Odds of Hypothetical Hands in Poker

[Chronos]

The thing about poker hands is that they are unique in this sense, you cannot have two pair and have a straight. You cannot have a full house and have a flush. The only combintation hands are the straight flushes. These new hands create too many (IMO) more combination hands. Two-pair/all-color, one of each/fullhouse etc.

What if you have wild cards? Suppose, for instance, I’m holding a 2h, a 5h, a 6h, a 9h, and a joker (assuming jokers wild). Could I say that the joker is another 9h, and thereby have both a flush and a pair of nines? Even if one can duplicate a specific card with a joker, though, I suspect that hand would be ranked just as the higher possibility, so it’d still lose to, say, a 10-high flush.

And I’ve pointed out in the past that “royal flush” is just a fancy name for “straight flush, ace high”. But I’ve heard that some house rules recognize Five of a Kind (possible with wild cards) as a valid hand. Assuming two jokers and nothing else wild, there are (I calculate) 64 possible Royal Flush hands, 78 possible Five of a Kind hands, and 540 non-royal Straight Flush hands. So in that case, it would not be meaningless to rule that Five of a Kind beats an ordinary Straight Flush, but a Royal Flush still beats Five of a Kind. It still wouldn’t necessarily be reasonable, though, because you could then point out that there are only six possible hands for Five Aces, so Five Aces (by that logic) ought to beat a Royal Flush.

[Cabbage]

@Chronos

What if you have wild cards? Suppose, for instance, I’m holding a 2h, a 5h, a 6h, a 9h, and a joker (assuming jokers wild). Could I say that the joker is another 9h, and thereby have both a flush and a pair of nines? Even if one can duplicate a specific card with a joker, though, I suspect that hand would be ranked just as the higher possibility, so it’d still lose to, say, a 10-high flush.

And I’ve pointed out in the past that “royal flush” is just a fancy name for “straight flush, ace high”. But I’ve heard that some house rules recognize Five of a Kind (possible with wild cards) as a valid hand. Assuming two jokers and nothing else wild, there are (I calculate) 64 possible Royal Flush hands, 78 possible Five of a Kind hands, and 540 non-royal Straight Flush hands. So in that case, it would not be meaningless to rule that Five of a Kind beats an ordinary Straight Flush, but a Royal Flush still beats Five of a Kind. It still wouldn’t necessarily be reasonable, though, because you could then point out that there are only six possible hands for Five Aces, so Five Aces (by that logic) ought to beat a Royal Flush.

One pretty standard thing about wild cards is that they can’t duplicate an already existing card. In particular, you can’t use a wild card to have a “flush with a pair of nine’s” or anything like that. In your example, the joker would be counted as the ace of hearts, since that’s the highest possible hand.

One exception to this (as you mentioned) is five of a kind. If you have all four aces plus a wild card, then you can count your wild card as a “generic” (no particular suit) ace, for five aces.

The rules I usually play by are that any five of a kind beats any straight flush (including a royal straight flush), but I expect there are variations on this.

[JustAnotherGeek]

Ok, how geeky is this:

I am running a routine in MatLab that should calculate for us the odds of the different hands. Assuming I have everything done correctly, it should be done in about an hour. Since I am starving, I left the program to run at work so I could go home for dinner. Since I am lazy, I ain’t goin back tonight for the results. But, in any event, some time tomorrow (whenever I decide to get my lazy butt up), we will have the (hopefully) definitive results.

@Gangster Octopus

The thing about poker hands is that they are unique in this sense, you cannot have two pair and have a straight. You cannot have a full house and have a flush. The only combintation hands are the straight flushes. These new hands create too many (IMO) more combination hands. Two-pair/all-color, one of each/fullhouse etc.

True. I have it set up (I hope) such that in order to qualify for a new hand (Color or Four Suits), it cannot qualify for any existing hands. I figure that they will probably end up only modifying the High Card category. (But, of course, I will have some numbers tomorrow.)

[Mathochist]

@JustAnotherGeek

True. I have it set up (I hope) such that in order to qualify for a new hand (Color or Four Suits), it cannot qualify for any existing hands. I figure that they will probably end up only modifying the High Card category. (But, of course, I will have some numbers tomorrow.)

This is blatantly wrong. Just think how many all-colors are also pairs, or how many four-suits are also 4-of-a-kind.

What we need for raw data is a program to go through and assign a number from 1 to 1024 to a hand, saying which of the models (pair, two pair, 3 of a kind, straight, flush, full house, four of a kind, straight flush, all-one-color, one-of-each) it matches. Then, go through each hand and increment one of 1024 counters depending on which hands it matches.

