A pair in the flop: what are the odds?

I’m watching the final table at MTT American Million, watching flop after flop after flop with a pair in the flop.

It’e enough to make me wonder, how often does this happen?

I’ve tried googling for this information, and there’s a lot of different not-quite-what-I’m-looking-for articles about the odds of flopping a pair (made with any combination of your hole cards and the cards on the board). I’m trying to find out what the odds of two cards of the same rank appearing together in the flop.

It seems like it happens quite a bit, but I’d like to get an accurate count of the combinatorials of possible flops that have two cards of the same rank in it, expressed as a probability.

Here is a link to the best thread on this forum. You will find your answer and so much more here: The Juicee Basics
(it’s about 5.9%)


I don’t see the answer to my question in the thread you linked to. There are 5 posts in the thread, and it deals almost entirely with the odds of starting hole cards.

For that matter, not that it’s on topic for this thread, the big chart doesn’t make sense to me.

Why would A5 be 60% when A6 is 59% in the left-most column? Why isn’t the chart symmetrical? Following A5 in the top row, I see the odds of winning are 62%, but down the left column, it’s only 59%. Huh?

But I see nothing telling me what the odds are of a flop coming up XXY XYX or YXX for any given values X and Y. That’s my question. Take a standard 52-card deck, deal out the first 18 cards at random (for the player’s hole cards), burn a card, then deal out the next three cards for the flop. What are the odds that any of the 2 of the 3 cards in the flop will be of the same rank?

Based on my limited observation, if the odds are merely 6%, I’m seeing some very aberrant cards being dealt. (Yes, I know about confirmation bias, but seriously I’m seeing it hand after hand after hand here.)

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The answer to your question about pairs in the flop is not in this chart. The 5.9% is in the answers to the quiz in that thread (6th question).

And regarding the chart, the differences are between suited and off-suit cards. The odds off winning with A7o are not the same as winning with A7 suited.

Hope this helps.


Thank you, @Maya.

So, in the quiz, is it not showing up right for me or something? What I saw is:

Question 6) What is the probability that a flop will arrive containing a pair (e.g., K-9-9)?
about 5.9 percent
about 16.7 percent

It’s like there’s two answers, (for mulitple choice?) and no answer key, so I don’t know which is the correct answer (apart from you and @Grateful_ed telling me it’s 5.9%). I’d still like to know how the odds are calculated, so I can run the math myself and get a deeper understanding (and so I can start calculating this stuff on my own without having to ask.)

As to the chart, I still don’t understand why A5o is a percent more likely to win than A6o is. What’s up with that?

The 2 possible answers are given as multiple choices in the quiz, but the correct answers are given at the end of the quiz, at the bottom of the page. They are blurred, but if you click on them you will be able to see them.

Good observation about A6 and A5. A7, A6, A5 and A4 all have very close odds (to the decimals) and they rounded them up in this chart I guess.

Thanks again. I didn’t even notice the blurry patch beneath the quiz. What markup is used to do that?

It doesn’t give you the math behind your specific question, but it does put you on the right track on how to calculate the odds yourself.

You’re basically drawing 3 cards out of 52 and want to know the odds of 2 of them being of the same rank :slight_smile:

No not a round up, A5 is better than A6 because it has ability to make a straight :slight_smile:


Blur Spoiler. You can find it in the toolbar on top of this reply window, the last icon on the right that looks like a settings icon.

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Thanks Juicee :slight_smile:
Totally missed that.

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With A5 you can make a straight using both cards - A2345 - which you can’t with A6. That’s why A5 has a higher %.


Nice, so this does help me to start to understand the math.

What I get from that article on stackexchange makes sense as far as it goes: Pick any card at random, 1 in 52. For the second card to be of the same rank, there are 3 of those cards in the remaining 51, which works out to 1-in-17, which is indeed 5.9%.

But that’s not quite right, is it? That 5.9% would be the odds of the first card in the flop pairing with the next card in the flop. But we get THREE cards, so that means we get TWO chances to pair the first card, since the flop gives us 3 cards, don’t we?

