Odds of drawing a three/four card straight in five cards

Okay, so I generally understand this Binomial Coefficient/combin stuff - at least when other people do it, apparently - though I'm not having luck on my own. Basically (ignoring overlap from flushes altogether), I know there are 10,240 combinations of five card straights based on the 10 possible high cards and irrelevancy of suits: combin(10,1)*combin(4,1)^5

However, I want to know how many combinations there are for three-card straights and four-card straights. I know there are 11 possible high cards for a four-card straight and 12 possible for a three-card. Once again, since I'm ignoring flushes, the suit is irrelevant, so the math is largely the same thusfar. However, where I'm having trouble is accounting for the extra, non-straight card or two in my equation (one extra for four-card and two extra for three-card). I've come up with a couple different ways of doing it but distrust my results. Could someone give me a hand?

Thanks in advance!

I imagine it's something like this:

Four-Card Straight: combin(11,1)*combin(4,1)^5*combin(X,1)

Three-Card Straight: combin(12,1)*combin(4,1)^5*combin(Y,2)

...but I don't know what X and Y would be. It's easy enough to simply subtract 3 or 4 from 52 but I know that's not right, because specific cards would bump them up into a higher straight. Worse still is that, depending on where the straight falls numerically there is a variable number of cards that can apply to: only one if the straight starts/ends with the the lowest/highest number but two if the straight is in the middle (a three-card straight is even worse since there are twice as many cards to consider).

Four Card Straight when dealing five cards to the hand works something like this:

http://people.math.sfu.ca/~alspach/comp2/

Dealing a 3-card straight to a 5-card hand would be based upon this.
NOTE: straight-flushes have to be added back to the derived figures, since this math DOES
take into account flushes and str-fl.

Another way to look at it. Deal the 4-card straight... the 5th card cannot make a 5-card straight.
A-2-3-4 and J-Q-K-A have only ONE rank that cannot be dealt (the 5 and 10 respective), all other ranks
of straights cannot have TWO ranks dealt. (Ex: 5-6-7-8 cannot have a 4 or 9 dealt as 5th card).

HTH

Thanks. I'm on top of four now, just have to figure out three. This has got me thinking in the right direction, I've just got to wrap my head around the weirdness of the extra cards (whereas, with a four-card straight, of say 3-4-5-6, you couldn't have a 2 or 7, with a three-card one, 3-4-5, you couldn't have a A & 2, 6 & 7, or 2 & 7). You got me on the road, now there are simply more I must account for. Thanks again!

N/M, I realized I was making it harder than I need to, no need to worry about it becoming a five-card straight, only a four-card one (because, obviously it's impossible for it to be a five-straight if it can't be a four-card one)...

This is one reason why I usually don't play the poker machines. I can't imagine even asking such a question, let alone what the answer would do as far as helping someone win. Slots are where it's all at. None of any of the what-ifs, theories, or calculations that drive vp players nuts and keeps them up at night. The only way I'll let a machine keep me up all night is by tracking the results of other players on a building progressive. But I AM impressed that there are some who strive to answer such questions properly.

Quote: tournamentking

This is one reason why I usually don't play the poker machines. I can't imagine even asking such a question, let alone what the answer would do as far as helping someone win. Slots are where it's all at. None of any of the what-ifs, theories, or calculations that drive vp players nuts and keeps them up at night. The only way I'll let a machine keep me up all night is by tracking the results of other players on a building progressive. But I AM impressed that there are some who strive to answer such questions properly.

Yeah, screw those games that can be analyzed for an exact return based on combinatorics or simulation. Slots! Where you don't even KNOW the base return before you start looking at the progressive!

Quote: AcesAndEights

Quote: tournamentking

This is one reason why I usually don't play the poker machines. I can't imagine even asking such a question, let alone what the answer would do as far as helping someone win. Slots are where it's all at. None of any of the what-ifs, theories, or calculations that drive vp players nuts and keeps them up at night. The only way I'll let a machine keep me up all night is by tracking the results of other players on a building progressive. But I AM impressed that there are some who strive to answer such questions properly.

