So last night I'm playing Texas No Limit Hold 'Em (cash game, not tournament). I have pocket Kings. Flop comes K - 8 - 4. There are three of us in the hand. Well, guess what, one of the other guys had pocket 8's, and the other guy had pocket 4's. In other words, all three of us flopped three of a kind!! What are the odds of this happening? First time I've ever seen it happen. And yes, I won a monster pot.

The formulas on that page show how to calculate probability, which is what the equation is, a probability. Here's the setup: Suit is moot (it does not matter whether they're spades, hearts, etc.). There are 13 different numbers in the deck each occurring 4 times (= 52 cards). Two cards are dealt to each of three persons. Then three "common cards" are dealt such that each of the three persons is playing on the "common cards." (correct me if I'm wrong; I'm not a poker player) Question: What is the probablity that: A) All three persons will be dealt two idential numbers (pairs), AND B) Of the three common cards, one each will match each of the three pairs. Note: In the whole dealing of nine cards ((3x2)+3) only three different numbers appeared. That means ten numbers never appeared.

Hey, I'll take a stab, let me know if my answer makes sense. I am getting approximately 1 in 9.38 million. I did this in three steps. First is the probability you each get a different first pocket card, second is that you all match your pocket card, and third is that the flop matches all three of your pairs. The probability you all get different first pocket cards is (52/52)(48/51)(44/50) = 352/425 or an 83% chance. The probability you all match your pockets is (3/49)(3/48)(3/47) = 9/36,848 or a .024% chance. Combining this with the first, this gives about a 1/4900 chance that three players will be dealt three different pocket pairs. The probability the flop matches all three pairs (in any order) is (6/46)(4/45)(2/44) = 1/3,795 or a .052% chance. Multiply these together and we get 12/112,559,125 or 1 in 9,379,927.0833.

This is fun. I'm working some numbers and I'll get you a reply sometime throughout the night. I'll be up most of the night trading. Talk to ya later.

For the first card to be unique (when there are no other cards having been dealt) the probability is 1:1=1. We can leave this part of the equation out (anything times 1=1). For the second card to be unique relative to the first card, the probability is [(12/13)(51/52)]= 0.905325444 (about a 90% chance) because there are 51 cards left and we are looking for 1 of 12 possible numbers. The reason I did 51/52 is because there are 3 of one card left in the deck while there are 4 of 12 other cards left in the deck. For the third card to be unique relative to the first two cards, the probability is [(11/13)(50/52)]= 0.813609467(.905325444)= 0.736581352 because there are 50 cards left and we are looking for one of 11 possible numbers. Now each person has one card. The above says that the odds of dealing three unique cards irrespective of suit is about 73%. Sounds right to me. For the fourth card to match the first card (not just any already dealt card but the specific card which was dealt to the first person), the probability is [(3/49=0.061224)/3]. Our running total is now (0.061224 x 0.736581352)/3=0.015032 or about 1.5%. For the fifth card to match the second card, the probability is (3/48)/2 or .0625/2=0.03125 which means our running total is now 0.03125 x 0.015032 = 0.00047. For the sixth card to match the third card, the probability is (3/47)/1 or 0.063829787/1 = 0.063829787 which means our running total is now 0.063829787 x 0.00047 = 0.0000299846. Now each person has two cards. For the first âupâ card (not a poker player and not sure if I named it properly, but you hopefully you know what I mean) to match any one pair, the probability is 2/46=0.043478 which puts our running total at 0.043478 x 0.0000299846 = 0.0000013037 For the second âupâ card to match either other pair, the probability is 2/45= 0.044444444 which puts our running total at 0.044444444 x 0.0000013037 = 0.000000057941229 For the third âupâ card to match the final pair, the probability is 2/44=0.045454545 which puts our total at 0.045454545 x 0.000000057941229 = 0.000000002633692 or about a 1: 379,695,086 probability that any one hand will have this result (which is 1/.000000002633692) That was fun and quicker then I thought. Iâm sure many people will argue the syntax and formulas as is common in these type equations. I would love to hear arguments from others on what they think is proper math.

Agreed I don't understand this. There are 51 cards left in the deck at this point, 48 of which are a different number than the first card, therefore the probability of a unique card is 48/51. Same here, 50 cards left, 44 of which are of a different numbers than the first two since we have eliminated two numbers with 4 cards each (52-8=44). So a unique card to the third person is 44/50. Sounds reasonable but I don't agree. 87% is reasonable too. Agreed. 3/49=.061124. Why divide by two? There are 3 cards that could match Player Three's card, there are 48 cards left, so 3/48. Still going with previous logic and 3/47 here. This is to only match the first person's pair. There are 6 cards that could come up here to match any one of the pairs so 6/46. Assuming a pair was matched on the first card of the flop, there are now 4 possible to make a pair, so 4/45. Now only 2 left, so 2/44. Still standing by my answer.

Think of it this way... the cards are numbered 1 through 13 and there are no suits (hearts, spades, etc.) After the first card is dealt and we are looking for the next unique card, there are 4x12 + 3x1 cards left. We are looking for a card from the 4x12 but the 3x1 must be considered as possible cards to be dealt. make sense?