在 Dart 中模拟二十一点游戏[关闭]

Question

我尝试在DartPad 中编写一个程序来模拟一系列二十一点游戏，其中玩家遵循与庄家完全相同的策略，即击球直到分数超过16。现在根据文章here 和here房子应该比玩家至少有 5% 的优势。我如何解释这个优势？还有为什么完全像庄家一样玩会使赔率相等，有利于玩家。这是前面提到的程序：-

/*
 * PROGRAM SIMULATIING BLACKJACK
 * 
 * PLAYER KEEPS HITTING UNTIL THE SCORE EXEEDS 16
 * 
 * RESULTS:
 * 
 * PLAYER WINS ~ 43.8% OF TIMES
 * MATCH IS DRAW ~ 12.4% OF TIMES
 * DEALER WINS ~ 43.8% OF TIMES
 * 
 * PROGRAM DOESN'T TAKE INTO ACCOUNT THE CARDS ALREADY DISTRIBUTED
 * I.E. DISTRIBUTES CARDS FROM A FULL DECK OF CARDS TO BOTH THE 
 * PLAYER AND THE DEALER.
 * 
 * */

import 'dart:math';

class Player {
  int score;
  bool hardHand, busted;

  Player() {
    this.score = 0;
    this.hardHand = false;
    this.busted = false;
  }

  void hit() {
    int randomNumber = Random().nextInt(13) + 1;

    if (randomNumber > 10) {
      this.score += 10;
    } else if (randomNumber == 1) {
      this.score += 11;
      this.hardHand = true;
    } else {
      this.score += randomNumber;
    }

    if (this.score > 21) {
      if (this.hardHand) {
        this.hardHand = false;
        this.score -= 10;
      }
    }

    if (this.score > 21) this.busted = true;
  }
}

void main() {
  int turns = 100000;
  int wins = 0, loses = 0, draws = 0;

  for (int i = 0; i < turns; i++) {
    Player player1 = new Player();
    Player dealer = new Player();

    while (player1.score < 17) {
      player1.hit();
    }

    while (dealer.score < 17) {
      dealer.hit();
    }

    //print("score: ${player1.score} busted: ${player1.busted}");

    if (player1.score > dealer.score) {
      wins++;
    } else if (player1.score == dealer.score) {
      draws++;
    } else {
      loses++;
    }
  }

  double winPercent = (wins / (wins + loses + draws)) * 100;
  double drawPercent = (draws / (wins + loses + draws)) * 100;
  double lossPercent = 100 - winPercent - drawPercent;

  print("WIN Percentage: $winPercent");
  print("DRAW Percentage: $drawPercent");
  print("LOSS Percentage: $lossPercent");
}

Answer 1

当我在提供的链接中阅读它们时，您的代码似乎偏离了二十一点规则。

1. 如果你破产了，你就输了。庄家在那之后不需要做任何事情，即使他们也破产了，你也输了。所以，你需要规则以if (player1.busted) { loses++; } else ...开头。

2. 如果你没有爆牌，但庄家爆牌，你就赢了。所以if (dealer.busted) { wins++; } else ...。

3. 你算平局，但得分相同意味着庄家获胜，所以if (dealer.score >= player1.score) { loses++; } else { wins++; }。

4. 最后，你算“硬牌”，但错过了有两个 A 的可能性，所以我可能会数个 A 而不是只有一个布尔值。这实际上不太重要。

解决这个问题，我得到类似的东西：

import 'dart:math';

class Player {
  int score = 0;
  int acesHigh = 0;

  void hit() {
    int randomNumber = Random().nextInt(13) + 1;

    if (randomNumber > 10) {
      score += 10;
    } else if (randomNumber == 1) {
      score += 11;
      acesHigh++;
    } else {
      score += randomNumber;
    }

    if (score > 21) {
      if (acesHigh > 0) {
        acesHigh--;
        score -= 10;
      }
    }
  }
  
  bool get busted => score > 21;
}

void main() {
  int turns = 100000;
  int wins = 0, losses = 0;

  for (int i = 0; i < turns; i++) {
    Player player1 = new Player();
    while (player1.score < 17) {
      player1.hit();
    }
    if (player1.busted) {
      losses++;
      continue;
    }

    Player dealer = new Player();
    while (dealer.score < 17) {
      dealer.hit();
    }
    if (dealer.busted) {
      wins++;
      continue;
    }
    
    // print("score: ${player1.score} busted: ${player1.busted}");

    if (player1.score > dealer.score) {
      wins++;
    } else {
      losses++;
    }
  }

  assert(wins + losses == turns);
  double winPercent = (wins / turns) * 100;
  double lossPercent = 100 - winPercent;

  print("WIN Percentage: $winPercent");
  print("LOSS Percentage: $lossPercent");
}

这通常给我大约 40.8% 的胜利和 59.2% 的失败。

房子总是赢。