python一盘游戏中，两人轮流掷骰子5次，并将每次掷出的点数累加，5局之后，累计点数较大者获胜，点数相同则为平局。根据此规则实现掷骰子游戏，并算出50盘之后的胜利者（50盘中赢得盘数最多的，即最终胜利者）。

你可以使用Python来创建这个简单的两人对战骰子游戏。首先，我们需要定义一个函数来模拟掷骰子，然后记录每人的得分。接着是一个循环来运行50局比赛，每次都计算双方的累积分数并更新胜者。这里是一个基础的实现：

import random

# 模拟一次掷骰子
def roll_dice():
    return random.randint(1, 6)

# 初始化玩家A和B的得分
player_a_score = 0
player_b_score = 0

# 主循环 - 进行50局游戏
for game_round in range(50):
    # 玩家A行动
    player_a_point = roll_dice()
    player_a_score += player_a_point
    print(f"Game {game_round + 1}: Player A rolled {player_a_point}, score: {player_a_score}")

    # 判断是否轮到Player B
    if (game_round + 1) % 2 == 0:
        # 玩家B行动
        player_b_point = roll_dice()
        player_b_score += player_b_point
        print(f"Game {game_round + 1}: Player B rolled {player_b_point}, score: {player_b_score}")

    # 比较当前局的赢家
    winner = "Player A" if player_a_score > player_b_score else "Player B" if player_b_score > player_a_score else "Tie"
    print(f"Game result: {winner} wins this round.")

# 统计最终胜者
final_scores = [player_a_score, player_b_score]
max_score_index = final_scores.index(max(final_scores))
if max_score_index == 0:
    final_winner = "Player A"
elif max_score_index == 1:
    final_winner = "Player B"
else:
    final_winner = "Tie"

print(f"\nAfter 50 games, the ultimate winner is: {final_winner}")