side (payoff 3 instead of -6).

These values are then propagated up the tree so from Player 1’s perspective, the value of going left is 1 and of going right is -3. The other leaf nodes are not considered because Player 2 will never choose those. Player 1 then decides to play left to maximize his payoff.

We can see all possible payouts, where the rows are P1 actions and the columns are P2 actions after P1 actions (e.g. Left/Left means P1 chose Left and then P2 also chose Left).

P1/P2 Left/Left Left/Right Right/Left Right/Right
Left 5,-5 5,-5 1,-1 1,-1
Right -3,3 -3,3 6,-6 6,-6

Note that Player 1 choosing right could result in a higher payout (6) if Player 2 also chose right, but a rational Player 2 would not do that, and so the algorithm requires maximizing one’s minimum payoff, which means Player 1 must choose left (earning a guaranteed value of 1).

By working backwards from the end of a game, we can evaluate each possible sequence of moves and propagate the values up the game tree.

Subgame perfect equilibrium means that each subgame, which is a decision state in the above game tree, is a Nash equilibrium. The strategy of P1 choosing Left and P2 choosing Right after Left and Left after Right is a subgame perfect equilibrium.

Two main problems arise with minimax and backward induction.

### Problem 1: The Game is too Damn Large

In theory, we could use the minimax algorithm to solve games like chess. The problem is that the game and the space of possible actions is HUGE. It’s not feasible to evaluate all possibilities. The first level of the tree would need to have every possible action and then the next level would have every possible action from each of those actions, and so on. Even checkers is very large, though smaller games like tic tac toe can be solved with minimax. More sophisticated methods and approximation techniques are used in practice for large games. One simple method is to only go down the tree to a depth of “X” and then approximate the value of the states there.

### Problem 2: Perfect Information vs. Imperfect Information

What about poker? Real poker games like Texas Hold’em are very large and run into the same problem we have with games like chess, but in addition, poker is an imperfect information game and games like chess and tic tac toe are perfect information games. The distinction is that in poker there is hidden information – each player’s private cards. In perfect information games, all players see all of the information.

With perfect information, each player knows exactly what node/state he is in in the game tree. With imperfect information, there is uncertainty about the state of the game because the other player’s cards are unknown.

## Poker Tree

Below we show the game tree for 1-card poker. In brief, it’s a 1v1 game where each player starts with $2 and antes$1, leaving a single \$1 bet remaining. We’ll go into more details about the game in the next section.

The top node is a chance node that “deals” the cards. To make it more readable, only 2 chance outcomes are shown, Player 1 dealt Q with Player 2 dealt J and Player 1 dealt Q with Player 2 dealt K.

1-card poker game tree from University of Alberta

Player 1’s initial action is to either bet or pass. If Player 1 bets, Player 2 can call or fold. If Player 1 passes, Player 2 can bet or pass. If Player 1 passed and Player 2 bet, then Player 1 can call or fold.

Note the nodes that are circled and connected by a line. This means that they are in the same information set. An information set consists of equivalent states based on information known to that player. For example, in the top information set, Player 1 has a Q in both of the shown states, so his actions will be the same in both even though Player 2 could have either a K or J. The information known to Player 1 is “Card Q, starting action”. At the later information set, the information known is “Card Q, I pass, opponent bets”. All decisions must be made based only on information known to each player! However, these are actually different true game states.

Looking at the information set at the bottom where Player 1 passes and Player 2 bets, Player 1 has the same information in both cases, but calling when Player 2 has a J means winning 2 and calling when Player 2 has a K means losing 2. The payoffs are completely different!

Therefore we can’t simply propagate values up the tree as we can do in perfect information games. Later in the tutorial, we will discuss CFR (counterfactual regret minimization), which is a way to solve games like poker that can’t be solved using minimax.

## Tic Tac Toe Tree

Tree goes here

On the tic tac toe tree, from the initial state, there are up to 9 levels of moves. Each subsequent level has fewer possible actions since more spaces on the game board are taken as we go down the tree. The tree ends at points either where the game is over because one player wins or when all the spaces are filled and no one has won, resulting in a tie.

In tic tac toe, the sequence of actions prior to a certain game state are not important.

## Tic Tac Toe Python Implementations

While we’re mainly focused in this tutorial on poker and imperfect information games, we take a short detour to look more in-depth at minimax and Monte Carlo Tree Search (MCTS) through the lens of tic tac toe.

