# Probability of flopping a straight

I was thinking about calculating the probability of flopping straights, with certain hands. First, I thought about situations: obviously the difference between two cards can't be higher than 4. If the difference is 4 we need the three ranks between them so: `((12/50)*(8/49)*(4/48))*6 = 0.0195 = 1.95%` - would this be correct? My thought process: at the first card of the flop the first card can be any of the three ranks, thus 12, second card of the flop can only be two different ranks, thus 8, and last card of the flop only one card rank, thus 4. Multiplied with 6 because the different combinations. For example: my card 6T, flop can be 789, 798, 879, 897, 987, 978.

From now on I will use examples because it is easier to explain that way. If difference is 3: my cards 7T. Possible flops that would form a straight: 689, 698, 896, 869, 986, 968, J89, J98, 89J, 8J9, 98J, 9J8. With the same logic the calculation would be: `((16/50)*(8/49)*(4/48))*12 = 0.052 = 5.2% `, which seems wrong... What is wrong with my calculation - I'd like to calculate too for difference of 1 and 2.

• Why did you multiply by 6 in your first cacluation? Jan 18, 2021 at 8:55

I don't know what the factor of 6 is for in your first calculation. If you have a particular hole pair with a difference of four (say 6♣T⋄), then your analysis (without the factor of 6) for making a straight on the flop is correct. `12/50 + 8/49 + 4/48 = 0.0033`

Another way of looking at this is that with the 50 cards remaining in the deck, there are 19600 flops possible, and there are 4^3 or 64 different combinations with a 7, an 8, and a 9. `64/19600 = 0.0033`

Then it's not so much the "difference" between the cards, its the possible positions to hit. 6♣T⋄ differ by 4, while A♣3⋄ differ by 2. But the odds to hit a straight are the same because there are only 3 positions in each that can match. (7, 8, 9 for the first one; 2, 4, 5 for the second).

For lesser differences, the percentages depend on the specific hole cards and the odd become more complex to calculate directly. As an example, with a 5 and a 7, you can hit a straight on both sides (like 3-7 or 5-9). But if you have an A and 3, there's still a difference of 2, but the only possibilities are hitting a 2, a 4, and a 5. The odds are the same as the first case. Even besides that issue, your example of `16/50 + 8/49 + 4/48` is not correct. The first figure of 16 is correct, but the second figure depends on the first. If the first card is "outside", then you have 8 outs remaining. But if the first card is "inside", then you have 12 outs remaining. So the number of cases to calculate goes up. That said, there are only 4 possibilities: 3, 4, 5, or 6 possible cards to hit.

Brute forcing it (looking at all 19600 flops), you get the following odds for flopping a straight:

possible matching ranks example hole cards straight flops chance of flopping straight
3 6♣ T⋄ or A♣2⋄ 64 0.33%
4 7♣ T⋄ or 2♣3⋄ 128 0.65%
5 8♣ T⋄ or 3♣4⋄ 192 0.98%
6 9♣ T⋄ or 4♣5⋄ 256 1.31%
• Thank you, as I see from your table, I have to multiply by 2, 3, and 4 depending in how many ways I can get a straight. Jan 18, 2021 at 16:08