每個撲克組合贏得攤牌的概率是多少?


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我在娛樂場上玩類似撲克的遊戲,該遊戲為桌上的每隻手提供動態變化的係數。

每個回合(盲注,翻牌圈,翻牌圈,翻牌圈)的賭徒都可以押注,每手牌或組合之一都會贏。例如,我認為這場比賽將由2號手贏得,我為此投入了賭注。賭場提供了該賭注的係數(例如,3.5),並且如果秒針實際獲勝,我將得到我的賭注乘以利潤係數。

組合也是如此:您打賭這種組合將是獲勝的組合。例如,當我押注兩對獲勝遊戲時,我押注任何兩對獲勝,但只能押兩對。如果碰巧是直牌在桌上,我會下注。

因此,有趣的部分是:您甚至可以盲目下注。這意味著沒有可用的信息:只是賭博。但。係數關閉。當然,賭場從每筆賭注中扣除利潤,但是有些賭注比其他賭注更有價值。他們為組合提供了這樣的係數:

High card wins: 100
Pair wins     : 5.80
Two pairs win : 3.10
Set wins      : 6.80
Straight wins : 5.70
Flush wins    : 8.70 
Fullhouse wins: 8.70
Quads win     : 80
Straight flush: 100
Royal flush   : 100

有趣的是,在河上看到高牌與看到皇家同花順是相當的。當然,一個孩子會知道,比起使用相同係數的皇家同花順,您最好選擇同花順100:1。但是,看到真實的計算結果會很有趣。我一直在考慮,但是什麼都沒想到,除了殘酷地打擊每個可能的六手游戲的每張桌子,但C(52, 12) * C(40, 5) 聽起來太殘酷

有什麼主意嗎?

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So, I did bruteforce the thing, but not the whole thing. I took 200000 random samples (a sample contains a table and 6 2-card hands) and found a maximum combination among each of them. Turns out, some coefficients are, in fact, better than others (and some of them marginally), yet none of them are good enough considering the expected value.

1709.4017094     
6.24024961    
3.26679952    
7.1857148     
6.02119461
9.2042892     
9.17010546  
116.75423234  
645.16129032           
nan

The nan thing means, that among 200000 samples the program never met any royal flush. So, speaking of the expected values:

HC: -0.94149994 
OP: -0.06451613 
TP: -0.06060606 
SE: -0.05555556 
ST: -0.05       
FL: -0.05434783
FH: -0.05434783 
QU: -0.31506849 
SF: -0.8450093 
RF:  nan

Which means, that betting on One Pair will lose you 6.45% of your money on average. The least money-losing bet turns out to be Straight and anything below One Pair and above Full-House is a dead bet.