Yeah, screw those games that can be analyzed for an exact return based on combinatorics or simulation. Slots! Where you don't even KNOW the base return before you start looking at the progressive!

I understand your point aces. What I was saying is analyze all you want and use those simulations until you're blue in the face. But how's it usually work out at the CASINOS afterwards...

My other point was that, with slots, math is not the driver, since they are as you identified. Therefore, I can get a good night's sleep every day I want one.

So if I understand this right, with a four-card straight, you count up all the possible combinations (where Os can be anything and Xs aren't available):

A-2-3-4-X-O-O-O-O-O-O-O-O = 8 possibilities
X-2-3-4-5-X-O-O-O-O-O-O-O = 7 possibilities
O-X-3-4-5-6-X-O-O-O-O-O-O = 7 possibilities
O-O-X-4-5-6-7-X-O-O-O-O-O = 7 possibilities
O-O-O-X-5-6-7-8-X-O-O-O-O = 7 possibilities
O-O-O-O-X-6-7-8-9-X-O-O-O = 7 possibilities
O-O-O-O-O-X-7-8-9-10-X-O-O = 7 possibilities
O-O-O-O-O-O-X-8-9-10-J-X-O = 7 possibilities
O-O-O-O-O-O-O-X-9-10-J-Q-X = 7 possibilities
X-O-O-O-O-O-O-O-X-10-J-Q-K = 7 possibilities
A-O-O-O-O-O-O-O-O-X-J-Q-K = 8 possibilities

Total: 79 possibilities (8x2 + 7x9)

Then 79 possibilities x 4^5 suits = 80,896 (a number I feel good about)

Therefore, for a three-card straight, we do the same thing:

A-2-3-X-O-O-O-O-O-O-O-O-O = 9 possibilities
X-2-3-4-X-O-O-O-O-O-O-O-O = 8 possibilities
O-X-3-4-5-X-O-O-O-O-O-O-O = 8 possibilities
O-O-X-4-5-6-X-O-O-O-O-O-O = 8 possibilities
O-O-O-X-5-6-7-X-O-O-O-O-O = 8 possibilities
O-O-O-O-X-6-7-8-X-O-O-O-O = 8 possibilities
O-O-O-O-O-X-7-8-9-X-O-O-O = 8 possibilities
O-O-O-O-O-O-X-8-9-10-X-O-O = 8 possibilities
O-O-O-O-O-O-O-X-9-10-J-X-O = 8 possibilities
O-O-O-O-O-O-O-O-X-10-J-Q-X = 8 possibilities
X-O-O-O-O-O-O-O-O-X-J-Q-K = 8 possibilities
A-O-O-O-O-O-O-O-O-O-X-Q-K = 9 possibilities

Total: 98 possibilities (9x2 + 10x8)

Then 98 possibilities x 4^5 suits = 100,352 (a number that I don't feel so good about)

100,352 is really not that much higher than 80,896, in fact there are 1,098,250 different combinations that can make a pair and it seems to me that a three-card straight should be easier to come up with: to turn one card into a pair there are only three other cards in the deck (the same number in three other suits) whereas to turn a single card into a three-card straight, there are eight other cards available for the second card (the card on either side of the first, available in all four suits), then eight options again for the third (the cards on either side of those two cards, in all four suits) -- alternately, there are eight cards that can be paired with the first card resulting in a one-card gap and then four cards in the deck that would fill it (which is less likely but seems like it should still be better odds than drawing a pair).

