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 10-12-2017, 02:42 PM #1 Geraint stranger   Join Date: Nov 2015 Posts: 12 combinatorics - Beginners math question Hi. So using the combinatorics formula; C = n! / x! * (n-x)! Can any1 show me how to calculate the number of combinations for a specific flop (AAK).
 10-12-2017, 02:57 PM #2 statmanhal Pooh-Bah   Join Date: Jan 2009 Posts: 4,047 Re: combinatorics - Beginners math question Sure. In general, if there are n ways for event 1 to happen and n2 ways for event 2 to happen, then there are n1 x n2 ways for both to happen. There are 4 aces and 4 kings. Your want the flop to have 2 of the 4 aces (event 1) and 1 of the four kings (event 2). Flop = AAK: 2 aces: C(4,2). ---- 1 king: C(4,1) Combos: C(4,2) x C(4,1) = 6 x 4 = 24 The above assumes nothing about player holdings. You divide this answer by the total number of flops C(52,3) to get the probability of an AAK flop, again assuming nothing about player holdings.
 10-12-2017, 03:16 PM #3 Geraint stranger   Join Date: Nov 2015 Posts: 12 Re: combinatorics - Beginners math question Thanks. I think I've got it. So basically split each card into an event. if each card if different, then it's 3 different events if a card is paired, then its 2 events if the flop is 3 of a kind, then it's 1 event
 10-12-2017, 04:28 PM #4 statmanhal Pooh-Bah   Join Date: Jan 2009 Posts: 4,047 Re: combinatorics - Beginners math question Yes, for a flop, that is correct in the context of your question. If the question was how many combos are there of flops of 3 aces OR 3 kings, the answer would be C(4,3) + C(4,3). Here we are using the fact the number of combos for 2 events that can't simultaneously occur is Combos(event 1) + Combos(event 2). If the events can occur simultaneously, you have to subtract out the number of combos that can be simultaneous. Example: How many combos are there of aces or hearts from a 52 card deck? Combos = 4 + 13 – 1. You should be able to figure out the -1 part, right? OP how to "calculate the number of AAK combinations if you're already holding A2, or AK? " Have you worked this out ok? Last edited by statmanhal; 10-12-2017 at 04:33 PM.
 10-12-2017, 06:15 PM #5 Geraint stranger   Join Date: Nov 2015 Posts: 12 Re: combinatorics - Beginners math question OP how to "calculate the number of AAK combinations if you're already holding A2, or AK? " Have you worked this out ok? I think so.. AAK hoding A2 Event 1 (AA) = C(3,2) = 3 Event 2 (2) = C(4,1) = 4 3*4 = 12 AAK holding AK Event 1 (AA) = C(3,2) = 3 Event 2 (K) = C(3,1) = 3 3*3 = 9 I am however struggling with how many rainbow, two suit, mono flops of AKQ when holding AA. Total (3 events) Event 1 (A) = C (2,1) = 2 Event 2 (K) = C (4,1) = 4 Event 3 (Q) = C (4,1) = 4 4*4*2 = 32 Rainbow Event 1 (A) = C (2,1) = 2 Event 2 (K, thats not the same suit as event 1) = C (3,1) = 3 Event 3 (Q, thats not the same suit as E1 or E2) = C (2,1) =2 2*3*2 = 12 2 suite Event 1 (A) = C (2,1) = 2 Event 2 (K, thats not the same suit as event 1) = C (3,1) = 3 Event 3 (Q, thats is the same suit as E1 or E2) = C (2,1) =2 2*3*2 = 12 Mono Event 1 (A) = C (2,1) = 2 Event 2 (K, thats is the same suit as event 1) = C (1,1) = 1 Event 3 (Q, thats is the same suit as E1/E2) = C (1,1) =1 2*1*1 = 2 I've done somthing wrong here. 12+12+2 does not equal my initial total of 32
 10-13-2017, 12:41 AM #6 statmanhal Pooh-Bah   Join Date: Jan 2009 Posts: 4,047 Re: combinatorics - Beginners math question 2 suit Q is odd suit Event 1 (A) = C (2,1) = 2 Event 2 (K, that is the same suit as event 1) = 1 Event 3 (Q, that is the not same suit as E1 or E2) =3 OR Q is not odd suit Event 1 (A) = C (2,1) = 2 Event 2 (K, thats not the same suit as event 1) = 3 Event 3 (Q, that’s is the same suit as E1 or E2) = C (2,1) =2 3*1*2 + 2 *3 *2 = 18 These kind of calcs can often be tricky. Common mistakes are leaving out combos or counting some twice. With practice it will become easier but not necessarily easy. There is usually more than 1 way to do these types of problems so a good check is doing them at least 2 ways.
 10-13-2017, 06:45 AM #7 Geraint stranger   Join Date: Nov 2015 Posts: 12 Re: combinatorics - Beginners math question Thanks. It does seem tricky, but i'll keep playing with it.

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