I'm trying to model a scenario where we have 5 standard decks of 52 cards, and are dealing a 5-card hand, each of the 5 cards coming from a discrete deck of cards.

This is the equivalent of:

1 Shuffle the deck

2 Deal 1 card

3 Make note of what the card is

4 Replace the card into the deck

5 Repeat four additional times for a total of 5 cards dealt in total

The easy part:

There are 52

^{5}= 380,204,032 possible results.

The not-so-easy part:

We now have a number of hands that would not otherwise be possible with just one deck of card (ie an Identical Pair which consists of 2 cards of both equal rank AND suit, or three of a kind, two of which are identical).

a) How do we go about working out the probabilities of these new hands, and b) other than the fact there are more possible results, are the number of "standard" poker hand combinations the same? (ie are there still 5,108 possible "flushes"? Intuitively I think "yes", but I've been wrong in probability situations before, so I want to be certain)

Thank you!

Quote:TechnoWizardHello everyone, this is my first post to this forum, so please be forgiving ;)

I'm trying to model a scenario where we have 5 standard decks of 52 cards, and are dealing a 5-card hand, each of the 5 cards coming from a discrete deck of cards.

This is the equivalent of:

1 Shuffle the deck

2 Deal 1 card

3 Make note of what the card is

4 Replace the card into the deck

5 Repeat four additional times for a total of 5 cards dealt in total

The easy part:

There are 52^{5}= 380,204,032 possible results.

The not-so-easy part:

We now have a number of hands that would not otherwise be possible with just one deck of card (ie an Identical Pair which consists of 2 cards of both equal rank AND suit, or three of a kind, two of which are identical).

a) How do we go about working out the probabilities of these new hands, and b) other than the fact there are more possible results, are the number of "standard" poker hand combinations the same? (ie are there still 5,108 possible "flushes"? Intuitively I think "yes", but I've been wrong in probability situations before, so I want to be certain)

Thank you!

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Welcome!

If you're only dealing one hand, yes, your method of shuffling and reusing the same deck seems equivalent.

I infer from some of the rest of the post that there may be two or more hands in play, for a comparison of relative hand rankings. In this case, reusing the single deck is NOT equivalent, as a card dealt to the first position cannot be in the hand at the second position, but your method may allow that.

A five of a kind flush is not a standard poker hand, but would be (remotely) possible in your game. I think the number of combinations changes based on that, and similar newly possible hands.

I'm writing a software program to emulate this approach, and while I have the dealing and displaying of the cards working fine, I'm now trying to work out the quantities of possible hands - for example, 3 of a kind, 2 of which are of identical suit. (JH, JH, JS, X, X) appearing in any order across the 5 cards dealt.

I'm not much on combinatorics, so I'm backing out here.

Best of luck!

Calculating the total possible of each hand type wouldn’t be too tough, except for the overlaps with flushes by pairs, two pairs, trips, boats, quads, and don’t forget 5 of a kinds! So before we begin, we should decide whether or not we are ok with double counting some hands, and if not then come up with a hierarchy so we know to which type each overlap belongs.

Edit to add: The calculations of the overlaps shouldn’t be too tough, but would it simplify gameplay to just eliminate flushes altogether?

Quote:miplet/games/poker/#infinite-decks

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Thank you! That will be very helpful, I believe.

It gets a little bit interesting when you get "Triplets, 2 Identical" or a "Full House Flush" (Full House with all cards the same suit), though.

Quote:unJonThis is a thread that needs US Paper Games

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????

Quote:charliepatrickOne possible approach is to look at 5-of-a-kind (call that "5"), then "41", then 32 then 311 etc. Only 11111 can be a straight, but any can be a flush with multiple decks. Your first hand is essentially infinite decks, since the 2nd card is independent from the 1st card etc. Normally you might only want to know flushes for hands Trips or lower, however it depends on your game.

link to original post

Initially, I thought of doing the same thing. That is, to assume the order of hands would be the same. Then it occurred to me that the math is different enough for some of the hands to the point of altering the order. (For example, consider the 5-card vs. 3-card hierarchies for single-deck.)

