How to calculate the odd board have 3 specific cards by the river - Gambling and Probability

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How to calculate the odd board have 3 specific cards by the river

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01-26-2024 , 11:34 PM

Arnakk2424

centurion

Join Date: Jun 2017 Posts: 145

Let say I hold 6d5d
Possible board are 50C5 = 2118760 boards
How I’m suppose to calculate the # of the boards with 4d3d2d

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01-27-2024 , 03:48 AM

heehaww

Pooh-Bah

Join Date: Aug 2011 Posts: 5,081

50C5 implies that you aren't counting order of the streets, in which case the answer is 47C2. We're counting the ways to choose the 2 irrelevant cards from the 47 that aren't a 4d/3d/2d.

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01-27-2024 , 12:45 PM

Arnakk2424

centurion

Join Date: Jun 2017 Posts: 145

Quote:

Originally Posted by heehaww

50C5 implies that you aren't counting order of the streets, in which case the answer is 47C2. We're counting the ways to choose the 2 irrelevant cards from the 47 that aren't a 4d/3d/2d.

not sure I understand
so 47c2 = 1081

1081 out of 2118760 meet the requirement ?

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01-27-2024 , 01:55 PM

heehaww

Pooh-Bah

Join Date: Aug 2011 Posts: 5,081

Yup.

But normally when one talks about "boards" they're distinguishing between flop/turn/river. In that case, there are (50C5)*5*4 = 42375200 boards and the number of those that contain 4d3d2d is 1081*20.

If we only needed the probability of 4d3d2d by the river, we wouldn't have to count boards: there are 50C3=19600 combos of places for those 3 cards to be in, compared to 5C3 combos of places within the board, so the probability is 10/19600 = 1/196

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01-27-2024 , 06:01 PM

Arnakk2424

centurion

Join Date: Jun 2017 Posts: 145

Quote:

Originally Posted by heehaww

I only need by the river because I'm trying to know how often hands hit straight flush using both hole cards in all in in or fold game.

Your result seems optimistic. I know from ProPokertools Odds Oracle 6d5d will hit a straight flush about 0.20% of the time but also include straight flush not using both hole card.

6d5d have 4 way to make straight flush 987d 874d 743d 432d using both hole cards.

I didn't do math in last 20 years.

Did you mean 1/1960 which is about 0.05% ?

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01-27-2024 , 07:52 PM

heehaww

Pooh-Bah

Join Date: Aug 2011 Posts: 5,081

Yes sorry 10/19600 = 1/1960

Also I just noticed your title asks about the probability, otherwise I'd have given it in my first post and wouldn't have mentioned the board arrangements in my last post. For calculating the probability, (47C2)/(50C5) is perfectly fine and gets the same answer as (5C3)/(50C3). Only if you were strictly asking a counting question would we possibly have to consider board arrangements.

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01-29-2024 , 01:33 PM

Arnakk2424

centurion

Join Date: Jun 2017 Posts: 145

Quote:

Originally Posted by heehaww

Yes sorry 10/19600 = 1/1960
(5C3)/(50C3)

Very easy to do the math

I can do (5C3)/(50C3)
*4 for suited connector
*3 one-gapper
*2 two-gapper
etc.

ty

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01-29-2024 , 01:59 PM

heehaww

Pooh-Bah

Join Date: Aug 2011 Posts: 5,081

Quote:

*4 for suited connector
*3 one-gapper
*2 two-gapper
etc.

I don't follow?

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01-30-2024 , 01:49 PM

Arnakk2424

centurion

Join Date: Jun 2017 Posts: 145

Quote:

Originally Posted by heehaww

I don't follow?

The plan is to know how often I can make a straight flush using both hole card.
In all-in or fold game you contribute to a jackpot and you get it making a straight flush.
A hand like 56s is a terrible call cEV but will hit jackpot 0.20% of the time (can hit straight flush 4 different way)
Jackpot is between 600BB and 750BB. That mean you have to add 1.2/1.5bb to the EV to the hand.

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01-30-2024 , 03:10 PM

#10

heehaww

Pooh-Bah

Join Date: Aug 2011 Posts: 5,081

Oh then now I see why you're multiplying by those numbers, but that's not quite right for two reasons:
1. You can get counterfeited out of the jackpot by making a higher SF that no longer uses both hole cards.
2. You're over-counting some of the same straight flush combos eg a board of 2347x is double-counted and 23478 is tripled-counted.

With 65, if the board comes 789 then the other two cards can't contain a T of that suit. If the board comes 478 then the other two cards can't be 9T.

P(jackpot with 65) = P(234xx) + P(347xx and no 2) + P(478xx and no 3 and no 9T) + P(789xx and no 4 and no T) = 1/1960 + [2(46C2)–1 + 45C2]/(50C5)
≈ .001954 ≈ 1/512

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01-30-2024 , 09:23 PM

#11

Arnakk2424

centurion

Join Date: Jun 2017 Posts: 145

You are good at math
I was aware a few hands I had to remove few hands get counterfeit jackpot but I was thinking it was negligeable.
Will have to do a little spreadsheet
thanks for your help you really aiming me to the right direction to know if that weird games beatable

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02-08-2024 , 12:58 PM

#12

Arnakk2424

centurion

Join Date: Jun 2017 Posts: 145

Quote:

Originally Posted by heehaww

Base on a 750BB jackpot

Suited connectors 54-JT
.001954 ≈ 1/512
+1.46bb in EV

One gapper QT-53 , QJ, 43
.001454
+1.09bb in EV

two gapper KT-52, KQ, KJ, 42, 32
.000954
+0.72bb in EV

other AT+,A5-A2, K9, Q8, J7, T6, 95, 84, 73, 62,
.000454
+0.34 in EV

I voluntary ignore the fact some hands cannot be counterfeit (exemple AK) because those hands gonna be +EV anyway.

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