So the real question: is this math wrong or am I wrong? :-)

Quote: baronkohinar

So if I understand this right, with a four-card straight, you count up all the possible combinations (where Os can be anything and Xs aren't available):

A-2-3-4-X-O-O-O-O-O-O-O-O = 8 possibilities
X-2-3-4-5-X-O-O-O-O-O-O-O = 7 possibilities
O-X-3-4-5-6-X-O-O-O-O-O-O = 7 possibilities
O-O-X-4-5-6-7-X-O-O-O-O-O = 7 possibilities
O-O-O-X-5-6-7-8-X-O-O-O-O = 7 possibilities
O-O-O-O-X-6-7-8-9-X-O-O-O = 7 possibilities
O-O-O-O-O-X-7-8-9-10-X-O-O = 7 possibilities
O-O-O-O-O-O-X-8-9-10-J-X-O = 7 possibilities
O-O-O-O-O-O-O-X-9-10-J-Q-X = 7 possibilities
X-O-O-O-O-O-O-O-X-10-J-Q-K = 7 possibilities
A-O-O-O-O-O-O-O-O-X-J-Q-K = 8 possibilities

Total: 79 possibilities (8x2 + 7x9)

Then 79 possibilities x 4^5 suits = 80,896 (a number I feel good about)

Therefore, for a three-card straight, we do the same thing:

A-2-3-X-O-O-O-O-O-O-O-O-O = 9 possibilities
X-2-3-4-X-O-O-O-O-O-O-O-O = 8 possibilities
O-X-3-4-5-X-O-O-O-O-O-O-O = 8 possibilities
O-O-X-4-5-6-X-O-O-O-O-O-O = 8 possibilities
O-O-O-X-5-6-7-X-O-O-O-O-O = 8 possibilities
O-O-O-O-X-6-7-8-X-O-O-O-O = 8 possibilities
O-O-O-O-O-X-7-8-9-X-O-O-O = 8 possibilities
O-O-O-O-O-O-X-8-9-10-X-O-O = 8 possibilities
O-O-O-O-O-O-O-X-9-10-J-X-O = 8 possibilities
O-O-O-O-O-O-O-O-X-10-J-Q-X = 8 possibilities
X-O-O-O-O-O-O-O-O-X-J-Q-K = 8 possibilities
A-O-O-O-O-O-O-O-O-O-X-Q-K = 9 possibilities

Total: 98 possibilities (9x2 + 10x8)

Then 98 possibilities x 4^5 suits = 100,352 (a number that I don't feel so good about)

100,352 is really not that much higher than 80,896, in fact there are 1,098,250 different combinations that can make a pair and it seems to me that a three-card straight should be easier to come up with: to turn one card into a pair there are only three other cards in the deck (the same number in three other suits) whereas to turn a single card into a three-card straight, there are eight other cards available for the second card (the card on either side of the first, available in all four suits), then eight options again for the third (the cards on either side of those two cards, in all four suits) -- alternately, there are eight cards that can be paired with the first card resulting in a one-card gap and then four cards in the deck that would fill it (which is less likely but seems like it should still be better odds than drawing a pair).

So the real question: is this math wrong or am I wrong? :-)

Baron,

I'm pretty sure you're underestimating the possible combinations for both 3 card straights and 4 card straights by something in the many millions. For example, in the first 4 card straight, the 5th card can be anything EXCEPT a 5 or one of the 4 cards in use. Without even counting the other 44 cards, the unique combinations of A-2-3-4 is 4x4x4x4, or 256. That number x 44 other cards or 11264 possibilities, and that's with a closed-ended straight. For an open-ended 4 card straight, like the 2-3-4-5, it would be 4x4x4x4, or 256 again, x40, or all cards except the 4 used, the 4 aces, and the 4 6's, x39, the same as I just said except for the card just picked, which is 399360.

I proved in developing my own game that I didn't understand the shortcuts to doing this (haha at me) so I'm still learning about combinatorials and the rest. But, long-form, that's how you figure these things to my present understanding. Good luck and I commend you for tackling the learning, since I'm doing the same thing and am finding it very worthwhile.

While I don't discount the likelihood that I did something wrong, it's my understanding that there are combin(52,5) possible five-card hands which is: 2,598,960

My case is rested. Now I'll get some.

lol, don't worry I wasn't planning to use this for poker, just because I'm curious :-)

My new theory is as follows: my original formulas were correct and I think I just figured out my X and Y...

