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What are the odds of this happening?

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  • What are the odds of this happening?

    click to enlarge image


    I was dealt pocket 10’ in middle pos. The table folded to me, so I raised, I had about 9G in chips and early stage of tourney. Blinds are 20/40. Armanisergio 2nd spot behind the button pushed all in. I took a chance thinking coin flip against 2 live over cards. He had (edit he had AA not A4) A4 and flopped an A. I lose a proportion of my stack. I was prepared to gamble here, if I lose the hand at least I’m not out of the tourney.

    Now “Armanisergio” has doubled up.

    Bang! I am dealt pocket 10’s again, back to back. I rasied again to about 320, “Macmarttigan” shoves all and “Armanisergio” pushes over the top all in. Again I have a decision to make with my 10’s. I figured this time that Armanisergio had a decent hand like J’s or Q’s. I certainly did not put him on AA again.

    If you zoom into the screen shot, you will see that I asked, “Hmmm are my 10’s good again”? I decided not to call him this time.

    I have some questions that I would like to ask.

    Q1) What are the odds of a player being dealt a pocket pair on a 9 handed table in a tournament?

    Q2) What are the odds of 3 players being dealt a pocket pair on a 9 handed table in a tournament?

    Q3) What are the odds of 3 players being dealt a pocket pair pre-flop and each player making a set (3 of a kind) on the flop?

    Q4) What are the odds of the middle pair (in this case 10’s), hitting quads?

    I would like to know statistically what the odds of this type of hand occurring. I would deeply appreciate some feedback.

    Thank you.
    Last edited by Aus-Redda; Sat Apr 28, 2012, 09:56 AM.

  • #2
    Smart ass answer #1: It happened so the probability is 100%.

    Smart ass answer #2: It happened so get over it.

    Actual answers:
    Q1 – 1.4:1
    Q2 – 14:1
    Q3 – 350:1
    Q4 – Flop 400:1; Turn 100:1; River 18:1
    Last edited by TrumpinJoe; Sun Apr 29, 2012, 07:09 PM. Reason: correction

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    • #3
      TrumpinJoe, can you tell me how you arrived at those odds?
      Bracelet Winner

      Comment


      • #4
        his odds are way out lol!

        Originally posted by Ovalman View Post
        TrumpinJoe, can you tell me how you arrived at those odds?
        Thank you all for replying to this post. I did some revision on my mathematics (I love mathematics….I just hate thinking, ha-ha). I think that I may have figured it out. I will try to answer my own questions. If I am incorrect, please feel free to comment.

        Q1) What are the odds of a player being dealt a pocket pair on a 9 handed table in a tournament?
        Q2) What are the odds of 3 players being dealt a pocket pair on a 9 handed table in a tournament?
        Q3) What are the odds of 3 players being dealt a pocket pair pre-flop and each player making a set (3 of a kind) on the flop?
        Q4) What are the odds of the middle pair (in this case 10’s), hitting quads?

        Read more: What are the odds of this happening? - PokerSchoolOnline Forum http://www.pokerschoolonline.com/for...#ixzz1ti7ESWX0

        Assuming the following is true and correct.
        Courtesy (http://people.richland.edu/james/lec.../ch04-not.html)
        (The only difference in the definition of a permutation and a combination is whether order is important.)
        A combination of n objects, arranged in groups of size r, without repetition, and order being important is:
        nCr = C(n,r) = n! / ( (n-r)! * r! )


        Answers to questions:
        (I apologize for the confusion relating to number of players seated and table size, mathematically, this is irrelevant, I was referring to specific outcomes).

        Q1) What are the odds of a player being dealt a pocket pair?

        Answer:

        52! / ( (52-2)! * 2!) = 1326

        There are 1326 ways of being dealt 2 cards out of a field of 52 cards. There are 6 ways (considering suits where c=clubs, h=hearts, d=diamonds and s=spades) that a player can be dealt a pair out of those 1326 combinations.
        Example: AA can be dealt as: Ac Ah or Ac Ad or Ac As or Ah Ad or Ah As or Ad As.

        Therefore 1326/6 = 221. The odds of being dealt any pair pre-flop are 1 in 221.

