anyone know the odds

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

I believe it's about 50:1 or about 2%.

Actually I think it's more like 4.5%, or one out of about 22. I might have made a mistake, but I don't think so.

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.

How does the calculation go?

Steve

Thought odds of one suited board is over 2000/1

The probability of exactly 4 suited cards:

[(13 take 4)*(52-13)*(4 take 1)]/(52 take 5)

I think you need to multiply the numerator by 5 to account for the 5 different positions that those 39 differently-suited cards could be dealt to. Other than that, everything else looks spot on to me!

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)

You are right that the odds of a particular, one suited board (spades, for instance) is over 2000/1. But the odds of any suit is 4 times that, or one in 505 hands.

DOH!!!!!!!!! thought i was going crazy. only have to work out odds of 4 cos doesnt matter about 1st suit

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

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 non-spade 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

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

"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 N-1 x N-2 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.

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

It's been a long time since my probability classes as well, but technically, I believed I used permutations (52P5) instead of combinations (52C5). Permutations consider all the different orderings and combinations do not. For instance, 52P5 = 52! / (52-5)! where 52C5 = 52P5 / 5!. Also, the great thing about probability is that there are usually many ways to work it all out. Here's how I did it:

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 52-13 = 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!

wow....thank you all for such informative responses...makes me wanna dust off the ol textbooks!