Note the table in the link provided by Miplet. The Full House, Flush, and Straight should be listed in reverse order for multi-deck from what we’re used to seeing for single deck. As the Flush is less likely than the Full House, we should remove the number of same-suit Full Houses from the Full House figure and add it to the Flush figure (assuming the Wiz added the totals of conflicting hands to whichever is higher on the list he created).

Also, 5 of a Kind and Straight Flush should be switched in the list. This won’t affect the totals, as they aren’t in conflict, but it may help with assigning payouts…

Quote:camaplQuote:charliepatrickOne possible approach is to look at 5-of-a-kind (call that "5"), then "41", then 32 then 311 etc. Only 11111 can be a straight, but any can be a flush with multiple decks. Your first hand is essentially infinite decks, since the 2nd card is independent from the 1st card etc. Normally you might only want to know flushes for hands Trips or lower, however it depends on your game.

link to original post

Initially, I thought of doing the same thing. That is, to assume the order of hands would be the same. Then it occurred to me that the math is different enough for some of the hands to the point of altering the order. (For example, consider the 5-card vs. 3-card hierarchies for single-deck.)

Note the table in the link provided by Miplet. The Full House, Flush, and Straight should be listed in reverse order for multi-deck from what we’re used to seeing for single deck. As the Flush is less likely than the Full House, we should remove the number of same-suit Full Houses from the Full House figure and add it to the Flush figure (assuming the Wiz added the totals of conflicting hands to whichever is higher on the list).

Also, 5 of a Kind and Straight Flush should be switched in the list. This won’t affect the totals, as they aren’t in conflict, but it may help with assigning payouts…

link to original post

I've been working through the list of hands. The so-called "standard" hands have the same possible quantities - that is, a FH with the 3 matching cards being different suits and the two matching cards being different suits ( [K♠] [K♣] [K♥] [2♥] [2♣] ) - there are still 3744 "Standard" Full Houses. The probability of a FH changes, obviously, because there are 52

^{5}possible combinations of cards.

Next, I worked out a "Full House, Triple-2-Identical" an example of which would be [K♠] [K♠] [K♥] [2♥] [2♣]

The combinations in this case are C(13,1) * C(4,2) * C(12,1) * C(3,1) = 13 * 6 * 12 * 3 = 2808. Which seems correct to me. Remember we already know the number of possible hands, so we aren't worrying about the order in which the cards get dealt - could someone check my math?

What's weird: I'm now stuck working out a "Quad-2I": [K♠] [K♠] [K♥] [K♦] [2♥]

The number of possible "Standard Quads" is C(13,1) * C(4,4) * C(12,1) * C(4,1) = 624. Intuitively, the number of possible Q2I hands should be less than a standard quad, given two cards must have matching suits; but I can't for the life of me figure out how to get to that result - I keep getting a result that is equal to or greater than 624.

Quote:camaplAre you just trying to verify the total figures, or do you actually need to know the intermediate figures (such as the number of trips comprised of two suits)?

link to original post

If you're asking about checking my math, I'm basically asking someone to run through a possible solution on the Triple-2-I that I shared and see if you come up with the same answer independently from me. If you do, it's likely that I've got it right.

HAND | PERMUTATIONS | REDUCTION |
---|---|---|

Royal flush | 480 | n/a |

Straight flush | 4,320 | n/a |

Five of a kind | 13,312 | n/a |

Four of a kind | 798,720 | n/a |

Straight | 1,224,000 | n/a |

Flush | 1,477,200 | 3,172 (52 Five of a kinds, 3,120 Four of a kinds) |

Full house | 1,591,200 | 6,240 Flushes |

Three of a kind | 17,503,200 | 68,640 Flushes |

Two pair | 26,254,800 | 102,960 Flushes |

Pair | 175,032,000 | 686,400 Flushes |

Nothing | 156,304,800 | n/a |

Except for non-winning hands (Nothing), rows are sorted from least likely (Royal Flush) to most likely (Pair). Totals in the Permutations column have been adjusted for Flush and Full House to account for overlaps in the totals of the higher, less likely hand, so as to conform with the rest of the table.