        Q2) What are the odds of 3 players being dealt a pocket pair?

        Answer:
        To answer this question, I have determined that each pair is not related to any other pair.

        Therefore, I arrived at 3*221 = 663. The odds of 3 players each being dealt a different pocket pair pre-flop are 1 in 663.

        Q3) What are the odds of 3 players being dealt a pocket pair pre-flop and each player making a set (3 of a kind) on the flop?

        Answer:
        It gets a bit tricky here and I hope my mathematics will hold up.

        Mathematically 52-6 =46
        Now we have to calculate (using the nCr formula) combinations of 1 of 2 cards being dealt from the remaining field of 46 cards and then multiply that by 3 in order for each player to hit 1 out of 2 cards to make a set (3 of a kind).

        46! / ( (46-2)! * 2!) = 1035

        Therefore: I arrived at 663*1035 = 686,205.

        The odds of three pairs pre-flop, each seeing the flop and flopping a set are 1 in 686,205


        Q4) What are the odds of the middle pair (in this case 10’s), hitting quads?

        There are 43 cards left. There is only 1 (10) remaining in the deck.

        The odds that trip 10’s makes quads = 686,205 * 43 = (29506815)

        1 in 29,506,815

        Please feel free to comment on this or correct my mathematics.

        Comment


        • #5
          Originally posted by Aus-Redda View Post
          Answers to questions:
          (I apologize for the confusion relating to number of players seated and table size, mathematically, this is irrelevant, I was referring to specific outcomes).

          Q1) What are the odds of a player being dealt a pocket pair?

          Answer:

          52! / ( (52-2)! * 2!) = 1326

          There are 1326 ways of being dealt 2 cards out of a field of 52 cards. There are 6 ways (considering suits where c=clubs, h=hearts, d=diamonds and s=spades) that a player can be dealt a pair out of those 1326 combinations.
          Example: AA can be dealt as: Ac Ah or Ac Ad or Ac As or Ah Ad or Ah As or Ad As.

          Therefore 1326/6 = 221. The odds of being dealt any pair pre-flop are 1 in 221.
          Those are the odds of being dealt a SPECIFIC pocket pair pre flop. You have 13 potential pp you can be dealt to fulfill your criteria (what are the chances of a player being dealt [ANY] pp pre flop).

          You must then adjust all your numbers after this...
          Double Bracelet Winner

          Comment


          • #6
            Originally posted by JDean View Post
            Those are the odds of being dealt a SPECIFIC pocket pair pre flop. You have 13 potential pp you can be dealt to fulfill your criteria (what are the chances of a player being dealt [ANY] pp pre flop).

            You must then adjust all your numbers after this...
            I see your point and it is valid.
            I appreciate the correction.

            The odds I stated were based on being dealt a specific pair.

            The odds on being dealt a random pair are 1 in 17.

            I will try to recalculate this on the odds of 3 players being dealt a random pair pre-flop. Then based on the flop, calculate the odds of this hand occurring.
            Last edited by Aus-Redda; Thu Sep 20, 2012, 05:05 PM.

            Comment


            • #7
              Originally posted by JDean View Post
              Those are the odds of being dealt a SPECIFIC pocket pair pre flop. You have 13 potential pp you can be dealt to fulfill your criteria (what are the chances of a player being dealt [ANY] pp pre flop).

              You must then adjust all your numbers after this...
              I appreciate your comment and I hope other readers of this post will appreciate the difference.

              example 1) If I sat down at a table and said....come on dealer how about AA this time the odds are 1 in 221 that I will get that hand. The reason for this is, the player is asking for a specific outcome.

              As you correctly pointed out. there are numereous ways a player can be dealt "ANY PAIR".

              Therefor doing the maths....the odds of being dealt a random pair are now reduced to 1 in 17. this can be explained as follows.

              The odds of being dealt a specific pair 221 divided by the number of ranks 13 that could make a pair. 221/13 = 17.

              I hope this makes sense.
              Last edited by Aus-Redda; Thu Sep 20, 2012, 04:48 PM.

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              • #8
                This also rases another question about raising with middle order pairs!
                Last edited by Aus-Redda; Thu Sep 20, 2012, 04:57 PM.

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