Then we can easily count up how many hands satisfy each model. There are only about 2.5M hands, so it shouldn’t take that long to find which each matches.

That gives the basic probabilities, which let us rank all the hands. Then we count how many of the rarest hand there are, then how many of the second rarest which aren’t also the rarest, and so on…

[JustAnotherGeek]

@Mathochist

This is blatantly wrong. Just think how many all-colors are also pairs, or how many four-suits are also 4-of-a-kind.

What we need for raw data is a program to go through and assign a number from 1 to 1024 to a hand, saying which of the models (pair, two pair, 3 of a kind, straight, flush, full house, four of a kind, straight flush, all-one-color, one-of-each) it matches. Then, go through each hand and increment one of 1024 counters depending on which hands it matches.

Then we can easily count up how many hands satisfy each model. There are only about 2.5M hands, so it shouldn’t take that long to find which each matches.

That gives the basic probabilities, which let us rank all the hands. Then we count how many of the rarest hand there are, then how many of the second rarest which aren’t also the rarest, and so on…

Hang on a sec, Math. I said that I have it set up so that the hands are mutually exclusive. I.e. if a hand qualifies for four of a kind, it is not considered to be “Four Suited.” By the definition given, a four of a kind must be a four suited; I have modified the definitions such that a hand is considered for either Red/Black or Four Suits only if it does not even qualify for a pair.

My program should spit out 13 different numbers; number of hands of:

“Old Fashioned” High Card
High Card
Red / Black
Four Suit
Pair
2 Pair
Three of a Kind
Straight
Flush
Full House
Four of a kind
Straight Flush (but not Ace high)
Royal Flush

in all 52C5.

After I make the numbers available, I will happily send to anyone my coding for your programming pleasure.

[ultrafilter]

How do you know in which bucket to place the hands that qualify as multiple categories without already knowing which ones are the least likely?

[Mathochist]

@JustAnotherGeek

Hang on a sec, Math. I said that I have it set up so that the hands are mutually exclusive. I.e. if a hand qualifies for four of a kind, it is not considered to be “Four Suited.” By the definition given, a four of a kind must be a four suited; I have modified the definitions such that a hand is considered for either Red/Black or Four Suits only if it does not even qualify for a pair.

But why do you assume the new categories are on the bottom of the ordering? Just as “pair” means “pair but not three” because pair is lower than three, you’re assuming that four-suit is lower than pair.

What I’m saying is we need to reconstruct the ordering of the hands. How do we do that? First we consider all their probabilities with duplications, then we sort them from rarest to commonest, then we find the probability that a given hand matches one model – and no higher models – in the new ordering. Your algorithm automatically puts the new models at the bottom of the stack, which isn’t justified at all.

[JustAnotherGeek]

@Mathochist

<snip>
Your algorithm automatically puts the new models at the bottom of the stack, which isn’t justified at all.

Not my algorithm, my hunch. If you will notice in the post where I first mention that I am actually running the program, I use the phrase “I figure.” As I’ve said, I will post numbers once I get them.

On the new hands: You can define them as you like. I have defined them as I like. The way I have defined them is to not have one five-card combination be described by two different hands in order to continue with the mutual exclusivity. If you want to take non-exclusivity into account, you are free to do so.

[Mathochist]

@JustAnotherGeek

On the new hands: You can define them as you like. I have defined them as I like. The way I have defined them is to not have one five-card combination be described by two different hands in order to continue with the mutual exclusivity. If you want to take non-exclusivity into account, you are free to do so.

Geek, be true to your nickname and read what I actually wrote. I do make the hands mutually exclusive at the end.

Look, just consider the classical poker hands. Any full house is automatically a pair as well. So, why is it counted as a full house rather than a pair? Because we’ve ordered the hands and only count the highest hand a given set of five cards satisfies.

What you’ve described throws the new hands in at the bottom of the order. Just as a deal which is both a full house and a pair gets counted as a full house (the rarer hand), so should a hand which is both a pair and a four-suit get counted as whichever is rarer. you simply count it as a pair, while I establish which is rarer first and only then decide which it should count as.

[JustAnotherGeek]

@Mathochist

What you’ve described throws the new hands in at the bottom of the order. Just as a deal which is both a full house and a pair gets counted as a full house (the rarer hand), so should a hand which is both a pair and a four-suit get counted as whichever is rarer. you simply count it as a pair, while I establish which is rarer first and only then decide which it should count as.