This is something like the “birthday problem” in probability, which I’ve read about numerous times, but still struggle to understand/explain.

So in a similar, but simpler, probability problem, if I were to flip a coin two times, what are the odds of it being heads both times or tails both times? 50%. But what are the odds of getting two heads or two tails after 3 flips? 100%. (Unless we’re not counting trials where it’s heads or tails all three times, I suppose, in which case it’d be 4/6.)

So this makes me think that the odds of the first card in the flop pairing with either the second or third card in the flop should be something more like:

3/51 + 3/50, which would be about 11.8%, or about double the 5.9% you’re telling me, isn’t it?

But wait, that’s just the odds of the FIRST card in the flop paring with the second or third card: XXY or XYX.

We can also get a pair from the second card and the third card matching each other in rank, if the second card from the flop doesn’t match the first: YXX.

So that would give us an additional ~5.9% (ignoring the slight change in the odds from removing the first couple cards from the deck as they’ve already been drawn). I guess to be precise, the odds of cards 2 and 3 in the flop pairing with each other are 3/49, or about 6.1%.

We don’t have to consider further combinations involving the third card, as they’re already accounted for in the considerations of the first two cards.

We could add in the odds of all three flop cards being of the same rank: (3/51) * (2/50) = 0.24% but it adds a nearly negligible increase to the overall odds of the flop containing a Pair, and anyway 3 of a kind is a drastically different hand than a pair is (since it makes quads and full houses much more likely) so I don’t really want to add that quarter-percent chance into this.

So that makes it more like 18% of the time that any one of the three cards in the flop will pair with either of the other two. Which is still not that much of the time, but I can buy something that is 18% likely happening in streaks a lot more frequently than something that’s only 6% likely happening in streaks. (Of course any probabilistic outcome will happen in streaks given true randomness, but it’s more likely to observe streaks when the odds of the thing happening are higher.)

And then, another thing I’m still not sure about is how drawing the first 19 cards at random for the 9 starting seats hole cards + the burn card will affect the above odds. Does the randomness over quadrillions of trials just cancel itself out, leaving the odds still at 5.9%? I have no idea to work out that probability.

As everything is dealt at random, I don’t think the order should affect how the odds are calculated.
18 hole cards then 3 on the flop, or 3 on the flop then 18 hole cards, the probabilities shouldn’t be affected.
I’m not 100% sure but if I remember my probability lessons correctly, the result would be the same.

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@Maya, I just re-edited my previous post, please go back and look at it again, I think I’ve found something interesting.

Let’s assume there are 50 unseen cards and that 10 of them are “good.” This would mean 1 in 5 of the remaining cards are good.

If there are 5 other players in the hand, each with 2 hole cards, we would expect 2 of these 10 cards to be “good” ones. This leaves 8 good cards of the remaining 40 undealt cards, and this is also 1 in 5.

So, with a uniform distribution, the fact that some of the “good” cards have been dealt to other players doesn’t change the probabilities in the remaining deck.


Very clever reasoning, and it makes sense to me. Thanks for that insight!

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I found this article that I think you and @JuiceeLoot should have a look at:


It gives different odds. It says that the probability that the flop would contain a pair is 17%.
And 5.9% is the probability that you would be dealt a pocket pair. I think these numbers make more sense.


Right. I don’t think that 5.9% is correct.

When the first flop card comes out it is one of 52 cards.
When the second flop card comes out there are 48/51 cards that do not make a pair.
When the third flop card comes out there are six cards that would make a pair and 44/50 cards that do not match one of the first two cards.

So the probability that there is NO pair on the flop is 52/52 * 48/51 * 44/50 = 82.8%.

Probability that the board has all 3 cards the same rank is 52/52 * 3/51 * 2/50 = 0.2%

So the probability that the flop contains exactly one pair is 1 - 82.8% - 0.2% = 16.9%


Brilliant @love2eattacos :+1:

Thank you for that!!

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