I was wondering if anyone can help me work out the math involving straight odds. Like the odds of hitting a one gappers and two gappers. I'm not sure what to do about when hitting one card depends on which cards come out and how that affects the odds.

I want to work out the odds of getting a straight on the flop, or the turn, or the river when dealt one gappers and two gappers. I know I can find the answers online, I'm using this site to check my work (nvm I can't post links due to how recently I made an account, but I found a source online), but I'm curious about how for one gappers and two gappers the card that comes out affects which cards will be helpful in completing the straight.

For example: A player is dealt 2,5. the cards that will help them complete a straight are A,3,4 and 6. If an ace comes out then only the 3 or the 4 can continue to help for the straight (if trying to hit a straight on the flop), however if a 3 comes out, then an A,4 or 6 will continue to help make the straight.

Quote:HUGEFISHI'm just realizing now, I meant to post this in the Math forum. I do apologize if I posted this in the wrong forum.

I want to work out the odds of getting a straight on the flop, or the turn, or the river when dealt one gappers and two gappers. I know I can find the answers online, I'm using this site to check my work (nvm I can't post links due to how recently I made an account, but I found a source online), but I'm curious about how for one gappers and two gappers the card that comes out affects which cards will be helpful in completing the straight.

For example: A player is dealt 2,5. the cards that will help them complete a straight are A,3,4 and 6. If an ace comes out then only the 3 or the 4 can continue to help for the straight (if trying to hit a straight on the flop), however if a 3 comes out, then an A,4 or 6 will continue to help make the straight.

Welcome to the forum, hugefish. Where you posted this is fine.

I'm not qualified to give you the math formulas to calculate this question of yours, but there are others here who may help.

You might find this calculator to be useful, though. Set up your situation with your hole cards, give other players hands you think they're on if you like, or just do your hand + the flop, turn, river. It will give you odds on the hands at each point. It doesn't break out straight gutshot hits from other hands, though, so may not fully answer your question.

https://wizardofodds.com/games/texas-hold-em/calculator/

From here on, I'm just thinking out loud - read it if you like, but don't depend on it. The math guys love to straighten me out because they're bright and helpful like that (not a dig - I love to learn things). But you're probably better off ignoring my rambling.

In general calculation of it, though, you would treat all cards as unseen to find your odds. So if you start with 2,5 , there are 16/50 cards left that will help you fill it (A,3,4,6 x 4) and 34 that won't. You'll see 3 on the flop. 1 or more of the flop MUST help or the straight Calc is dead right there.

Since you get 3 shots at help, and only need 1 to continue, it's 16/50 for the 1st one, + 16/49 for the 2nd if the first one didn't help, + 16/48 if the first 2 didn't help. (I'm not sure how to weight these factors, sorry).

If the 1st one helped, then the 2nd becomes either 12/49 (for a 3 or 4 caught already) or 8/49 (for an A or 6 caught) to get more help. If that didn't help, the third card will 12/48 or 8/48 of the time. If it did help, then that drops to 8/48 (outside fill) or 4/48 (gutshot) to flop the straight.

Then you figure what you need from there to fill the straight on the turn and/or river.

There are subsets of what helps, as you obviously understand. Here's one.

If you ONLY get a 3 or 4, there will be 12/47 cards that will help on the turn/river, and you must go runner-runner. So it will be 12/47 x 8/46 if you catch the other 3/4, (8/46 to catch either the A or 6 (outside straight) on the river.)

You get a small percentage. Express it as a decimal.

Same ONLY 3 or 4 on flop, 12/47 for turn, but you catch an A or 6. You've lost a rank of outs, so now need 4/46 river (the missing 3 or 4) to fill the straight.

You get a small percentage. Add that decimal to the one above.

Keep calculating for each possible state with a 2 gapper hand, adding the percentages. Catching 1 that's either the A or 6 (figure 1 of those, then double it), Catching 2 on the flop, broken further into whether you're left with an outside or gutshot straight that EITHER the turn or river will fill. Got to count flopping the straight, too.

This (2,5) is a single hole card example. It would be different for 1 gap, 3 gap, runner-runner hole cards, with subsets for holdings on the ends (like A,J or A,3) that would be less likely. You would correct for that at the end of each calculation, by (using your example) 2,5 is the same state for each 2 gapper up to K,10, so 9 straight draw hands get full value of possible help. But A,J and A,4, (2 hands), have 1 less rank (4 cards) on the end that can help. This part is important in evaluating your TOTAL chances of receiving a 2 gapper hand and making a straight, but if the problem is, GIVEN you have a particular 2 gap hand, can be ignored unless you have a terminus card.

I'm sure there are easier ways to calculate this, but a spreadsheet doing it brute force like this helps ME think stuff like this out.

1. In the next five cards, you must get three cards that are different ranks than each, different than what you already hold, and that as a combination will give you a straight. If you get those, then the other 2 cards may be anything at all.

2. First you must count the number of rank combinations that will give you a straight

- EX: 76; there are 4: T98, 985, 854, 543

- EX: Q9; there are 2: KJT, JT8

- EX: AK: there is 1: QJT

There will always be 1,2,3 or 4 combinations, depending upon the number of gaps and how close you are to the ends of the ranks.

3. For each rank combination that will give you a straight, the probability of those three ranks appearing in the next three cards is:

(1*4^3)*combin (47,2)/combin(50,5) = 0.032653

where, of course, combin(n,k) = n!/ (k!*(n-k)!)

Explanation of terms:

(1*4^3) is the number of combinations for getting the three specific ranks. The term 4^3 takes into account that for each of the three ranks there are four possible suits that you can get.

combin (47,2) This term accounts for the number of combinations of the two other cards, given that you have the 5 cards that give you a straight. We use 47 because that is the number of cards that remain after your five straights cards are removed from the deck.

combin(50,5) represents the total number of card combinations on the board (the next five cards) given that you know the two cards in your own hand.

So if you hold a 7-6, you have four ways to make a straight, and the probability of at least one of them hitting is approximately 4* 0.032653= 0.1306.

I says approximately, because (for example) when you hold a 7-6 and the cards come 9-8-5-4-A, you have made both a 98765 straight and a 87654 straight. This is one instance of making two straights and should be counted as one instance, but our technique of multiplying 0.032653 *4 double counts this as two instances. So there is a small correction to what we have done to adjust the probability downward so as to account for the 5 card combinations in which you hit multiple 5 card straights.

If the above explanation is good enough for you, then great. However, if you do want to understand the math for properly accounting for hands that make multiple 5 card straights, then please say so, and I guess I will show that to you. It's more complicated and does not change the above values very much.