Edit to add the link provided by Miplet in a prior post:

https://wizardofodds.com/games/poker/#infinite-decks

Edit to add: I tried to right justify the 2nd column, but my 1/2-a’d attempt didn’t prevail… Sorry if it’s difficult to read!

Quote:TechnoWizardQuote:camaplAre you just trying to verify the total figures, or do you actually need to know the intermediate figures (such as the number of trips comprised of two suits)?

link to original post

If you're asking about checking my math, I'm basically asking someone to run through a possible solution on the Triple-2-I that I shared and see if you come up with the same answer independently from me. If you do, it's likely that I've got it right.

link to original post

Merely asking because coming up with totals including overlaps and subtracting out flushes was pretty straightforward (without having to calculate non-/partial flushes). If you don’t otherwise need the partials, then I probably wouldn’t bother with them… Hope the table above helps!

Quote:camaplSorting the table on WoO named Infinite Deck yields the following:

HAND PERMUTATIONS REDUCTION Royal flush 480 n/a Straight flush 4,320 n/a Five of a kind 13,312 n/a Four of a kind 798,720 n/a Straight 1,224,000 n/a Flush 1,477,200 3,172 (52 Five of a kinds, 3,120 Four of a kinds) Full house 1,591,200 6,240 Flushes Three of a kind 17,503,200 68,640 Flushes Two pair 26,254,800 102,960 Flushes Pair 175,032,000 686,400 Flushes Nothing 156,304,800 n/a

Except for non-winning hands (Nothing), rows are sorted from least likely (Royal Flush) to most likely (Pair). Totals in the Permutations column have been adjusted for Flush and Full House to account for overlaps in the totals of the higher, less likely hand, so as to conform with the rest of the table.

Edit to add the link provided by Miplet in a prior post:

/games/poker/#infinite-decks

Edit to add: I tried to right justify the 2nd column, but my 1/2-a’d attempt didn’t prevail…

(Link removed as I’m too much of a “Noob” to post links, apparently)

I looked at the table and I can’t understand why it shows 480 Royal Flushes. There are only 4 possible Royal flushes in a draw of 5 cards regardless of the number of decks being drawn from.

Quote:camaplSorting the table on WoO named Infinite Deck yields the following:

HAND PERMUTATIONS REDUCTION Royal flush 480 n/a Straight flush 4,320 n/a Five of a kind 13,312 n/a Four of a kind 798,720 n/a Straight 1,224,000 n/a Flush 1,477,200 3,172 (52 Five of a kinds, 3,120 Four of a kinds) Full house 1,591,200 6,240 Flushes Three of a kind 17,503,200 68,640 Flushes Two pair 26,254,800 102,960 Flushes Pair 175,032,000 686,400 Flushes Nothing 156,304,800 n/a

Except for non-winning hands (Nothing), rows are sorted from least likely (Royal Flush) to most likely (Pair). Totals in the Permutations column have been adjusted for Flush and Full House to account for overlaps in the totals of the higher, less likely hand, so as to conform with the rest of the table.

Edit to add the link provided by Miplet in a prior post:

/games/poker/#infinite-decks

Edit to add: I tried to right justify the 2nd column, but my 1/2-a’d attempt didn’t prevail…

(Link removed as I’m too much of a “Noob” to post links, apparently)

I looked at the table and I can’t understand why it shows 480 Royal Flushes. There are only 4 possible Royal flushes in a draw of 5 cards regardless of the number of decks being drawn from.

Quote:TechnoWizardQuote:camaplSorting the table on WoO named Infinite Deck yields the following:

HAND PERMUTATIONS REDUCTION Royal flush 480 n/a Straight flush 4,320 n/a Five of a kind 13,312 n/a Four of a kind 798,720 n/a Straight 1,224,000 n/a Flush 1,477,200 3,172 (52 Five of a kinds, 3,120 Four of a kinds) Full house 1,591,200 6,240 Flushes Three of a kind 17,503,200 68,640 Flushes Two pair 26,254,800 102,960 Flushes Pair 175,032,000 686,400 Flushes Nothing 156,304,800 n/a

Except for non-winning hands (Nothing), rows are sorted from least likely (Royal Flush) to most likely (Pair). Totals in the Permutations column have been adjusted for Flush and Full House to account for overlaps in the totals of the higher, less likely hand, so as to conform with the rest of the table.