Math, I see what you are saying, but we are talking about the definitions, not the ranking. I have defined a “Four Suited” hand as one which does not contain a pair. That does not rank the hand, it defines it. I understand what you are saying, but I disagree with your example. The hand “a Pair” by its definition cannot also contain another pair (or three of a kind). Granted, a “Full House” contains both a pair and a three of a kind, but it qualifies as the hand of neither.

I think that you are looking at the definitions for the hands as non-exclusive sets. I.e. The Full House would fall into the set of “hands that have two identically numbered cards” and “hands that have three identically numbered cards.” I have defined “Three of a Kind” to be a set of cards that contains exactly three identically numbered cards and the remaining two are not identically numbered. I have defined “Four Suits” to be a hand that does not qualify for the standard poker hand categories and contains all four suits. The definitions, not the decision of the player make the categories mutually exclusive. (The player does not get to choose “Three of a Kind” when he has a “Full House.” In fact, in many casinos, this act results in the loss of the hand, even if the “Three of a Kind” beats the rest of the hands on the table.)

Anyway, I now am back at work, the program seems to have run correctly, and I will post the results momentarily. If anyone wants my code (it ain’t pretty), email me. I am pretty sure my email is in my profile.

[Gangster_Octopus]

Let’s just stick with the four suit for now. Keep in mind that four suit actually has a fifith card that can be any suit (of course).

So here is what yoiu can have wtha four suiter:
High Card
One pair
Two pair
three of a kind
Straight
Full house

So you actually can duplicate every hadn except the flush hands with a four suiter. Now while it might seem that all you have to do is rank them this way, it isn’t necessarily clear to me that, for example a three of a kind four suiter is rarer than a two of a kind four suiter. And a straight four suiter might actually be rarer than full house full suiter. IAGin I just don’t know, but you can see it duplicates many hands that already exist.

[JustAnotherGeek]

The moment you have been waiting for is here!

::trumpets blare::

::confetti is thrown::

::angels sing::

Alright, most folks on this board probably couldn’t care less, but for the couple of you who subscribed to this thread, here ya go:

Classic Poker Hands

Hand...............Number..................Odds
High Card:.........1,302,540...............1:2.00
One Pair:..........1,098,240...............1:2.37
Two Pair:............123,552...............1:21.0
Three of a Kind:......54,912...............1:47.3
Straight:.............10,200...............1:255
Flush:.................5,108...............1:509
Full House:............3,744...............1:694
Four of a Kind:..........624...............1:4,165
Straight Flush:...........36...............1:72,193
Royal Flush:...............4...............1:649,740

Improvised Poker Hands

Hand...............Number..................Odds
One Pair:..........1,098,240...............1:2.37
High Card:...........919,440...............1:2.83
Four Suit:...........306,480...............1:8.48
Two Pair:............123,552...............1:21.0
Color Flush:..........76,620...............1:33.9
Three of a Kind:......54,912...............1:47.3
Straight:.............10,200...............1:255
Flush:.................5,108...............1:509
Full House:............3,744...............1:694
Four of a Kind:..........624...............1:4,165
Straight Flush:...........36...............1:72,193
Royal Flush:...............4...............1:649,740

Some discussion on the “Improvised” hands:

The two new hands must not contain any other kind of hand. E.g. a “Color Flush” that has a pair in it is “One Pair.” Yes, I know that it ranks the hand lower. Thems is the breaks, kid. Also, if you’ll notice, “High Card” (meaning no pair, no flush, etc.) is now harder to get than a “One Pair.” So “One Pair” officially sucks.

As you can see, this now really mucks up the order of ranking. If you really want to add in a new hand, I would suggest adding in only the “Color Flush.” Given my definitions, this should combine the “Four Suit” back into the “High Card.” This would make the chances of getting only a high card smaller than that of getting only one pair.

If, as Mathochist is suggesting, we redefine “Color Flush” to include the possiblity of containing, for instance, a pair or two pair, it would only take an additional ~50,000 hands to drop its ranking below “Two Pair.” Perhaps, if someone gets ambitious, they could run my code to see how many “One Pair” hands are also a color flush. Then they could recalculate the odds and rankings.

[Dag_Otto]

@JustAnotherGeek

The player does not get to choose “Three of a Kind” when he has a “Full House.” In fact, in many casinos, this act results in the loss of the hand, even if the “Three of a Kind” beats the rest of the hands on the table.

What casinos? In every card room I’ve played at, the rule is ‘Cards Speak’.

Now if it was done a some sort of angle shooting, I could see the floor being called for a decision, but the usual rule is best hand wins.

