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