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Subject: The Inevitable Poker Thread rss

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K Septyn
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I was requested to pull out my poker-math spreadsheet and see if I could get some numbers for the Badger Deck. I believe the way I build the functions holds up under these conditions.

These numbers assume a few things. First, 10 suits with 22 ranks each (0-10 numbered plus 11 lettered/court cards). Second, the Ace is always high due to the presence of a One card in the ranks. Finally, a "rainbow" hand is 5 unmatched cards of 5 different suits--if there's a pair or higher, it counts as that type of hand and not a rainbow. (This is to avoid adding 6 boring and/or finicky hands like "rainbow two pair")


Total
Straight Flush 180
Five of a Kind 5,544
Flush 263,160
Rainbow Straight 544,320
Four of a Kind 970,200
Straight 1,255,500
Full House 2,494,800
Three of a Kind 55,440,000
Two Pair 93,555,000
Rainbow (Plain) 795,795,840
One Pair 1,316,700,000
High Card 1,835,541,000

Total Hands 4,102,565,544


4.1 billion possible poker hands. That'll keep you busy.

As usual when the number of ranks and suits change around, the flushes and straights go all cattywampus when compared to their normal positions. So does High Card vs One Pair, but at least they're in the normal order here.

Straight Flush is king of the heap with Five of a Kind right below it, which feels right given the contents of the deck. The Rainbow Straight is normally right below a Straight Flush in a standard 5-suited deck, but I don't think it's too far out of whack where it is.

A plain Rainbow hand looks almost worthless, but given the gap between it and Two Pair, it's probably a winner most of the time.

75% of your hands are going to be either High Card or One Pair, and 96% are going to be Rainbow or lower. That's not all too far from standard poker, with about 97% of the hands at Two Pair or lower.
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Chad Smith
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Ok, so I'm a little confused about the Rainbow...

If I am reading this correctly, there are more possible Pair hands than Rainbow hands (by a lot). That, and the location on the chart, makes me think a Rainbow would "beat" a pair.

However, the way you described the Rainbow, it makes it sound like a pair would negate a rainbow.

Meaning, if I had a 3 of Hearts, 5 of Spades, 6 of Stars, 7 of Diamonds, and 9 of Clubs - no two cards from the same suit... that would be a Rainbow.

But if I had a 3 of Hearts, 5 of Spades, 6 of Stars, 7 of Diamonds, and 7 of Clubs - again no two cards from the same suit... but this hand would have to be played as a Pair - and not a Rainbow.

Am I understanding that correctly?

What is confusing about that is - in any other Poker hand, if a hand can be called more than one way, it is always called by the *highest possible hand*. For example, a Royal Flush is also just a straight, but you play it as a Royal Flush. And 4-of-a-Kind could also be called 2 pairs, but you play it as a 4-of-a-Kind.

To me, that would mean a Rainbow hand that also happens to have a pair in it, would be counted as a Rainbow, since that hand is more rare, and would logically be more "valuable" than a Pair.

Does that make sense? What am I missing?
 
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Vic DiGital
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I think what he's trying to do is eliminate many of the sub-combos you can make within a rainbow. Such as the Rainbow with a Pair in your example. Rather than making that an established hand you can shoot for (A "Rainbow With a Pair" beats a plain Rainbow), it just keeps it simple by making a only a Rainbow or a Rainbow straight. So no "Rainbow Two Pair" or "Rainbow Full House" or "Rainbow Four of a Kind"

However, I AM confused as to why he put the Pair beating a plain Rainbow, when just purely looking at the probabilities he provided, the Rainbow is significantly harder to get than a Pair. If it's a Rainbow, it should just automatically beat a Pair.

I think maybe what should be done is WITHIN the confines of a Rainbow, let's say two players both reveal Rainbow hands, THEN you look to see if there are any poker sub-ranks that would break a Rainbow tie. So a Rainbow with a pair of Queens would beat a Rainbow with a pair of 5's. Since all other Two-pairs, Full Houses and Four-of-a-Kinds beat a plain Rainbow, you'd always just take that value instead of looking at it as a Rainbow. The only time you'd ever have to do the internal Rainbow ranking is for pairs within it.
 
