How should I rank these alternative poker hands?

May 13, 2011 12:26 PM

I'm trying to rank some non-proper poker hands within the conventional poker hand-ranking framework. I would like to rank them conventionally, i.e. on their probability of occurring in a straight, five-card deal. The non-proper hands are: (1) "Four-card straight" (four cards in a row); (2) "Four-card flush" (four cards of the same suit); (3) "Four-card straight flush" (four cards in a row of the same suit); (4) "Same-color flush" (all 5 red cards or all 5 black cards); (5) "Straight same-color flush" (a straight composed of all black cards or all red cards); (6) "Four-card straight same-color flush" (a four-card straight composed of all black cards or all red cards). My probability skills are "OK" (the odds for the four-card flush and same-color flush are straightforward), but some of them (particularly the straights) seem too tricky for me.

Here's how I posit the rankings, from low to high:

* High card
* 1 pair
* Same-color flush (4)
* Four-card flush (2)
* Four-card straight (1)
* 2 pair
* Four-card straight same-color flush (6)
* Four-card straight flush (3)
* 3 of a kind
* Straight
* Flush
* Straight same-color flush (5)
* Full House
* 4 of a kind
* Straight Flush
* 5 of a kind (if wild cards available)

More specific questions: 1) is it harder to get a Four-card straight than a Four-card flush? (I think so, but not sure); 2) Is 3 of a kind harder to get than a Four-card straight flush?; 3) Is a Full House harder to get than a Straight same-color flush?

But my general question is: Based on probability in a straight 5-card deal, where should those 6 non-proper hands rank within the conventional poker hand structure?

Can you tell it's Friday? ^_^

posted by mrgrimm to sports, hobbies, & recreation (9 answers total)

So there's a bit of a problem here, based on the fact that some hands are going to count in two of your categories. (For example, it's possible to have a full house which also has four cards of the same color.) So what does that count as? This sort of ambiguity also arises when you introduce wild cards.

But we can try to compute some of the probabilities anyway. For example, for the straight same-color flush: the low card can be any of ace through ten; there are two ways to choose the color; then 2⁵ = 32 ways to choose the suits of the cards once the color is chosen. Thus there are (10)(2)(32) = 640 of these. But 40 of them are straight flushes, leaving 600. (That correction matters, because there are 624 four-of-a-kinds.) So straight same-color flush should be ranked between straight flush and 4 of a kind... except that seems a little silly somehow, as it basically just expands the definition of "straight flush".
posted by madcaptenor at 12:42 PM on May 13, 2011

So there's a bit of a problem here, based on the fact that some hands are going to count in two of your categories. (For example, it's possible to have a full house which also has four cards of the same color.) So what does that count as? This sort of ambiguity also arises when you introduce wild cards.

I don't want to confuse things more, but I'll try to clarify by appealing to convention. Treat a hand that counts in two categories as the higher hands. (This already occurs in poker with Full House/3 of a Kind, i.e. "there are 54,912 possible three of a kind hands that are not also full houses" etc.)

So when determining the number of possibilities for each hand and ranking them by that number, count a Full House with 4 cards of the same color as a Full House, and not a Four-card flush?

Or basically what you just did with the straight same-color flush, i.e. eliminating the actual straight flushes from the count of possible hands.

Does that make sense?

So straight same-color flush should be ranked between straight flush and 4 of a kind... except that seems a little silly somehow, as it basically just expands the definition of "straight flush".

Well, the whole exercise is a little "silly" I suppose, but I couldn't find an easy answer anywhere.

I certainly don't think it's silly to differentiate between a real straight flush and the goofy one I made up. The idea is to expand the number and type of poker hands to create a twist game.

A same-color straight flush (that is not a regular straight flush) is not that easy to get, but I certainly wouldn't have pegged it above 4 of a kind or even a Full House. Based on your math, it seems like it should be ... (it certainly seems like 640 - 40 is the correct answer, but again, IANAM).
posted by mrgrimm at 1:54 PM on May 13, 2011

What I'm saying I'm afraid of is the following:
- there are 9 hands which are A but not B
- there are 2 hands which are A and B
- there are 10 hands which are B but not A.

So let's say that we ignore the 2, and rank A ahead of B because there are less As. But now there are 11 = 9 + 2 As and 10 Bs. So B should really be ranked ahead of A! But now there are 12 = 10 + 2 Bs and 9 As. So A should really be ranked ahead of B! And this repeats forever.

This doesn't happen with the standard set of poker hands and the standard ranking. But it might happen with your unconventional hands. That being said, there's no reason not to do the computation anyway, other than that I don't feel like it and am hoping someone else will come along and do it.
posted by madcaptenor at 1:58 PM on May 13, 2011 [1 favorite]

Thanks for the explanation, madcaptenor. As soon as I posted, I realized your case, which is that one hand might end up ranked higher than another hand solely based on preference.

