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August 5th, 2018, 09:29 PM  #1 
Newbie Joined: Aug 2018 From: USA Posts: 1 Thanks: 0  Probability Applied To Poker
I apply probability theory to Texas Hold Em Poker for fun, but I found a problem that has me stumped. I'm not sure exactly what is the correct way to solve this: In a game of Texas Hold Em Poker, you know that you've won...unless one (or more) of your opponents were dealt one (or more) of the 9 specific cards that can beat you. Of the 52 cards in a deck, only 45 remain unseen (2 were dealt to you faceup, 5 were dealt to the board faceup, 6 were dealt unseen to your opponents). None of the faceup cards are the 9 cards that can beat you. What is the probability that 1 (or more) of the 6 opponent's cards are 1 (or more) of the 9 specific cards needed to beat you? This is either a really easy problem that I'm overthinking, or one that is beyond my current understanding. Any guidance would be appreciated. Thanks! 
August 5th, 2018, 11:32 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,302 Thanks: 1974 
Calculate the probability that the 39 cards remaining in the deck include all of the 9 specific cards, then subtract that probability from 1.


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