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K Septyn
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This space reserved for an explanation when I get a chance to do so.

* * * * *

I have time now.

chad78 wrote:
What is confusing about that is - in any other Poker hand, if a hand can be called more than one way, it is always called by the *highest possible hand*. For example, a Royal Flush is also just a straight, but you play it as a Royal Flush. And 4-of-a-Kind could also be called 2 pairs, but you play it as a 4-of-a-Kind.


Yes, the cards speak for themselves. So if you erroneously think you have Two Pair, but it turns out it's "a pair of ones and another pair of ones" (thank you, Bugs) it's a Four of a Kind regardless of what you think.

I get what you're saying about a Rainbow Pair being a Rainbow and not a (plain) Pair. I just think the definition makes for simpler hand ordering. Here's the table when you break out all the special Rainbow combinations:


Straight Flush 180
Five of a Kind 5,544
Flush 263,160
Four of a Kind 388,080
Rainbow Straight 544,320
Rainbow Fours 582,120
Rainbow Full House 1,164,240
Straight 1,255,500
Full House 1,330,560
Rainbow Trips 23,284,800
Three of a Kind 32,155,200
Rainbow Two Pair 34,927,200
Two Pair 58,627,800
Rainbow Pair 442,411,200
Rainbow (Plain) 795,795,840
One Pair 874,288,800
High Card 1,835,541,000


This is my first time looking at this particular ordering, so you get to hear my initial thoughts. We've added five new hands to the list, which isn't a decision I like much. Since we're already messing with the normal hand order, adding two new ranks instead of seven is more appealing to me.

You could argue for making the Rainbow version of a hand more valuable than the plain version, but then you're putting the person who gets the rare plain Four of a Kind, which is harder to get than the Rainbow version. A Rainbow Straight also gets absorbed into the regular Straight ranking, and that doesn't feel right either. And if you make an exception for Rainbow Straight and maybe Rainbow Pair, why not the others? Then you're back to 17 type of hands.

If you go with your idea of "the better hand wins", and assuming that all Rainbow Pairs are counted as Rainbows, the bottom of the table looks like this:


One Pair 874,288,800
Rainbow (pair+plain) 1,238,207,040
High Card 1,835,541,000


Now the tables are reversed! A Rainbow that has a pair in it should count as One Pair, since that's the better hand. But then we're back to my original table.

* * * * *

Some of my definition of a Rainbow comes from the (semi) simplification of hands and hand order, but some is also from a little-used definition of a Flush.

A Flush is 5 cards of the same suit, right? Yes, but it can also be considered 5 unmatched cards of a single suit. Then the Rainbow definition of 5 unmatched cards of 5 unmatched suits follows as the half-inverse of a Flush. (Five of a Kind is the true inverse of a Flush.)

I've done some of the math for duplicate-ranked suits in a poker deck, and it gets ugly. If you go with the simple definition of a Flush as "all the same suit", you have to figure out where hands like "Flush Two Pair" fit into the scheme of things...and so do the players.

* * * * *

Translating poker to expanded decks is tricky. It seems like it should be simple, but it's more like trying to move chess to a different tessellation. Sure, Glinksi's Hexagonal Chess does a really good job of it, but McCooey's seems to be a good translation too. Or the expansion of Scrabble into Super Scrabble--the designer tried simply using two sets of letter tiles, but playtesting showed that wasn't the best solution.

Maybe playtesting would show that a Rainbow Pair is a desired hand, and all the others are so rare that they'll never really be a problem. Personally, I think poker with a deck this big would be fun for about 5 hands, then it would be a game of "why are we doing this again?"
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K Septyn
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VicDigital wrote:
However, I AM confused as to why he put the Pair beating a plain Rainbow, when just purely looking at the probabilities he provided, the Rainbow is significantly harder to get than a Pair. If it's a Rainbow, it should just automatically beat a Pair.


No, I'm confused. In the original table I posted, a Rainbow does beat a Pair. I'm not sure where you're looking....