I'm not convinced that case exists, however. And we won't know until somebody does it. ^_^

That being said, there's no reason not to do the computation anyway, other than that I don't feel like it and am hoping someone else will come along and do it.

yeah, that. I'm not sure I can do it though ... I won't hold my breath, but DON'T LET ME DOWN, ASKME!
posted by mrgrimm at 2:45 PM on May 13, 2011

madcaptenor: In your example there are 11 hands which are A and 12 hands which are B, so A should be ranked higher than B. It doesn't matter what the overlap is. Unless you're scoring B-and-not-A specifically.
posted by aubilenon at 3:26 PM on May 13, 2011

aubilenon: yes, I was scoring B-and-not-A specifically. The usual assumption in poker is that a hand which is more than one thing counts only as the higher one.
posted by madcaptenor at 5:37 PM on May 13, 2011

Okay here is what I get. Hopefully someone will check my math because I dashed this off fairly quickly.

Four card flush (2):
13 choose 4 combinations * 4 suits * 48 off cards - 5148 regular flushes/straight flushes = 132132

Same color flush (4)
26 choose 5 combinations * 2 colors - 40 straight flushes = 131520

Four card straight (1)
11 possible runs * 4^4 suit arrangements * 48 off cards - 10240 regular straights/straight flushes = 124928

Four-card straight same-color flush (6)
11 runs * 2^4 suit arrangements * 2 colors * 48 off cards - 600 straight same-color flushes = 16296

Four-card straight flush (3)
11 runs * 4 suits * 48 off cards - 80 straight flushes or same-color straight flushes w/last card opposite suit = 2032

Straight same-color flush (5)
10 runs * 2^5 suit arrangements * 2 colors - 40 straight flushes = 600

You can check a poker probability list to see where these slot in exactly. But overall you are really undervaluing the straight flushes.
posted by shadow vector at 10:01 PM on May 13, 2011 [1 favorite]

I rather like a challenge! Code here: https://github.com/vasi/pokerfeet

Results:

StraightFlush : 40
StraightColorFlush : 600
FourKind : 624
StraightFlush4 : 1952
FullHouse : 3744
Flush : 4792
Straight : 9440
StraightColorFlush4 : 11840
ThreeKind : 54912
Straight4 : 84160
Flush4 : 105672
ColorFlush : 85144
TwoPair : 120120
Pair : 1013760
HighCard : 1102160

These numbers don't include combinations that are already taken by a better hand, so StraightColorFlush is 600 instead of 640.

Note that we do in fact have a reversal of the sort madcapternor is talking about. Flush4 has a particularly strong overlap with ColorFlush, so whichever one we decide is better will end up with more combinations! My algorithm just iteratively removes the rarest remaining hand. So at the time it decided between Flush4 and ColorFlush, there were actually 120368 ColorFlush hands left.

Also note that my numbers differ from shadow vector's. Eg: for StraightFlush4, his calculation "11 * 4 * 48" actually double-counts the straight-flushes, and then they're only subtracted once, so that's 40 extra. And there's 80 same-color straight flushes that should have been subtracted, instead of 40, because each straight flush could have either end changed to the other same-color suit. Don't worry, I got the same answer as you for that one when I did it by hand, and only figured out what was wrong after staring at the printout for quite awhile.
posted by vasi at 2:08 AM on May 14, 2011 [1 favorite]

You all rule. Thanks for helping me out, b/c I'm not sure I even knew how to figure out some of them ... and saved code to boot. I'm going to try to figure it out for a learning exercise. Thanks double for that!

except that seems a little silly somehow, as it basically just expands the definition of "straight flush".

Apologies again, madcaptenor, for the repeated misunderstandings. Again, as soon as I posted that last message, I realized what you meant. I agree if we're going strictly by probability it would be mostly silly to differentiate between the Straight Flush and the Same-Color Straight Flush (but I still think it might be necessary to differentiate between A-high straight flush, and A-high same-color straight flush).

That's why I wanted to get the odds breakdown in the first place. I think if I were going to introduce the Same-Color Straight Flush into a game, I'd rank it below 4 of a kind ... even though it's slightly harder to get.

What I am very thankful to have confirmed are: the ranks of the Canadian Straight and Flush (4 cards), i.e. CanStraight over CanFlush, and both higher than 2 pair.

And the ColorFlush vs. Flush4 is fascinating as well. I actually think I would rank ColorFlush above Flush 4 just for symmetry's sake.

For the record, I'm calling the game "NAFTA Woodshed" and designating the same-color flushes as "Mexican Flushes." (Hopefully no one finds that offensive?)

The game is 5-card stud, high/low declare. Ranks are, high to low:

High Card
Pair
Two Pair
Canadian Flush (4-card flush)
Mexican Flush (5 cards of same color)
Canadian Straight (4-card straight)
Three of a Kind
Mexican-Canadian Straight Flush (4-card straight of same color)
Straight
Flush
Full House
Canadian Straight Flush (4-card straight of same suit)
Mexican Straight Flush (5-card straight of same color)
Four of a Kind
Straight Flush
Five of a Kind (if wild cards)

Thanks again. I quite appreciate the help.
posted by mrgrimm at 11:28 AM on May 16, 2011