Edit to add the link provided by Miplet in a prior post:

/games/poker/#infinite-decks

Edit to add: I tried to right justify the 2nd column, but my 1/2-a’d attempt didn’t prevail…

(Link removed as I’m too much of a “Noob” to post links, apparently)

I looked at the table and I can’t understand why it shows 480 Royal Flushes. There are only 4 possible Royal flushes in a draw of 5 cards regardless of the number of decks being drawn from.

link to original post

It’s a permutation table so it counts up all the different ways you could be dealt a royal flush.

Take spades:

You could be dealt any of 5 cards first (10s, Js, Qs, Ks, As)

You could be dealt any of 4 remaining cards second

3 options for 3rd card, 2 for the 4th card and 1 remaining card for the 5th.

5*4*3*2*1 = 120 permutations of a spade flush. Repeat for other suits to get 120 * 4 = 480 Royal flushes.

You have to use permutations to get the % right since your denominator is also in permutations of the 52^5 different hands you could get.

Quote:unJonQuote:TechnoWizardQuote:camaplSorting the table on WoO named Infinite Deck yields the following:

HAND PERMUTATIONS REDUCTION Royal flush 480 n/a Straight flush 4,320 n/a Five of a kind 13,312 n/a Four of a kind 798,720 n/a Straight 1,224,000 n/a Flush 1,477,200 3,172 (52 Five of a kinds, 3,120 Four of a kinds) Full house 1,591,200 6,240 Flushes Three of a kind 17,503,200 68,640 Flushes Two pair 26,254,800 102,960 Flushes Pair 175,032,000 686,400 Flushes Nothing 156,304,800 n/a

Except for non-winning hands (Nothing), rows are sorted from least likely (Royal Flush) to most likely (Pair). Totals in the Permutations column have been adjusted for Flush and Full House to account for overlaps in the totals of the higher, less likely hand, so as to conform with the rest of the table.

Edit to add the link provided by Miplet in a prior post:

/games/poker/#infinite-decks

Edit to add: I tried to right justify the 2nd column, but my 1/2-a’d attempt didn’t prevail…

(Link removed as I’m too much of a “Noob” to post links, apparently)

I looked at the table and I can’t understand why it shows 480 Royal Flushes. There are only 4 possible Royal flushes in a draw of 5 cards regardless of the number of decks being drawn from.

link to original post

It’s a permutation table so it counts up all the different ways you could be dealt a royal flush.

Take spades:

You could be dealt any of 5 cards first (10s, Js, Qs, Ks, As)

You could be dealt any of 4 remaining cards second

3 options for 3rd card, 2 for the 4th card and 1 remaining card for the 5th.

5*4*3*2*1 = 120 permutations of a spade flush. Repeat for other suits to get 120 * 4 = 480 Royal flushes.

You have to use permutations to get the % right since your denominator is also in permutations of the 52^5 different hands you could get.

link to original post

I might be tired, but I’m a bit lost here - as I understand it, a probability is n(desirable)/total possible - so probability of a RF is 4/52^5

Quote:TechnoWizardQuote:unJonQuote:TechnoWizardQuote:camaplSorting the table on WoO named Infinite Deck yields the following:

HAND PERMUTATIONS REDUCTION Royal flush 480 n/a Straight flush 4,320 n/a Five of a kind 13,312 n/a Four of a kind 798,720 n/a Straight 1,224,000 n/a Flush 1,477,200 3,172 (52 Five of a kinds, 3,120 Four of a kinds) Full house 1,591,200 6,240 Flushes Three of a kind 17,503,200 68,640 Flushes Two pair 26,254,800 102,960 Flushes Pair 175,032,000 686,400 Flushes Nothing 156,304,800 n/a

Except for non-winning hands (Nothing), rows are sorted from least likely (Royal Flush) to most likely (Pair). Totals in the Permutations column have been adjusted for Flush and Full House to account for overlaps in the totals of the higher, less likely hand, so as to conform with the rest of the table.