 
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Chad Smith
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Ok - I think we're talking past each other a little bit.

Here is the line I originally had problems with:

VlcDigital wrote:
Finally, a "rainbow" hand is 5 unmatched cards of 5 different suits--if there's a pair or higher, it counts as that type of hand and not a rainbow


What it says here is "if there is a Pair or higher" then it is "Not a Rainbow".

But a Rainbow is HIGHER than a Pair.

What I don't want is for a Pair to negate a Rainbow. Two Pairs, sure. Two pairs are harder to get than a Rainbow. But a Rainbow that has a single Pair and no other special combination of cards should be a Rainbow.

If you change it to: "Finally, a Rainbow hand is 5 unmatched cards of 5 different suits--if there's Two Pair or higher, it counts as that type of hand and not a Rainbow."

Then we are all set.
 
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Vic DiGital
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Septyn wrote:
VicDigital wrote:
However, I AM confused as to why he put the Pair beating a plain Rainbow, when just purely looking at the probabilities he provided, the Rainbow is significantly harder to get than a Pair. If it's a Rainbow, it should just automatically beat a Pair.


No, I'm confused. In the original table I posted, a Rainbow does beat a Pair. I'm not sure where you're looking....



In your table, yes, you have the Rainbow above the Pair, which is what I was acknowledging. My confusion came from your explanation at the top which contained "if there's a pair or higher, it counts as that type of hand and not a rainbow," which to me (and Chad, also, I think) read as "it's a Rainbow UNLESS it has a pair in it, in which case, it would be considered a Pair and no longer a Rainbow". That didn't make sense because the hand was being penalized for having a lower-ranked Pair in the midst of its Rainbow. I THINK it was meant to say/mean "If a Rainbow has TWO PAIRS in it, then the lower-ranked Rainbow ranking is ignored in favor of the higher-ranked Two Pair designation."

For this, yes, I totally agree that this reduction of the number of Rainbow sub-suits is the best way to go. For a game I created using poker as the combat mechanism and a five-suited deck of cards (the 5 Dimension deck), I had all the Rainbow sub-rankings included, and it just totally bogged everything down. While it makes logical and statistical sense to have all the sub-rankings, it was ultimately counter-intuitive. This current, simplified rankings is very intuitive. The plain Rainbow and the Rainbow Straight fall exactly where you'd estimate them to go.

I'd abandoned that game idea, but with this new simplified ranking, and having the Badger Deck's unlimited number of suits, it makes my game (Vendetta Poker) something I'm excited to re-examine.
 
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K Septyn
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Now I see some of the confusion. It's all about what should be done with a Rainbow Pair.

The basic numbers look like this:

Rainbow Pair 442,411,200
Rainbow (Plain) 795,795,840
One Pair 874,288,800


When you count the Rainbow Pair as *just* a Pair (so by definition it ranks as a Pair, not by how it looks or feels), you get:

Rainbow (Plain) 795,795,840
One Pair 1,316,700,000


You get the apparent contradiction of a Rainbow with no pair beating a rainbow with a pair, since it "feels right" that the pair should make it worth more.

When you count the Rainbow Pair as just another Rainbow hand (again a definition thing), you get:

One Pair 874,288,800
Rainbow (pair+plain) 1,238,207,040


So in this case you could say a Rainbow with a pair beats a Rainbow that doesn't, which clears up the first problem but leads to a new one. Now you'll have a Rainbow (A-A-K-Q-J) lose to a Pair (2-2-5-4-3). (It also implies that a Rainbow Four of a Kind should beat a plain Four of a Kind, when that's not the case in terms of the numbers.)

So probably the best solution of the three is to bite the bullet and add Rainbow Pair as a hand of its own, and leave rainbows integrated into everything else except the Rainbow Straight.

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Chad Smith
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Septyn wrote:
Now I see some of the confusion. It's all about what should be done with a Rainbow Pair.

The basic numbers look like this:

Rainbow Pair 442,411,200
Rainbow (Plain) 795,795,840
One Pair 874,288,800


....