[JustAnotherGeek]

Ok, I ran some code to determine the number of previously defined “One Pair” hands that also qualify for “Color Flush.” If “Four Suit” was combined with High Card, and “Color Flush” could contain one pair, the first couple of categories would look like this:

Improvised Poker Hands 
Hand...............Number..................Odds 
High Card:.........1,225,920...............1:2.12 
One Pair:..........1,098,240...............1:2.47 
Two Pair:............123,552...............1:21.0 
Color Flush:.........122,380...............1:21.2
Three of a Kind:......54,912...............1:47.3 
<snip>

…but I didn’t check to see how many “Two Pair” also qualify as a “Color Flush.” If more than 1,172 do, then that means that a “Color Flush” is lower than “Two Pair.” Which would mean that you can’t count a Two Pair/Color Flush as a Color Flush. Which would mean that a Color Flush is more rare than a Two Pair, but then that means…

…so it’s a useless hand. I will run the code one more time to figure out haw many TP/CF combinations there are. I have a hunch that we will have to eliminate this combination. So, sorry, HeyHomie.

[JustAnotherGeek]

Last post (unless someone asks a question).

There are 3,432 combinations of cards that qualifiy as both a “Color Flush” and a “Two Pair” (if the former is not exclusively defined). Thus, unless one uses totally exclusive definitions for “Color Flush” (i.e. it cannot also qualify for One Pair or Two Pair), you will not be able to place it in the rankings. If you allow a Color Flush to contain one pair, but not two, it is more rare than a Two Pair. But, if you allow a Color Flush to have either a pair or two pair, it is more common than a Two Pair. (In which case, all CF/TP combinations would be counted as a Two Pair, making CF harder to get than a TP. Which means that all TP/CF would be counted as a CF…)

In short, there is statistical reason why there is no such hand as a “Color Flush.”

[JustAnotherGeek]

bah, sorry, I just reread my posts. In post 34, the number of One Pair hands (that are not also a Color Flush) should read as 1,052,480. The odds are right.

[ultrafilter]

@JustAnotherGeek

There are 3,432 combinations of cards that qualifiy as both a “Color Flush” and a “Two Pair” (if the former is not exclusively defined). Thus, unless one uses totally exclusive definitions for “Color Flush” (i.e. it cannot also qualify for One Pair or Two Pair), you will not be able to place it in the rankings. If you allow a Color Flush to contain one pair, but not two, it is more rare than a Two Pair. But, if you allow a Color Flush to have either a pair or two pair, it is more common than a Two Pair. (In which case, all CF/TP combinations would be counted as a Two Pair, making CF harder to get than a TP. Which means that all TP/CF would be counted as a CF…)

You’re horribly confused here. Whether a TP/CF hand counts as a two-pair or a color flush depends only on which is more likely. You shouldn’t be making any assumptions about which is more desirable in order to do the calculation.

[ultrafilter]

@JustAnotherGeek

There are 3,432 combinations of cards that qualifiy as both a “Color Flush” and a “Two Pair” (if the former is not exclusively defined). Thus, unless one uses totally exclusive definitions for “Color Flush” (i.e. it cannot also qualify for One Pair or Two Pair), you will not be able to place it in the rankings. If you allow a Color Flush to contain one pair, but not two, it is more rare than a Two Pair. But, if you allow a Color Flush to have either a pair or two pair, it is more common than a Two Pair. (In which case, all CF/TP combinations would be counted as a Two Pair, making CF harder to get than a TP. Which means that all TP/CF would be counted as a CF…)

You’re horribly confused here. Whether a TP/CF hand counts as a two-pair or a color flush depends only on which is more likely. You shouldn’t be making any assumptions about which is more desirable in order to do the calculation.

[JustAnotherGeek]

@ultrafilter

Whether a TP/CF hand counts as a two-pair or a color flush depends only on which is more likely. You shouldn’t be making any assumptions about which is more desirable in order to do the calculation.

Given that the rankings of the hands are based on their respective probabilities, The CF hand is meaningless, if not forced to be exclusive of all other hands. (I.e. by forcing some combination to the lower ranking.)

If a CF hand is defined as exclusive to all other hands (i.e. PR/CF = PR and TP/CF = TP), it would be ranked above TP.

If a CF is defined as exclusive to TP, but inclusive to one pair (i.e PR/CF = CF and TP/CF = TP), it still ranks above TP, but only just barely.

If a CF is defined as inclusive to both one and two pair (i.e PR/CF = CF and TP/CF = CF), it is ranked below TP. However, in that case, TP/CF is forced to be ranked lower than its higher possible ranking (of TP).

To recap, because CF can include too many other hands, unless it is artificially forced to be exclusive (or some other hand(s) is(are) forced), there will be conflicts in the ranking structure. (In that a hand will be artificially lowered in the ranking scheme.)