Edit to add the link provided by Miplet in a prior post:

/games/poker/#infinite-decks

Edit to add: I tried to right justify the 2nd column, but my 1/2-a’d attempt didn’t prevail…

(Link removed as I’m too much of a “Noob” to post links, apparently)

I looked at the table and I can’t understand why it shows 480 Royal Flushes. There are only 4 possible Royal flushes in a draw of 5 cards regardless of the number of decks being drawn from.

link to original post

It’s a permutation table so it counts up all the different ways you could be dealt a royal flush.

Take spades:

You could be dealt any of 5 cards first (10s, Js, Qs, Ks, As)

You could be dealt any of 4 remaining cards second

3 options for 3rd card, 2 for the 4th card and 1 remaining card for the 5th.

5*4*3*2*1 = 120 permutations of a spade flush. Repeat for other suits to get 120 * 4 = 480 Royal flushes.

You have to use permutations to get the % right since your denominator is also in permutations of the 52^5 different hands you could get.

link to original post

I might be tired, but I’m a bit lost here - as I understand it, a probability is n(desirable)/total possible - so probability of a RF is 4/52^5

link to original post

No. That’s apples to oranges. If you want to use 4 as the numerator, then your denominator needs to be the number of hand combinations, not permutations. That number is (52/5)/5!

5! = 5*4*3*2*1 = 120. That’s the number of ways you could deal a five card hand by changing the order of what card comes first, second, third, fourth and fifth.

I’m other words: being dealt A A K J 9 is the same thing as being dealt 9 J K A A.

Make sense?

Quote:unJonQuote:TechnoWizardQuote:unJonQuote:TechnoWizardQuote:camaplSorting the table on WoO named Infinite Deck yields the following:

HAND PERMUTATIONS REDUCTION Royal flush 480 n/a Straight flush 4,320 n/a Five of a kind 13,312 n/a Four of a kind 798,720 n/a Straight 1,224,000 n/a Flush 1,477,200 3,172 (52 Five of a kinds, 3,120 Four of a kinds) Full house 1,591,200 6,240 Flushes Three of a kind 17,503,200 68,640 Flushes Two pair 26,254,800 102,960 Flushes Pair 175,032,000 686,400 Flushes Nothing 156,304,800 n/a

Except for non-winning hands (Nothing), rows are sorted from least likely (Royal Flush) to most likely (Pair). Totals in the Permutations column have been adjusted for Flush and Full House to account for overlaps in the totals of the higher, less likely hand, so as to conform with the rest of the table.

Edit to add the link provided by Miplet in a prior post:

/games/poker/#infinite-decks

Edit to add: I tried to right justify the 2nd column, but my 1/2-a’d attempt didn’t prevail…

(Link removed as I’m too much of a “Noob” to post links, apparently)

I looked at the table and I can’t understand why it shows 480 Royal Flushes. There are only 4 possible Royal flushes in a draw of 5 cards regardless of the number of decks being drawn from.

link to original post

It’s a permutation table so it counts up all the different ways you could be dealt a royal flush.

Take spades:

You could be dealt any of 5 cards first (10s, Js, Qs, Ks, As)

You could be dealt any of 4 remaining cards second

3 options for 3rd card, 2 for the 4th card and 1 remaining card for the 5th.

5*4*3*2*1 = 120 permutations of a spade flush. Repeat for other suits to get 120 * 4 = 480 Royal flushes.

You have to use permutations to get the % right since your denominator is also in permutations of the 52^5 different hands you could get.

link to original post

I might be tired, but I’m a bit lost here - as I understand it, a probability is n(desirable)/total possible - so probability of a RF is 4/52^5

link to original post

No. That’s apples to oranges. If you want to use 4 as the numerator, then your denominator needs to be the number of hand combinations, not permutations. That number is (52/5)/5!

5! = 5*4*3*2*1 = 120. That’s the number of ways you could deal a five card hand by changing the order of what card comes first, second, third, fourth and fifth.