### Tic Tac Toe in Python

Below we show a basic Python class called Tictactoe. The board is initialized with all 0’s and each player is represented by a 1 or -1. Those numbers go into board spaces when the associated player makes a move. The class has 5 functions:

1. make_move: Enters the player’s move onto the board if the space is available and advances the play to the next player
2. new_state_with_move: Same as make_move, but returns a copy of the board with the new move instead of the original board
3. available_moves: Lists the moves that are currently available on the board
4. check_result: Checks every possible winning sequence and returns either the winning player’s ID if there is a winner, a 0 if the game has ended in a tie, or None if the game is not over yet
5. repr: Used for printing the board. Empty slots are represented by their number from 0 to 8, player 1 is represented with ‘x’, player 2 is represented with ‘o’, and a line break is added as needed after the first 3 and middle 3 positions.

We also have two simple agent classes:

1. HumanAgent: Enters a move from 0-8 and the move is placed if it’s available, otherwise we ask for the move again
2. RandomAgent: Randomly selects a move from the available moves

Finally, we need another function to actually run the game.

class Tictactoe:
def __init__(self, board = [0] * 9, acting_player = 1):
self.board = board
self.acting_player = acting_player

def make_move(self, move):
if move in self.available_moves():
self.board[move] = self.acting_player
self.acting_player = 0 - self.acting_player #Players are 1 or -1

def new_state_with_move(self, move): #Return new ttt state with move, but don't change this state
if move in self.available_moves():
board_copy = copy.deepcopy(self.board)
board_copy[move] = self.acting_player
return Tictactoe(board_copy, 0 - self.acting_player)

def available_moves(self):
return [i for i in range(9) if self.board[i] == 0]

def check_result(self):
for (a,b,c) in [(0,1,2),(3,4,5),(6,7,8),(0,3,6),(1,4,7),(2,5,8),(0,4,8),(2,4,6)]:
if self.board[a] == self.board[b] == self.board[c] != 0:
return self.board[a]
if self.available_moves() == []: return 0 #Tie
return None #Game not over

def __repr__(self):
s= ""
for i in range(9):
if self.board[i] == 0:
s+=str(i)
elif self.board[i] == 1:
s+='x'
elif self.board[i] == -1:
s+='o'
if i == 2 or i == 5: s += "\n"
return s

class HumanAgent:
def select_move(self, game_state):
move = int(float(input()))
#print('move', move)
#print('game state available moves', game_state.available_moves())
if move in game_state.available_moves():
return move
else:
print('Invalid move, try again')
self.select_move(game_state)

class RandomAgent:
def select_move(self, game_state):
return random.choice(game_state.available_moves())

if __name__ == "__main__":
ttt = Tictactoe()
#ttt = Tictactoe([0,0,-1,0,0,0,1,-1,1]) #Optionally can start from a pre-set game position
h = HumanAgent()
r = RandomAgent()
moves = 0
while ttt.available_moves():
print(ttt)
print('move', moves)
print('acting player', ttt.acting_player)
if moves % 2 == 0:
move = player1.select_move(ttt)
else:
move = player2.select_move(ttt)
ttt.make_move(move)
if ttt.check_result() == 0:
print('Draw game')
break
elif ttt.check_result() == 1:
print('Player 1 wins')
elif ttt.check_result() == -1:
print('Player 2 wins')
moves+=1


### Minimax Applied to Tic Tac Toe

We can apply the minimax algorithm to tic tac toe. Here we use a simplified version of minimax called negamax, because in a zero-sum game like tic tac toe, the value of a position to one player is the negative value to the other player.

We store already evaluated states in a memo dictionary containing each move and its value. When a state has not been seen before, we check the game state, which will either return the winning player, 0 for tie, or None for “game not yet over”.

If the game is over, then there is no move from this position and the value is simply the result of the game.

If the game is not over, we iterate through the available moves and recursively find a value for each possible move. As each move is evaluated, we store the best move and the value for that move and return the overall best after evaluating each move.

class NegamaxAgent:
def __init__(self):
self.memo = {} #move, value

def negamax(self, game_state):
if game_state not in self.memo: #already visited this state?
result = game_state.check_result()
if result is not None: #leaf node or end of search
best_move = None
best_val = result * game_state.acting_player #return 0 for tie or 1 for maximizing win or -1 for minimizing win
else:
best_val = float('-inf')
for i in game_state.available_moves():
clone_state = copy.deepcopy(game_state)
clone_state.make_move(i) #makes move and switches to next player
_, val = self.negamax(clone_state)
val *= -1
if val > best_val:
best_move = i
best_val = val
self.memo[game_state] = (best_move, best_val)
return self.memo[game_state]


### Monte Carlo Tree Search (MCTS) Applied to Tic Tac Toe

MCTS is a more advanced algorithm that finds the optimal move through simulation. This algorithm is used as part of some recent advances in AI poker agents.

