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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 preflop 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. 




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 




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, haha). 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 preflop 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.../ch04not.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! / ( (nr)! * 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! / ( (522)! * 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 preflop 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 preflop are 1 in 663. Q3) What are the odds of 3 players being dealt a pocket pair preflop 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 526 =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! / ( (462)! * 2!) = 1035 Therefore: I arrived at 663*1035 = 686,205. The odds of three pairs preflop, 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. 




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





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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 preflop. Then based on the flop, calculate the odds of this hand occurring. 




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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. 




This also rases another question about raising with middle order pairs!