I’m other words: being dealt A A K J 9 is the same thing as being dealt 9 J K A A.

Make sense?

link to original post

I certainly get that 9JKAA is the same as AAKJ9 which is the same as AKAJ9 etc. The main difference I've introduced is that 5 cards are being dealt from 5 decks, each card coming from a deck of 52 cards, which means there are 52^5 possible hands dealt (5 decks, 1 card from each). I'm obviously missing something here, I just don't know what I'm missing.

Quote:TechnoWizard

I certainly get that 9JKAA is the same as AAKJ9 which is the same as AKAJ9 etc. The main difference I've introduced is that 5 cards are being dealt from 5 decks, each card coming from a deck of 52 cards, which means there are 52^5 possible hands dealt (5 decks, 1 card from each). I'm obviously missing something here, I just don't know what I'm missing.

link to original post

Let me try it this way. You agree that 9JKAA is the same as AAKJ9 and AKAJ9. Let’s call them identical twin hands.

But you are counting them as different hands when you use 52^5 as the total number of hands. Each of those three identical twin hands counts towards that total of 52^5. If you want to eliminate those identical twin hands from the total then you divide 52^5 by 120.

Because every single hand actually has 120 identical twin hands.

So you are right that there are only 4 Royal flushes. But each one has 120 identical twin Royal flushes for a total of 480.

For example: you might deal out AKQJT all of hearts. Or you might deal out KAJTQ all of hearts. They are identical twins. But when you deal out all 52^5 ways to deal out hands you will deal out both of those, and in fact 120 heart Royal Flushes will show up, each time in a different order of cards.

Quote:unJonQuote:TechnoWizard

I certainly get that 9JKAA is the same as AAKJ9 which is the same as AKAJ9 etc. The main difference I've introduced is that 5 cards are being dealt from 5 decks, each card coming from a deck of 52 cards, which means there are 52^5 possible hands dealt (5 decks, 1 card from each). I'm obviously missing something here, I just don't know what I'm missing.

link to original post

Let me try it this way. You agree that 9JKAA is the same as AAKJ9 and AKAJ9. Let’s call them identical twin hands.

But you are counting them as different hands when you use 52^5 as the total number of hands. Each of those three identical twin hands counts towards that total of 52^5. If you want to eliminate those identical twin hands from the total then you divide 52^5 by 120.

Because every single hand actually has 120 identical twin hands.

So you are right that there are only 4 Royal flushes. But each one has 120 identical twin Royal flushes for a total of 480.

For example: you might deal out AKQJT all of hearts. Or you might deal out KAJTQ all of hearts. They are identical twins. But when you deal out all 52^5 ways to deal out hands you will deal out both of those, and in fact 120 heart Royal Flushes will show up, each time in a different order of cards.

link to original post

AHHHH! Okay, so 52^5 is not the correct way to determine the number of possible outcomes of dealing from an infinite deck of cards (or 5 cards from 5 separate decks).

Therefore, my denominator should be (52^5)/5! (Your earlier reply said (52/5)/5! (which isn't the same thing ;) )

Correct?

Or you could leave your denominator as 52^5 but use 480 Royal flushes instead of 4.

Quote:TechnoWizarderm... (52^5)/5! = 3168366.933 (which doesn't make sense either)

link to original post

Oh. Hmmm. I think that’s because not every hand has 120 identical twins. Because some hands have identical cards (same rank and suit). For instance there’s only 1 way to deal out five aces of hearts.

Yeah, that’s complicated and now I’m too tired. Best to stick with permutations I think and use 480 Royal flushes in numerator.

Quote:unJonQuote:TechnoWizarderm... (52^5)/5! = 3168366.933 (which doesn't make sense either)

link to original post

Oh. Hmmm. I think that’s because not every hand has 120 identical twins. Because some hands have identical cards (same rank and suit). For instance there’s only 1 way to deal out five aces of hearts.

Yeah, that’s complicated and now I’m too tired. Best to stick with permutations I think and use 480 Royal flushes in numerator.

link to original post

Nicely explained, unJon! Also, I agree that permutations must be used due to “duplicate” cards.