MCTS allows us to determine the best optimal move from a game state without having to expand the entire tree like we had to do in the minimax algorithm.

Let’s consider that we want to find the best tic tac toe move from some state of the game that we can pre-specify. Let’s go through the algorithm step by step.

The MCTSAgent class and its select_move function contain the core of the algorithm. The function begins by setting the root of the game tree as an MCTSNode class. A node is a decision point in the game tree, so the root node is the beginning of the tree. If we wanted to find out the best move from the beginning of the game, this would represent an empty tic tac toe board.

Each node is initialized with a game state, a parent node, a move, a set of child nodes, a counter for wins by each player, a counter for rollouts that have gone through this node, and a list of available moves from this node.

For some fixed number of rounds, we go through the following steps:

1. Selection:

2. Expansion: If we can add a child node, then we select a random move from the current game state and create a new node to represent the game with this new move, which becomes a child of the prior node. This is done through the add_random_child function in the MCTSNode class.

3. Simulation: Next we run a random playout from the newly expanded node until the game ends

4. Backpropagation: From the end of the game, we update each node that was passed through by updating the win counts for each player (one player gets +1 and one gets -1 or both get 0 in the case of a tie) and adding 1 to the number of rollouts that have passed through each of the nodes.

After running MCTS, we look at each child node from the root and evaluate its win percentage over all of the simulations. We then print a list of the moves in order of their winning percentage, along with how many simulations were run for each move.

class MCTSNode:
def __init__(self, game_state, parent = None, move = None):
self.parent = parent
self.move = move
self.game_state = game_state
self.children = []
self.win_counts = {1: 0, -1: 0}
self.num_rollouts = 0
self.unvisited_moves = game_state.available_moves()

move_index = random.randint(0, len(self.unvisited_moves)-1) #inclusive
new_move = self.unvisited_moves.pop(move_index)
new_node = MCTSNode(self.game_state.new_state_with_move(new_move), self, new_move)
self.children.append(new_node)
return new_node

return len(self.unvisited_moves) > 0

def is_terminal(self):
return self.game_state.check_result() is not None

def update(self, result):
if result == 1:
self.win_counts[1] += 1
self.win_counts[-1] -= 1
elif result == -1:
self.win_counts[-1] += 1
self.win_counts[1] -= 1
self.num_rollouts += 1

def winning_frac(self, player):
return float(self.win_counts[player]) / float(self.num_rollouts)

class MCTSAgent:
def __init__(self, num_rounds = 10000, temperature = 2):
self.num_rounds = num_rounds
self.temperature = temperature

def uct_select_child(self, node):
best_score = -float('inf')
best_child = None
total_rollouts = sum(child.num_rollouts for child in node.children)
log_rollouts = math.log(total_rollouts)

for child in node.children:
win_pct = child.winning_frac(node.game_state.acting_player)
exploration_factor = math.sqrt(log_rollouts / child.num_rollouts)
uct_score = win_pct + self.temperature * exploration_factor
if uct_score > best_score:
best_score = uct_score
best_child = child
return best_child

def select_move(self, game_state):
root = MCTSNode(game_state)

for i in range(self.num_rounds):
node = root

#selection -- UCT select child until we get to a node that can be expanded
while (not node.can_add_child()) and (not node.is_terminal()):
node = self.uct_select_child(node)

#expansion -- expand from leaf unless leaf is end of game

#simulation -- complete a random playout from the newly expanded node
gs_temp = copy.deepcopy(node.game_state)
while gs_temp.check_result() is None:
gs_temp.make_move(random.choice(gs_temp.available_moves()))

#backpropagation -- update all nodes from the selection to leaf stage
while node is not None:
node.update(gs_temp.check_result())
node = node.parent

scored_moves = [(child.winning_frac(game_state.acting_player), child.move, child.num_rollouts) for child in root.children]
scored_moves.sort(key = lambda x: x[0], reverse=True)
for s, m, n in scored_moves[:10]:
print('%s - %.3f (%d)' % (m, s, n))

best_pct = -1.0
best_move = None
for child in root.children:
child_pct = child.winning_frac(game_state.acting_player)
if child_pct > best_pct:
best_pct = child_pct
best_move = child.move
print('Select move %s with avg val %.3f' % (best_move, best_pct))
return best_move


Updated: