Heyho,

I got a question regarding probabilities in No Limit Hold'em:

I want to know/calculate the probabilities of the following scenarios:

- Royal Flush while using both hole cards
- Straight Flush while using both hole cards
- Quads while using both hole cards

For the flushes we would have to hit with both our hole cards, there is no other option.
For the Quads however something like AQ on QQQ84 would win but Q7 would not.

What are the odds for quads, straight flushes and the royal flush in this scenario?

I would be really grateful if someone could help me out here. I have tried to reseach this scenario/the math to calculate it; but so far I had no luck.

Thank you and best regards

You may want to post your question in the Probability Forum.

Hmmmm. On my laptop, this is the probability forum

I'll just do the royal flush.

There are 4*C(5,2) ways to select two possible RF cards dealt to hero.
Given that, there are C(3,3)*C*(47,2) possible boards for a RF

Prob = 4*C(5,2)/C(52,2) *C(3,3)*C(47,2)/C(50,5) = 0.00154%

There are other ways to get this answer.

"For the Quads however something like AQ on QQQ84 would win but Q7 would not."

This statement about Q7 on QQQ84 not winning quads is troubling to me especially since you are restricting the case to a player having a pair.

I think he's probably asking in relation to a BBJ which requires both of your hole cards to play. In the Q7 case, the 7 does not play but in the AQ case it does.

What I mean by that statement is:
You HAVE to use both your hole cards to make the best 5-card-combo out of all 7 cards. So in the example Q9 would be a winning hand, because all 4 Queens get used + the highest possible kicker.

Thank you very much for helping me out with the RF.

How and where can I calculate such a formula?
And what does the C(X,X) part mean?

Yes both cards have to play. But its not for a Bad Beat Jackpot. I was just wondering how the probabilities change in such a case. And I would like to be able to calculate these things on my own in the future.

What if your kicker is also the highest board card? E.g. we have AK and the board is AAAKK.

in these cases, it will count.

I'm getting 0.0831% for the quads with hole cards both playing, which doesn't seem too unbelievable given that it's 0.168% for quads in general.

First of all, a word on the C(n,r) thing used earlier. This reads "n choose r". It is the number of ways of choosing r objects from a total collection of n. The obvious example in hold'em is C(52,2); this is the number of possible combos of hole cards you can have. You can just type 52 choose 2 in google and it will give you the answer, but it wouldn't hurt to look up how this is calculated as well.

Now, we can split into the case where we have a pocket pair, and the case where we don't. First of all note that there are 6*13 = 78 pocket pairs possible. Let's just say we have some AA. How many boards are there (ignoring order) containing the other two aces? Well, when the board has two aces on it, there are 48 cards left in the deck and three places for them to go, so the answer is C(48,3).

Thus the total number of combos here is 78*C(48,3).

What about when we don't have a pocket pair? Well let's pick a card, say A , and decide that that's our quad card. So we have A x and the board is AAAxx . Let's run down the kickers and see how many possible combos there are where the kicker plays.

First of all, if we have a king, we don't care about the board; we can always play AAAAK as the best hand. Thus, given that we have a king, there are C(47,2) possibilities, as there are 47 cards left in the deck and 2 spaces for them. Since there are 4 possible kings, the total number of combos we get here is 4*C(47,2).

If we have a queen, we want to avoid any kings on the board, but anything else is fine. Thus in this case there are 4 extra cards we have to avoid, and only 43 "good" cards. This means there are 4*C(43,2) combos here.

Working down in the same way, you can see that, with the ace of spades in our hand as the quad card, the total number of good combos is

4*(C(47,2) + C(43,2) + C(39,2) + ... + C(7,2) + C(3,2)).

Now, there was nothing special about the ace of spades. If we chose any other card we could work down the kickers in the same way and would get the same result. Thus the total number of combos in the case where we don't have a pocket pair is going to be

52*4*(C(47,2) + C(43,2) + C(39,2) + ... + C(7,2) + C(3,2)).

Now, add this to the 4*C(47,2) from before. This is the number of combos. The overall possible number, given that we care about hole cards of course, is going to be C(52,2)*C(50, 5), since first we get 2 hole cards and then of the remaining 50 cards we get 5 for the board.

So just divide by that and you should get your probability.

I'm sure there's a more elegant way to do it though. Assuming this is even right; I'm pretty tired lol.

Repeat: Your statement about Q7 not winning quads is troubling to me

OP Condition: Quads while using both hole cards

If the board has a triplet, like QQQ84, then for getting a quads result, you only need a Q and any other card that's not really needed . That’s why I said that the restriction to use both hole cards for quads mean to me hero has to have a pair.

Pr(quads using both hole cards) = Pr(pair) * Pr(board has other pair of same rank)

= (1/17)*C(2,2)*C(48,3)/C(50,5) = 0.048%

OP confirmed that he really means that AQ on QQQ95 would qualify as "quads using both hole cards".

I think Rusty is on the right track. Think of this as a BBJ (or a high-hand jackpot) question.

We've had a similar question fairly recently but I am too busy to look for it or contribute to this thread at the moment.

In the Q7 case, the 7 doesn't play because in poker you play your best 5 card hand. With Q7 on a QQQ84 board, the best 5 card hand is QQQQ8 which does not use your 7. If you had AQ the best hand you could make would be QQQQA, which does use your A.

I agree.

