有效的随机选择设置位的方法

我目前的爱好项目提供蒙特卡罗模拟纸牌游戏与法语牌组(52张牌，从2到Ace)。

为了尽可能快地模拟，我使用在某些位置将多个卡表示为位掩码。下面是一些(简化的)代码:

public struct Card
{
    public enum CardColor : byte { Diamonds = 0, Hearts = 1, Spades = 2, Clubs = 3 }
    public enum CardValue : byte { Two = 0, Three = 1, Four = 2, Five = 3, Six = 4, Seven = 5, Eight = 6, Nine = 7, Ten = 8, Jack = 9, Queen = 10, King = 11, Ace = 12 }
    public CardColor Color { get; private set; }
    public CardValue Value { get; private set; }
    // ID provides a unique value for each card, ranging from 0 to 51, from 2Diamonds to AceClubs
    public byte ID { get { return (byte)(((byte)this.Value * 4) + (byte)this.Color); } }
    // --- Constructors ---
    public Card(CardColor color, CardValue value)
    {
        this.Color = color;
        this.Value = value;
    }
    public Card(byte id)
    {
        this.Color = (CardColor)(id % 4);
        this.Value = (CardValue)((id - (byte)this.Color) / 4);
    }
}

保存多张卡作为bitmask的结构:

 public struct CardPool
 {
    private const ulong FULL_POOL = 4503599627370495;
    internal ulong Pool { get; private set; } // Holds all cards as set bit at Card.ID position
    public int Count()
    {
        ulong i = this.Pool;
        i = i - ((i >> 1) & 0x5555555555555555);
        i = (i & 0x3333333333333333) + ((i >> 2) & 0x3333333333333333);
        return (int)((((i + (i >> 4)) & 0xF0F0F0F0F0F0F0F) * 0x101010101010101) >> 56);
    }
    public CardPool Clone()
    {
        return new CardPool() { Pool = this.Pool };
    }
    public void Add(Card card)
    {
        Add(card.ID);
    }
    public void Add(byte cardID)
    {
        this.Pool = this.Pool | ((ulong)1 << cardID);
    }
    public void Remove(Card card)
    {
        Remove(card.ID);
    }
    public void Remove(byte cardID)
    {
        this.Pool = this.Pool & ~((ulong)1 << cardID);
    }
    public bool Contains(Card card)
    {
        ulong mask = ((ulong)1 << card.ID);
        return (this.Pool & mask) == mask;
    }
    // --- Constructor ---
    public CardPool(bool filled)
    {
        if (filled)
            this.Pool = FULL_POOL;
        else
            this.Pool = 0;
    }
}

我想从第二个结构体CardPool随机抽取一张或多张牌，但我无法想象如何在不迭代池中的单个比特的情况下做到这一点。有没有任何已知的算法来执行这个?如果没有，你有办法快速完成吗?

:该结构并不打算从中抽取所有的牌。它经常被克隆，克隆数组是不可能的。我很想用位运算来抽取一张或多张牌。

更新2 :我写了一个类，它将卡片作为列表保存，以便进行比较。

public class CardPoolClass
{
    private List<Card> Cards;
    public void Add(Card card)
    {
        this.Cards.Add(card);
    }
    public CardPoolClass Clone()
    {
        return new CardPoolClass(this.Cards);
    }
    public CardPoolClass()
    {
        this.Cards = new List<Card> { };
    }
    public CardPoolClass(List<Card> cards)
    {
        this.Cards = cards.ToList();
    }
}

全甲板100万克隆操作的比较:—struct: 17毫秒-等级:73 ms

承认:差别没有我想的那么大。但是考虑到我额外放弃了查找预先计算值的简单操作，这就产生了很大的不同。当然，用这个类随机抽一张牌会更快，但我必须计算一个索引来查找，这只是把问题转移到另一个点。

我重复我最初的问题:是否有一种已知的算法可以从整数值中选择一个随机集合位，或者有人有一个比迭代所有位更快完成此操作的想法?

关于在List或Array中使用类的讨论没有任何意义，这不是我的问题，如果我使用类会更好，我可以自己详细说明。

Update3，查找代码:

CODE DELETED:这可能会引起误解，因为它并不是指那些性能受到线程主题影响的段落。

declare cards as an array of indices
for each bit in Pool
    push its index into cards
shuffle cards
when a card needs to be drawn
    pop an index from cards
    look up the card with Card(byte id)

ulong v = this.Pool;
// ulong a = (v & ~0UL/3) + ((v >> 1) & ~0UL/3);
ulong a = v - ((v >> 1) & ~0UL/3);
// ulong b = (a & ~0UL/5) + ((a >> 2) & ~0UL/5);
ulong b = (a & ~0UL/5) + ((a >> 2) & ~0UL/5);
// ulong c = (b & ~0UL/0x11) + ((b >> 4) & ~0UL/0x11);
ulong c = (b + (b >> 4)) & ~0UL/0x11;
// ulong d = (c & ~0UL/0x101) + ((c >> 8) & ~0UL/0x101);
ulong d = (c + (c >> 8)) & ~0UL/0x101;
ulong t = (d >> 32) + (d >> 48);
int bitCount = ((c * (~0UL / 0xff)) >> 56);
ulong r = Randomizer.Next(1, bitCount+1);
ulong s = 64;
// if (r > t) {s -= 32; r -= t;}
s -= ((t - r) & 256) >> 3; r -= (t & ((t - r) >> 8));
t = (d >> (s - 16)) & 0xff;
// if (r > t) {s -= 16; r -= t;}
s -= ((t - r) & 256) >> 4; r -= (t & ((t - r) >> 8));
t = (c >> (s - 8)) & 0xf;
// if (r > t) {s -= 8; r -= t;}
s -= ((t - r) & 256) >> 5; r -= (t & ((t - r) >> 8));
t = (b >> (s - 4)) & 0x7;
// if (r > t) {s -= 4; r -= t;}
s -= ((t - r) & 256) >> 6; r -= (t & ((t - r) >> 8));
t = (a >> (s - 2)) & 0x3;
// if (r > t) {s -= 2; r -= t;}
s -= ((t - r) & 256) >> 7; r -= (t & ((t - r) >> 8));
t = (v >> (s - 1)) & 0x1;
// if (r > t) s--;
s -= ((t - r) & 256) >> 8;
s--; // s is now the position of a random set bit in v