Calculating odds for a 2 run scenario with Combos

So I'm trying to calculate odds of hitting a set or better by the river when holding a pocket pair for the game of flip n go on GG Poker. This is an 8 seat all in game, everyone receives 3 cards visible to all players and we all burn 1 to keep 2 hole cards so you can verify if your rank are still in the deck or not. This means 28 cards are left in deck and the game is run twice.

Using my basic understanding of combos I think I can easily calculate for the first run

W 2 outs I get: [c(28,5)-c(26,5)] ÷ c(28,5) equals 0.330687831

W 1 out I get: [c(28,5)-c(27,5)] ÷ c(28,5) equals 0.178571429

So I'm trying to use the method of finding combos where I don't hit my outs and subtracting from total combos to find amount of combos where I hit at least 1 out. But how do you handle a 2 run scenario?

Is total combinations in 2 runs calculated like this:

c(28,5) × c(23,5)?

I tried this way so my total formula w 2 outs is: {[c(28,5) × c(23,5)] - [c(26,5) × c(21,5)]} ÷ [c(28,5) × c(23,5)] equals 0.595238095

Am I on the right path?

If that's incorrect I'd like to know the correct way of figuring out the odds. If I got it right though then my question is, how to tackle flush odds for this game? I'm definately lost here because it's a bit more complicated.

Thank you!

Your first calculations (for a single run) are correct, given that you see everyone's burn cards as well as their hole cards.

Your second calculation is also valid, under the same assumption.

The difficulty comes when you start talking about "a set or better". You can calculate the odds of a flush, but some boards will give you a set and a flush, so you can't just add the two probabilities together or you'll double-count those boards.

For flush odds specifically: you will need to treat the two runs separately, and sum over all cases for the first run (because this will affect the number of 'outs' for the second). You'll also need to consider that you could hit a flush in either suit.

To start you off: suppose you hold one heart and you can see another four hearts in other people's hole cards/burn cards. Then there are eight hearts out of 28 cards remaining, and you need (at least) four of them on the board for a heart flush.

The odds of selecting four hearts followed by a non-heart are:

``````(8/28) * (7/27) * (6/26) * (5/25) * (20/24)
``````

But the non-heart card can be in any of the five positions, so you need to multiply this by 5 to get the odds of 'four to a flush' in hearts on the board, which gives 0.0142 (i.e. 1.42%).

You would need to add to this the odds of a 'flush on the board' - a similar calculation, except with `(4/24)` instead of `(20/24)` and without multiplying by 5 (since all positions are taken by hearts), which gives just 0.00057 or 0.057%.

So the probability of hitting a flush (or possibly a straight flush) by the river on the first run with a pocket pair using your first suit is around 1.48%.

Then you would need to calculate the odds of hitting a flush with your other card (which will be different if the number of cards remaining in your second suit is different).

Then you would need to iterate the other cases: namely, three or fewer of each of your suits on the first board. So if you have a heart and a spade, the board could have three hearts and two spades, or one heart and three spades, or no hearts and no spades, etc. You'd need to calculate the odds of each, based on the 24 cards you can already see.

For each of those cases (there are 15, I think - 4*4 minus one, because three hearts and three spades on the same board isn't possible) you'll have a different number of 'outs' for each suit for the second run. You'll then need to calculate two probabilities:

• the probability that the case exists to start with (for example, the odds of two hearts and a spade on the first run)
• the probability that, in this case (out of the fifteen), you hit a flush on the second run (using a method similar to the one above)

You can then multiply the two together to get the chance of hitting a flush for each case on the second run. The 15 cases are all distinct, so you can sum the probabilities to find the total probability of hitting a flush on the second run (but not the first) You can then add this to the probability of hitting a flush on the first run to get the overall probability of hitting a flush.

Clearly this would be long and laborious but I don't believe there is a simpler way - it's a complex problem! It might also not be very meaningful: after all, if you hit a flush with a pocket pair in an 8-way pot then your flush is unlikely to be good...

And again, these probabilities aren't exclusive: if you hit a set, you could also have a straight, a flush, a full house (etc) at the same time. To calculate an exact probability you would need to define your question very clearly, but the mathematical solution would be very complicated indeed.

(If you're just after the answer rather than a mathematical method, I would suggest setting up a Monte Carlo simulation or similar to calculate the odds for a large but manageable number of random run-outs, but you would still need to define your question very clearly.)

• This is excellent, thank you so much! Right now I was hoping for some help just to approach the problem and you've provided just that. I'll be writing a simulator to check later but I've found working out the math usually helps me retain stats better to actually use them in play. Thanks again!
– Seb
Commented Apr 18, 2023 at 22:11