So probably the best solution of the three is to bite the bullet and add Rainbow Pair as a hand of its own, and leave rainbows integrated into everything else except the Rainbow Straight.
+1 This makes the most sense to me.
 
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Vic DiGital
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Septyn wrote:
One Pair 874,288,800
Rainbow (pair+plain) 1,238,207,040[/c]


I see what you're doing, but I don't think that logic works.

A plain rainbow and a Rainbow Pair (or Rainbow anything) aren't two different things that should be added together for these purposes. A Rainbow Pair by definition can't be made without it being a plain Rainbow first and foremost. So the smaller Rainbow Pair probabilities falls WITHIN the larger Rainbow probability. It would be like taking the probability sum of Full House, Three of a Kind and a Pair, adding all those together and saying that this combination of cards is more common than a standard Pair.

So I think you just use the Rainbow Straight, right where it is, and the plain Rainbow, right above One Pair, and the rankings are perfect. A Rainbow Pair would only need to be consulted when hands need to be ranked within two Rainbow hands. A Rainbow Pair would rank higher than a plain Rainbow, but when all is said and done, that Rainbow still counts as only a Rainbow, ranked right above a pair.

You COULD probably use Rainbows as unofficial tie-breakers. So if two players play matching hands of say 3 Kings, a seven, and a five, then if one of those hands happens to be a Rainbow, it would win that tie. Otherwise, that Rainbow Trips doesn't come into play. If player two has the Rainbow Trips, but his fifth card is a four instead of a five, then he loses that hand based on the values of the cards first, even though he's got a Rainbow.

You'll have to take my word for it that this IS an elegant, exciting solution to the multi-suited poker rankings problem. None of the playtests of my Vendetta Poker went smoothly because we had to consult a jumbled ranking chart every single time. Even putting the chart ON EACH CARD didn't speed things up.

Once the Badger Deck is available, I'm going to redo the game using his cards and post it in here somewhere for anyone to playtest. Maybe I'm deluding myself that these rankings are as intuitive and streamlined as I now think they are, lol!
 
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VicDigital wrote:
Septyn wrote:
One Pair 874,288,800
Rainbow (pair+plain) 1,238,207,040[/c]


I see what you're doing, but I don't think that logic works.

A plain rainbow and a Rainbow Pair (or Rainbow anything) aren't two different things that should be added together for these purposes. A Rainbow Pair by definition can't be made without it being a plain Rainbow first and foremost. So the smaller Rainbow Pair probabilities falls WITHIN the larger Rainbow probability. It would be like taking the probability sum of Full House, Three of a Kind and a Pair, adding all those together and saying that this combination of cards is more common than a standard Pair.
This should not be nearly as confusing as we are making it.

A Hand must always *only* be counted as its highest form.

All Full House Hands are *only* Full House hands, they are not any other thing. They are not counted as Pairs. They are not counted as Three-of-a-Kinds, and they are not counted as Two-Pairs. They are always *only* counted as Full Houses.

If a Rainbow is 5 cards of different suits, then it would not include any other higher hands. A Full House that is also a Rainbow would always only be counted as a Full House. Same with Straights, 5-of-a-kind, 4-of-a-kind, 3-of-a-kind, 2 pair, whatever.

The only real questions are:
* Where does a Rainbow fall? More rare than a Pair or less?
* Is a Higher hand (something more rare than *just* a Rainbow) considered more rare if it is a Rainbow version. And if so - in what cases would that apply with regards to scoring? (Would it be treated as "Trumps" in all cases - like in standard Poker if two players both had a Royal Straight, the one with the Spades would win)


I don't know the math behind the numbers above. But if I were going to try to figure out the math, I would start with the total number of possible hands, then figure out the *HIGHEST RANK* first. After subtracting all of those hands, I would move down the list until I was left with "High Card" (which would be any hand that didn't fit any other defition).

So, again, the higher ranked hands should not be included in number for the lower ranked hands. If it is called a "Pair" - that means it is *only* a Pair.
 
