If you are playing lazy pineapple (like hold em but with a third hole card, up to two hole cards play at showdown), and a flop comes all one suit, what are the odds a heads up opponent has that flush? Derivation would be appreciated. Thanks!

It would depend on your own cards, but if we're ignoring that then there are 49 cards left in the deck. We will count combinations of opponent hands which contain 1 or 0 of the key suit.

There are 39C3 combinations containing 0 cards of the key suit, and 10*39C2 combinations containing exactly 1 card of the key suit. There are 49C3 total 3 card hand combinations possible.

The probability that the opponent does NOT have a flush is (39C3+10*39C2)/(49C3) = 16549/18424

The probability that the opponent does have a flush is then, 1875/18424 =~10.2%

This does not account for the fact that people are more likely to be playing suited cards.

Direct approach is a little better for this one: [C(10,2)*39 + C(10,3)] / C(49,3) = same answer

thanks for quick replies!

Oops, for some reason I thought that would be more complicated and I guess I'm used to many similar problems being easier to solve the way I did it.

One thing that is important is that the people above who did this calculation did it on the assumption that they don't know your hole cards. In practice, you know what your own hole cards are. You could adjust the probability depending on whether or not you have 0,1,2 or 3 of the suit.

for completeness, would you be so kind as to solve when you have 0 and 1 of that suit?

This should be straightforward to mimic what heehaww did above when your holding is unknown.

Why don't you give it a try?


Hint: heehaww's answer in words is: the opponent needs to have two or three cards of the suit to have flopped a flush. Since there are 3 of the suit on board (the flop), that leaves 13-3=10 of the suit remaining.

C(10,2) is a way of saying the number of ways to choose 2 cards of 10 cards, so this is clearly the number of ways that opponent can have exactly two cards in the suit.

39, or C(39,1), is a way of saying the number of cards opponent can have for his third hole card. Of course, 13*3=39 is the number of cards in suits not appearing on the flop.

You multiply these together, C(10,2)*C(39,1), to give the total number of 3-card hands the opponent can have that contain exactly two of the suit when the flop is monotone.

Clearly, by extension, C(10,3) gives the number of 3-card hands that the opponent can have that contain exactly three of the suit when the flop is monotone.

C(49,3) is the total number of possible 3-card holdings that your opponent can hold given the flop contains three cards (of course, 52-3=49).

So the probability is simply [C(10,2)*C(39,1) + C(10,3)] / C(49,3).

If you don't know, and it's perfectly understandable if you don't, the C(x,y) notation is called the Choose function or Combinations. The formula for C(x,y) is given in a million places and is programmed in most modern calculators and computer languages.

C(10,2) = 45
C(10,3) = 120
C(39,1) = 39
C(49,3) = 18,424

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If you understand the above, deriving the respective probabilities for the cases in which you hold 0, 1, 2, or 3 of that suit should be straightforward.

Given your presumed holding (a number of cards in the suit in question and a number of cards in a different suit), think about how many suited cards the opponent's hand must have and how many cards the opponent is choosing from (suited and non-suited as the case may be).

Basically, you will find that many of the numbers in heehaww's formula (10,2,10,3,39,1,49,3) will change for each scenario of the make-up of your own hand. But it is quite straightforward and relies exclusively on simply counting how many suited cards and non-suited cards are possible in each case.

Give it a try if you want. You can post any progress (or lack of progress) here and everyone on this forum will be tremendously helpful and non-judgmental when it comes to helping others learn this stuff.

It's okay if you don't want to try, but you'll be surprised how easy it is. And I promise that you'll find many opportunities to utilize this type of thinking be it poker or other endeavors.