我如何使用組合數學在梭哈遊戲的前三張牌中計算至少兩個黑桃


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我通常對數學很好,但是概率計算似乎讓我很吃力。

我正在嘗試制定一個公式來確定擊中至少2張黑桃的機率,因為該玩家已經獲得了前三張牌。

概率約為 15.05%。我已經通過運行5百萬張模擬3張牌並比較2或3黑桃命中了多少次來證實了這一點。這與此處的在線計算器相符:https://www.ohrt.com/odds/index.php?t[]=52&d[]=3&o[]=13&c[]=2&&s=any&p=9

根據我的推理,得出概率的公式應為:

combin(13,2)*combin(50,1)/combin(52,3)

或"從13個黑桃中,取2張,從剩餘的牌中取1張,然後除以52張卡組中的3張卡組合的總數"。

但這給我的結果是17.65%。

我正在嘗試為任何數量的出局或發牌(類似於在線計算器)開發一個通用公式,但這是我堅持了一段時間的測試用例。

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I got about 15.06%.

We want to sum the probabilities of drawing three spades in three cards, plus drawing two spades in three cards. The reason why we need to treat the cases separately is because while there is only one arrangement of suits when you draw three spades, there are three arrangements of suits when you only draw two spades.

As such, we have (13/52)(12/51)(11/50)+3(13/52)(12/51)(39/50)=15.0588%.


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Ok, Andrew Chin's answer was not exactly what I was asking for, he did make me rethink my approach, and I was able to come up with a general solution that works for my example and others.

if:

d = number of cards dealt

r = number of required outs

u = number of unknown cards remaining

t = number of outs

then the general formula to compute the probability of getting at least the number of required outs when 'd' cards are dealt is (note the expressions in parenthesis are using combination notation, not fractions. the combin function in excel will evaluate that):

general probability for  # of outs.

For the example case in my question, plugging in the numbers:

probability of at least 2 spades dealt in 3 cards

When yields 15.058% when evaluated. The top part of the fraction sums up all possible 3 card hands with 2 or 3 spades, and the denominator expression are the number of possible 3 card hands in a 52 card deck.