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Flush in 7-card stud

  1. Dec 25, 2011 #1
    I'm trying to approximate the probability of drawing a flush in texas hold-em. The attached image is the expression I think should work. However, it's inconsistent with the odds given for 7-card stud, given here: http://wizardofodds.com/games/poker/

    I think the probabilities should be the same. Does anyone know what I did wrong?

    The link to the expression in wolfram alpha is as follows: http://www.wolframalpha.com/input/?...)-4)+(13+choose+7)(4+choose+1))/(52+choose+7)

    Thank you!
     

    Attached Files:

  2. jcsd
  3. Dec 26, 2011 #2
    Maybe I'm missing something, but why should the odds/probabilities be the same? They're two different games. There are no community cards in 7 card stud.
     
  4. Dec 26, 2011 #3
    My thinking was that either way the probability is defined by the number of combinations of 7 cards that include 5 cards of the same suit over total possible combinations of 52 cards.
     
  5. Dec 26, 2011 #4
    Yes, but in stud games each player has their own face up cards. In Texas Hold 'em there's community face up cards, and the number of face up cards is independent of the number of players. This affects the number of cards dealt and how a flush or any hand is put together.
     
    Last edited: Dec 26, 2011
  6. Dec 26, 2011 #5
    Sorry, I don't follow your last sentence. Cards dealt to other players are private and don't affect the player's probability of getting any particular hand. So the absolute odds of one player getting any particular hand are the same regardless of the number of players. Please correct me if I'm wrong.
     
  7. Dec 26, 2011 #6
    So you're saying that number of cards dealt from a deck don't affect the probabilities? One reason that Texas Hold'em is popular is that it is easier to get hand like a flush. Given 10 players, the number of cards of a given suit that are out will likely be greater in stud games than in Hold'em. Besides you've already said your theory doesn't agree with your sources.
     
    Last edited: Dec 26, 2011
  8. Dec 26, 2011 #7
    Can you frame what you're saying in terms of a probability quotient? What I'm saying is that the probability of a single player's holdem hand being a flush is equal to the number of combinations of 7 cards chosen from 52 that have 5 of the same suit, over 52c7. What part of this would you change and why? Thank you.
     
  9. Dec 26, 2011 #8

    D H

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    We've gone a bit off-topic from the original question.

    The basic odds for seven card stud and Texas Hold'em are one and the same. However, it is important to note that these basic odds are for a game that is neither seven card stud nor Texas Hold'em. These are instead the odds for a one player game in which a person is dealt a hand of seven cards (all at once) from a standard 52 card deck.

    As far as the original question is concerned, here is how I would calculate the odds for a flush: You can get a flush as
    • Five cards from suit X plus 2 cards from other suits,
    • Six cards from suit X plus one other card from another suit, or
    • All seven cards from suit X.

    One way to get case (a) is to be dealt some card, then dealt four cards in a row of the same suit as the first card, and then two final cards of a different suit. The probability of this outcome is 1*(12/51)*(11/50)*(10/49)*(9/48)*(39/47)*(48/46). Call this particular hand "XXXXYY". There are 21 ways to arrange the string "XXXXYY". Those other 20 arrangements will have the same probability as that for XXXXYY, so altogether, the probability of case (a) is 21*1*(12/51)*(11/50)*(10/49)*(9/48)*(39/47)*(38/46). Cases (b) and (c) are treated similarly, with probabilities 7*1*(12/51)*(11/50)*(10/49)*(9/48)*(8/47)*(39/46) and 1*1*(12/51)*(11/50)*(10/49)*(9/48)*(8/47)*(7/46). Add the probabilities for these three mutually exclusive cases together and you get (12*11*10*9)/(51*50*49*48)*(21*39*38+7*8*39+8*7)/(47*46), or 4089228/133784560.

    Note that this is not the same as the odds given in the site referenced in the OP. It gives a probability of 4047644/133784560. The discrepancy is easily explained: My calculation includes all flushes, including straight flushes. 41584 of those flush hands are straight flushes (counting a royal flush as just the best possible straight flush. Those straight flushes are much better than run of the mill flushes and should not be counted as such.
     
  10. Dec 26, 2011 #9
    Thanks, DH, that is very helpful.
     
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