J1 sale con doble X, J2 juega X:N, J3 se acuesta con doble N, J4 pasa

Escenario

J1 sale con doble X, J2 juega X:N, J3 se acuesta con doble N, J4 pasa

Resultados

Luego de correr 168 simulaciones con el escenario donde J1 sale con doble X, J2 juega X:N, J3 se acuesta con doble N, J4 pasa; la ocurrencia fue 1/168, es decir, 0,59%

¿Crees que es común este escenario?

Código python

scenario_counts_per_pair = {}

# Helper function to check if a double is present in a hand (re-defining for self-contained cell)

def has_double_in_hand(hand, target_number):

"""

Checks if the double piece (target_number, target_number) is present in the hand.

"""

return (target_number, target_number) in hand

# Helper function to check if a piece (num1, num2) or (num2, num1) is present in a hand (re-defining for self-contained cell)

def has_exact_piece_in_hand(hand, piece):

"""

Verifica si una ficha de dominó específica (ej. (3,5) o (5,3)) está presente en una mano.

Considera que (a,b) es lo mismo que (b,a).

"""

normalized_piece = tuple(sorted(piece))

for hand_piece in hand:

if tuple(sorted(hand_piece)) == normalized_piece:

return True

return False

# Helper function to count occurrences of a number in a hand (re-defining for self-contained cell)

def count_number_in_hand(hand, target_number):

"""

Counts the total occurrences of a specific number across all pieces in a player's domino hand.

"""

counter = 0

for piece in hand:

if piece[0] == target_number:

counter += 1

if piece[1] == target_number:

counter += 1

return counter

# Iterate through all possible pairs of distinct numbers (X, N)

for X in range(7):

for N in range(7):

if X == N: # Ensure X is distinct from N

continue

count_current_scenario = 0

# Iterate through each simulation

for simulation_hands in all_simulations_player_hands:

p1_hand = simulation_hands[0] # Player One's hand

p2_hand = simulation_hands[1] # Player Two's hand

p3_hand = simulation_hands[2] # Player Three's hand

p4_hand = simulation_hands[3] # Player Four's hand

# Condition 1: Player One has double X

p1_has_double_X = has_double_in_hand(p1_hand, X)

# Condition 2: Player Two has the piece (X,N)

p2_has_X_N_piece = has_exact_piece_in_hand(p2_hand, (X, N))

# Condition 3: Player Three has double N

p3_has_double_N = has_double_in_hand(p3_hand, N)

# Condition 4: Player Four has neither X nor N

p4_has_no_X = (count_number_in_hand(p4_hand, X) == 0)

p4_has_no_N = (count_number_in_hand(p4_hand, N) == 0)

p4_has_neither_X_nor_N = p4_has_no_X and p4_has_no_N

# If all four conditions are met, increment the counter

if p1_has_double_X and p2_has_X_N_piece and p3_has_double_N and p4_has_neither_X_nor_N:

count_current_scenario += 1

# Store the count for the current pair (X, N)

scenario_counts_per_pair[(X, N)] = count_current_scenario

print("Conteo de ocurrencias del escenario espec\u00edfico (J1 tiene doble X, J2 tiene X:N, J3 tiene doble N, J4 no tiene X ni N) para cada par (X, N):")

for pair, count in scenario_counts_per_pair.items():

print(f" Par (X={pair[0]}, N={pair[1]}): {count} veces")

total_global_scenario_count = sum(scenario_counts_per_pair.values())

print(f"\nConteo total global del escenario espec\u00edfico (sumando todos los pares X, N): {total_global_scenario_count} veces")