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Poker - 52 cards
4 suits of 13 (rank) cards each

Poker probability

Rank - ace (1) is both high and low so
A23456789TJQKA

Suited - Holding 2 cards of the same suit

Connectors - 2 sequential (e.g. 56)

A straight is 5 sequential cards

A flush is 5 cards of the same suit and it beats a straight so need to subtract it out. In the calculations for subtracting a flush assume holding suited connectors. In most cases that is what you play - mid suited connectors as then you have a chance of going both ways. I know how to do the calculation for a flush draw. Flopping a flush with the straight or straight draw is not much of a difference.

Outer - how many cards can make your draw. For an open ended straight 5678 there are 8 outs - four 4 and four 9. 3_567_9 also has 8 outs (double gutter). 56_89 only has 4 outs - not nearly as attractive.

You get 2 cards, bet, share 3 cards (flop), bet, add 1 shared card (turn), bet, add 1 last shared card (river), and bet. In the end it is the best 5 cards out of seven wins. This question only deals with 2 cards + 3 (flop).

When someone bets they are giving you certain pots odds. It is important to compare the odds of your hand to the pot odds. What is the chance of flopping (3 more cards) an 8 outer straight draw or a full straight?

Use holding 56 suited as an example that is the xx
y is the ways to make a straight

Ways to make an open ended straight draw (4 sequential)
Z is other card

   A  2  3  4  5  6  7  8  9  T  J  Q  K  A   
1 Z1     Y  Y  X  X    Z2 Z3 Z4 Z5 Z6 Z7 
2 Z1 Z2     Y  X  X  Y    Z3 Z4 Z5 Z6 Z7
3 Z1 Z2 Z3     X  X  Y  Y    Z4 Z5 Z6 Z7

Ways to make a double gutter - still 8 outs

   A  2  3  4  5  6  7  8  9  T  J  Q  K  A   
1     Y     Y  X  X     Y   
2        Y     X  X  Y     Y 

Ways to make as full straight

   A  2  3  4  5  6  7  8  9  T  J  Q  K  A   
1     Y  Y  Y  X  X    
2        Y  Y  X  X  Y    
3           Y  X  X  Y  Y    
4              X  X  Y  Y  Y  

Number of 3 card flops

Two cards are in your hand
$\binom{50}{3} = 19,600$

A) 3 cards of different rank - open ended draw

Number of straight draws $\binom{3}{1} = 3$

With suits number of ways to make 2 cards $\binom{4}{1}^2 = 16$

Number of other cards 4 suit $\binom{7}{1}\binom{4}{1} = 28$

Number of other cards 3 suit $\binom{4}{1}\binom{3}{1} = 12$

Number of flushes $-\binom{3}{1} = -3$

B) double gutter draw

Number of straight draws $\binom{2}{1} = 2$

With suits number of ways to make 3 cards $\binom{4}{1}^3 = 64$

Number of flushes $-\binom{2}{1} = -2$

D) full straigth

Number of full straight draws $\binom{4}{1} = 4$

With suits number of ways to make 3 cards $\binom{4}{1}^3 = 64$

Number of flushes $-\binom{4}{1} = -4$

Full calculation

$$\frac { (\binom{3}{1} * \binom{4}{1}^2 * (\binom{7}{1} * \binom{4}{1} + \binom{4}{1} * \binom{3}{1})) -\binom{3}{1} + (\binom{2}{1} * \binom{4}{1}^3) -\binom{2}{1} + (\binom{4}{1} * \binom{4}{1}^3) -\binom{4}{1} } { \binom{50}{3} } = 0.1171$$

Been working on this for a while and I think this is correct. Parts match up with numbers I have found on the Internet.

Is the correct calculation?

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  • 1
    $\begingroup$ Down vote what is the problem? $\endgroup$ – paparazzo Feb 12 '16 at 14:11
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    $\begingroup$ You probably need to refine the question. Do you want the chance of flopping a straight? Or of flopping a straight draw? Is the 65 hand suited? Or does the (x) mean unsuited? It's not clear exactly what you want. $\endgroup$ – soakley Feb 12 '16 at 19:01
  • $\begingroup$ @soakley Either - "What is the chance of flopping (3 more cards) a straight draw (4 sequential) or a full straight". (x) is just the in the matrix below to show how many ways to make draw or straight. And the flush calcs assume suited - I think. I will update the question. $\endgroup$ – paparazzo Feb 12 '16 at 19:21
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    $\begingroup$ You should probably avoid - or at least explain more of - the poker jargon, and explain more of what's going on. It will increase the number of people able to answer and also greatly improve the value of the question to people looking for understanding of how to solve similar problems. It looks like this is Texas Hold 'em, but you just say poker, as if that explains everything, and use terms like "suited connectors" as if it was obvious what that was. Many people here will be fairly good at calculating these kinds of things but won't know all the terms. Please try to make the title simpler $\endgroup$ – Glen_b Feb 14 '16 at 0:56
  • $\begingroup$ @Glen_b What would you suggest for a title? $\endgroup$ – paparazzo Feb 14 '16 at 1:34
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I will show the calculation for straights with a starting hand of 65 suited and then just give a complete breakdown for the rest. For the 65 hand, how many straight possibilities are there to make 23456?

$${ 4 \choose 1}{4 \choose 1}{4 \choose 1} {38 \choose 0}= 64$$

The same applies to 34567, 45678, and 56789. So there are 256 possibilities. But 4 of these are straight flushes, so we end up with 252.

Now to get what you want for the straight draws and others, you need to have more categories. I will give my version and you can judge whether the detail is enough for you.

Sorry about the lack of formatting.

Straight Flush: 4

4 of a kind: 2

Full House: 18

Flush: 161

A. Flush with open-ended straight draw: 21

B. Flush with double inside straight draw: 2

C. Flush with inside straight draw: 38

D. Other flushes: 100

Straights: 252

A. Straight with open-ended straight flush draw: 18

B. Straight with inside straight flush draw: 18

C. Other straights: 216

3 of a kind: 308

2 pair: 792

Flush draws: 2109

A. Open-ended straight flush draw: 99

B. Inside straight flush draw: 216

C. Open-ended straight draw: 114

D. Double inside straight draw: 6

E. Inside straight draw: 222

F. Other flush draws: 1452

Straight draws: 4482

A. Open-ended: 1530

B. Double inside: 108

C. Inside: 2844

Other hands: 11472

Total = 4+2+18+161+252+308+792+2109+4482+11472 = 19600

If you want definitions for some of these, let me know.

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  • $\begingroup$ The numbers add up to 19600. Is the answer to my question 252 + 1530? How do you get the 1530 please? $\endgroup$ – paparazzo Feb 13 '16 at 3:08
  • $\begingroup$ I will try again - how to you get that 1530? $\endgroup$ – paparazzo Feb 15 '16 at 18:09
  • $\begingroup$ These were done by computer. If I get to it, I will try to reproduce the result manually. $\endgroup$ – soakley Feb 15 '16 at 23:01
  • $\begingroup$ So you did them in like R with statistical probability? $\endgroup$ – paparazzo Feb 15 '16 at 23:03

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