Four-Card Straight: combin(11,1)*combin(4,1)^5*combin(X,1)

X = 79/11 because that's the average number of cards that are available to complete the hand.

So, combin(11,1)*combin(4,1)^5*combin(79/11,1) = *drumroll please* 80,896 which, if we recall, is the correct number based on the math from the page that 98Clubs linked me to. So far so good.

So onward to the Three-Card Straight: combin(12,1)*combin(4,1)^5*combin(Y,2)

Three cards, assuming I figured out all the available cards correctly in my last post, is a little more complicated and my original math was wrong since I, effectively, did combin(Y,1) on the average instead of combin(Y,2) which I should have done since there are two cards - meaning my calculations were indeed off. So, anyway, again assuming I'm correct, Y should be, once again, the average number of cards available to complete the hand, in this case 98/12.

Therefore, combin(12,1)*combin(4,1)^5*combin(98/12,2) = 359,594 2/3 possible combinations

This number looks more correct to me, though possibly still off, now that I think about it, because how can you have 2/3 of a combination? Anyway, I'm sure I'll get there eventually, or maybe a math wizard will descend upon a cloud and figure it out for me? Only time will tell. Certainly better off than I was yesterday in any event ;-)

If you look at the 2nd half of my explanation... and take Alspach's formula...

there are 10 3-card straights that cannot have the rank above or below it, and the 3-card suit can be any combination.
That works out to 10 * 4^3 with 49 cards remaining, but only 41 can be drawn, so the FOUR cards are 10 * 64 * 41.
The fifth card cannot make a 5-card flush, nor can it make a 4-card straight, (NOTE: two of the cards that make a 4-card straight, also make a 5-card flush!: don't count them twice) so with 48 cards remaining, there are
(48 - 6 - 9) = 33 cards. For this partial answer, 10*64*41*33 = 865920. (typo corrected)

The other two 3-card straights A23 and QKA have one rank not allowed. So the FOURTH card is 2 * 4^3 * (49-4) = 128*45
The FIFTH card as above cannot make a 4-straight or a 5-flush (48 - 2 - 9) = 37 cards.
This partial solves as 128*45*37 = 213120.

Adding both partials 865920 + 213120 = 1079040

I think this is correct. Please note that its possible to have a 3-straight with a pair.

Though what you say makes sense, if there are 1,079,040 three-card straights as you say and 1,098,250 hands with a pair as found multiple places online, that totals 2,177,290. With 2,598,960 possible hands total (though I realize there is some overlap in three-card straights with pairs), that leaves somewhere around 421,670 combinations left (again, plus the number of three-card straight/pairs, which we shouldn't count twice) to account for everything else, which to me seems a little low - certainly enough to accommodate the rarer combinations such as three/four of a kind and five-card straights, but what about hands with only a high card? Surely we don't have enough combinations left to account for those?

ummm... no... some of the three card straights have a pair as the last two cards, one cannot sum the two answers.
There are even hands like 3-4-4-4-5... in short each of the 11 ranks (12 if A23 or QKA) can be paired, if the pair is one of the 3 ranks of the 3-straight, its possible to have a 3-oak. The answer for a 3-straight should not include a 3-oak, but allow a pair. Therefore my answer is a little too high, but does include all possible 3-straights.

For ranking purposes, the paired three-straights must be subtracted from One Pair hands, meaning fewer One Pair hands.

Imagine how complex this gets if you allow 4-Flushes, 3-Flushes, and 4-Str-Flushes.

Obviously you can't sum them directly (which I said in my last post) but I was using it illustratively as a rough example. Really, though, where I think our disconnect is is that I'm looking for the number of three-card straight combinations that *aren't* a higher hand (in this case, as I'm ignoring flushes, it should not also be: a pair, two pair, and three of a kind) which appears to be considerably less. Sorry if I didn't make that clear before.