This is a discussion on anyone know the odds within the online poker forums, in the General Poker section; what are the odds that there will be four suited cards on the board? seems to happen quite a lot 


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anyone know the odds
what are the odds that there will be four suited cards on the board? seems to happen quite a lot

#4




I should clarify that I calculated the probability that the board will have at least 4 of a single suit. And I confirmed that my original figures are indeed correct. Interestingly, one out of every 500 or so hands would have a board consisting entirely of one suit.

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re: Poker & anyone know the odds
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For me, it came out like this (using your "take" terminology): (13 take 4)*(4 take 1)*(39 take 1)*(5 take 1) / (52 take 5) 
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Could someone explain how these calculations are arrived at? BTW, I am quite familiar with 'normal' probability, such as 52C4 = The number of combinations of 52 items taken 4 at a time  and that kind of thing. I'm fine with the maths but can't see how those formulae are derived.
Excuse me if I'm just being thick! :) Steve 
#12




re: Poker & anyone know the odds
I don't generally bother with combination terminology, since these calcs are usually pretty simple and easy to come up with in a spreadsheet...
First assume we're looking at a particular suit, say spades. We need exactly 4 spades, and exactly 1 of any other suit. So we have the following: (13/52) x (12/51) x (11/50) x (10/49) x (39/48) = .002146 But we need to remember that the nonspade can be in any of the 5 card positions, so we need to multiply .002146 x 5 = .010729 Now we must realize that each of the four suits is equally likely, so the probability of getting exactly four of any suit is .010729 x 4 = .042917, or 1 in 23.3 It's even easier to calculate the probability of all 5 being the same suit, as follows: (13/52) x (12/51) x (11/50) x (10/49) x (9/48) x 4 = .001981, or 1 in 504.85 
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Yep, I follow that (just.) Thank you, Sir!
I've not come across the 13 take 4 terminology before. I'm assuming that means 'The number of combinations of 13 cards taken 4 at a time?' or 'Any 4 from 13 in any order' Would that be right? Steve 
#14




"13 take 4" is the same as "13 combinations taken 4 at a time". The key is that order doesn't matter, since it doesn't matter where on the board the cards appear. This differentiates combinations from permutations, in which case order does matter. What we're really saying here is "Of these 13 cards, give me 4 of them in any order."
In mathematical terms, it equals (13!)/(9! x 4!), where "!" means "factorial". For any number N, N factorial, or N! is [N x N1 x N2 x ... x 3 x 2 x 1]. So back to our "13 take 4" we have (13x12x11x10x9x8x7x6x5x4x3x2x1)/[(9x8x7x6x5x4x3x2x1)x(4x3x2x1)]. Simplifying, you get (13x12x11x10)/(4x3x2). Hope this helps. 
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Very good walk through ck! It was actually nice to get a statistics refresher as I was trying to remember pieces of this. Very much appreciated 
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First, consider the total number of ways to select exactly four cards of any single suit. For a single suit, there are 13P4 ways to select and order 4 cards of that 13 card suit. Since there are 4 suits, we multiply that by 4 to consider all suits. So there are 4*13P4 ways to select and order 4 cards of any single suit. For the remaining single card, it can be 5213 = 39 different possible cards and that card can be placed into any of 5 positions. So there are 39*5 ways to select and order that remaining card. Therefore, there are 4*13P4*39*5 total ways to select and order the cards. Second, to arrive at a probability, we just divide that total by the total number of ways to select and order 5 cards from a total of 52, which is 52P5. So, the probability of a board with exactly 4 cards of a single suit is: 4*13P4*39*5 / 52P5 = 4*13*12*11*10*39*5 / 52*51*50*49*48 = 13384800 / 311875200 = 0.0429 = 4.29% Note that this is the same answer that others have gotten using different methods, so that is definitely a plus! BTW, I'm glad you asked the question because it forced me to refresh all that 52P5 and 52C5 notation for myself as well! 