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Vic DiGital
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chad78 wrote:
VicDigital wrote:
Septyn wrote:
One Pair 874,288,800
Rainbow (pair+plain) 1,238,207,040[/c]


I see what you're doing, but I don't think that logic works.

A plain rainbow and a Rainbow Pair (or Rainbow anything) aren't two different things that should be added together for these purposes. A Rainbow Pair by definition can't be made without it being a plain Rainbow first and foremost. So the smaller Rainbow Pair probabilities falls WITHIN the larger Rainbow probability. It would be like taking the probability sum of Full House, Three of a Kind and a Pair, adding all those together and saying that this combination of cards is more common than a standard Pair.
This should not be nearly as confusing as we are making it.

A Hand must always *only* be counted as its highest form.

All Full House Hands are *only* Full House hands, they are not any other thing. They are not counted as Pairs. They are not counted as Three-of-a-Kinds, and they are not counted as Two-Pairs. They are always *only* counted as Full Houses.

If a Rainbow is 5 cards of different suits, then it would not include any other higher hands. A Full House that is also a Rainbow would always only be counted as a Full House. Same with Straights, 5-of-a-kind, 4-of-a-kind, 3-of-a-kind, 2 pair, whatever.

The only real questions are:
* Where does a Rainbow fall? More rare than a Pair or less?
* Is a Higher hand (something more rare than *just* a Rainbow) considered more rare if it is a Rainbow version. And if so - in what cases would that apply with regards to scoring? (Would it be treated as "Trumps" in all cases - like in standard Poker if two players both had a Royal Straight, the one with the Spades would win)


I don't know the math behind the numbers above. But if I were going to try to figure out the math, I would start with the total number of possible hands, then figure out the *HIGHEST RANK* first. After subtracting all of those hands, I would move down the list until I was left with "High Card" (which would be any hand that didn't fit any other defition).

So, again, the higher ranked hands should not be included in number for the lower ranked hands. If it is called a "Pair" - that means it is *only* a Pair.

(I have to laugh at how intensely we're discussing the finer points of rainbows)

The debate isn't over whether or not the Rainbow sub-ranks (pairs, trips, full house, straight) are legitimate ranks (they are), but whether or not a game wants to take on the complexity of including ALL the Rainbow sub-ranks. There's no way to include every single one of them without having to always consult a chart every time you draw a card that might shift you over into Rainbows. Including everything is the most accurate way of accounting for everything, but it definitely slows any game down to a halt.

For myself, I'm going to lean towards simplicity. Maybe even as simple is eliminating all Rainbow ranks from the list (except for plain Rainbow right above One Pair), and simply make a Rainbow Pair/Trips/Straight/Etc, the highest version of that rank. A Rainbow straight always beats a regular straight. But all you're having to mentally calculate is the fact that your straight is going to beat that Full House.


Or maybe it should just be divided into FULL Rainbow Poker and BASIC Rainbow Poker and you can choose the level of Analysis Paralysis you want to inject into your poker.

I'm in favor of double rainbows, all the way.

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K Septyn
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The Inevitable Poker Thread: The Inevitable Sequel
With the extended version of the Badger Deck available, I thought I'd update the table for any masochists that want to play poker with a deck of 320 cards.

The same assumptions apply with a couple tweaks. 10 suits of 32 ranks each (0-20 + 11 face cards), where Ace is always high. Rainbow hands are counted with their normal counterparts with the exception of Rainbow Straight and Rainbow Pair. A (plain) Rainbow is still 5 unmatched cards of 5 different suits.


Straight Flush 280
Five of a Kind 8,064
Rainbow Straight 846,720
Straight 1,953,000
Flush 2,013,480
Four of a Kind 2,083,200
Full House 5,356,800
Three of a Kind 178,560,000
Two Pair 301,320,000
Rainbow Pair 2,174,860,800
One Pair 4,297,939,200
Rainbow 6,088,763,520
"Runt" 14,044,023,000

Total Hands 27,097,728,064


You have an 0.045% chance to get anything better than a three of a kind. If you get one, I suggest going all-in.
 
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