My way: 13 * {[C(48,3) - 12*4 - C(12,2)*4*6]*10 + 12*4*18 + C(12,2)*4*6*14} / C(52,7) / C(7,2)

If it's quads+trips then there are C(7,2)-3 = 18 ways to group the 7 cards so that both hole cards play.
Else if the other three cards make a pair lower than the singleton, there are C(4,2)+2*4 = 14 ways.
Else there are C(4,2)+4 = 10 ways.

First of all: Thanks for all the replies!
I'm starting to understand the calculations

So for the Straight Flushes I first have to calculate how many possible starting combos can even make a straight flush, right? And afterwards I can deduct the possibility for the royal flush, correct?

manual counting :

9*4+3+2+1= 42
times all the suits: 42*4= 168

Or is it:

4*C(13,2)-4*C(5,2)= 272

So 168 or 272 combos that can can make a straight flush but not a royal flush?

I somehow can't edit my post above:

I think I found the correct way to calculate the possible hands:

C(13,1)*C(4,1)-C(5,2)= 168 possible combinations


I am currently stuck at this point:

(C(13,1) * C(4,1) *4 -(4* C(5,2)))/C(52,2) *(C(6,3)*C(47,2)/C(50,5))

The bold part is incorrect, because I can not get any3 of the 6 cards. How can I narrow this part down correctly?

Can't edit again, so sorry for the triple post.

I tried many different things, but I can't get it to work.

I would do the following:

1. Tally how many straight flushes are possible in one suit (say for spades)

2. Then for each possible straight flush, tally how many ways are there for Hero to have two hole cards being part of that straight flush

3. Then tally how many ways can the board "fill out" given that Hero has made a particular straight flush (remember to exclude making a higher straight flush)

4. Put it all together and multiply by four (number of suits).

I don't know if that is helpful or not.

There are at least 3 approaches for these kinds of problems:

a) What whosnext just said, going in order of preflop to postflop.
b) Going in reverse order: postflop to preflop.
c) The approach I showed in my earlier post for quads: start with all 7 cards and then think about the 5:2 split later.

Before I begin a problem like this, first I quickly try to imagine what those three approaches would look like and predict which one will be easiest. Right now, thinking about straight flushes, I predict (c) is the easiest route. What jumps out to me about (a) is that you'd have to split it into multiple preflop hand types (how connected they are and whether they're mid-range or high/low). And the drawback of (b) is you'd have to split it into multiple board types. Having to split the problem into multiple cases tends not to be fun.

So I'll go with option (c) and this time I'll explain my calculations.

Since we're doing the Royal separately, there are 9 possible SF's for each suit.

Each SF in a given suit is 5 specific cards and 2 extra cards. With SF's the thing to watch out for is double-counting, which isn't an issue with a royal flush. If 5 royal flush cards are present, then the extra 2 cards can be any of the other C(47,2) combos. But for the Q high SF, you can't say C(47,2) because the K would make it a K high, and we will already be counting the K high separately. So for the extra cards there are only C(46,2) combos.

All told, there are (4 suits)*(9 SF's)*C(46,2) = 37260 ways for a SF to be dealt among the 7 cards.

Now is when we consider how the 7 cards are split between the board and hole cards, to ensure that both hole cards are used.

There are C(7,2) ways to divide the cards. We know that 2 of the 5 important cards need to be hole cards. So if a SF is present in the 7 cards, the chance it's using both hole cards is C(5,2) / C(7,2) = 10/21

So the answer is 10/21 * 37260 / C(52,7) ≈ 1 in 7540

For the royal flush it's 10/21 * 4*C(47,2) / C(52,7) = 1 in 64974 exact

Also wanna take this moment to say welcome @mellyor and @math42! It's refreshing to see new faces because this place has been a bit of a ghost town of late.

Not that whosnext & statman aren't good people to be trapped in an elevator with, but I'd like for this subforum not to be an elevator :P

Just skimmed. Did you take into account boards that contain two pairs or a full house can create two different quads (except for xxxKK)?

He/she didn't have to, because their calculation went in order from preflop to postflop. (Edit: not saying this reasoning applies to other problems, but here it does in combination with the point below.)

For the pocket pairs, they multiplied 13*6*C(48,3). This double-counts the two-pair boards, but those should be double-counted (once for each of the possible quads). The important thing is that no quads or 7-card hands were double-counted in that product.

Am I thinking about this correctly?

It seems like the answers above all pertain to the probability of a single player making the hand in question in NLHE.

However, if the question becomes a question about the percentage of deals for which any player at an N-player NLHE table makes the hand in question, then the "double counting" phenomenon becomes relevant.

The issue kicks in for quads (as David points out) and also for straight flushes (which I remember wrestling with in a previous thread).

Right?

Yes and yes.

If this isn't about Hero, but N players, then I would advise against calculating in order of preflop to postflop. I'd focus on the board first and then the preflop hands. As for the 3rd approach I showed earlier ITT, I imagine it would get weird for N players.

P(one-pair board)*P(quads | opb) + P(two-pair board)*P(quads | 2pb) + P(trips board)*... + P(boat board)*...

We'd have to avoid overlap when calculating P(quads | 2pb) and P(quads | boat board), either by being selective with our adding or by using inclusion-exclusion.

When it comes to straight flushes, the only possible overlap is on boards where two qualifying straight flushes are possible. Then it's the same thing, be aware that two players can have a SF and don't double-count that case.