Identical Random Shuffles?

[kiddj]

I know the number of possible shuffles of a 52 card deck is 52!. How many of these shuffles are, for all intents and purposes, basically the same? An example would be if you kept the card order the same but shifted the suits.

Is there a thread, link, or study on this? (I already ran a search.)

[TomCollins]

52!/4!

[kiddj]

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52!/4!

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Is it really that simple? (If it is, that's great: thanks!)

A smaller scale example would be if you held Q[img]/images/graemlins/heart.gif[/img]K[img]/images/graemlins/heart.gif[/img] in HE and the flop came 2[img]/images/graemlins/diamond.gif[/img]7[img]/images/graemlins/club.gif[/img]9[img]/images/graemlins/spade.gif[/img]. There are 6 flops total that basically give you the exact same hand. If you take this for any QKs, you get 24 equal hands. Another example would be if you had 44 and the flop came KT7 rainbow. There are 144 combinations of this setup. Does 52!/4! cover this?

[TomCollins]

[ QUOTE ]
[ QUOTE ]
52!/4!

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Is it really that simple? (If it is, that's great: thanks!)

A smaller scale example would be if you held Q[img]/images/graemlins/heart.gif[/img]K[img]/images/graemlins/heart.gif[/img] in HE and the flop came 2[img]/images/graemlins/diamond.gif[/img]7[img]/images/graemlins/club.gif[/img]9[img]/images/graemlins/spade.gif[/img]. There are 6 flops total that basically give you the exact same hand. If you take this for any QKs, you get 24 equal hands. Another example would be if you had 44 and the flop came KT7 rainbow. There are 144 combinations of this setup. Does 52!/4! cover this?

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This situation is much different than the above described one. You wanted to know how many shuffles, which has nothing to do with flops. Each of your cases have unique answers, because they are unique.

[kiddj]

[ QUOTE ]
[ QUOTE ]

A smaller scale example would be if you held Q[img]/images/graemlins/heart.gif[/img]K[img]/images/graemlins/heart.gif[/img] in HE and the flop came 2[img]/images/graemlins/diamond.gif[/img]7[img]/images/graemlins/club.gif[/img]9[img]/images/graemlins/spade.gif[/img]. There are 6 flops total that basically give you the exact same hand. If you take this for any QKs, you get 24 equal hands. Another example would be if you had 44 and the flop came KT7 rainbow. There are 144 combinations of this setup. Does 52!/4! cover this?

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This situation is much different than the above described one. You wanted to know how many shuffles, which has nothing to do with flops. Each of your cases have unique answers, because they are unique.

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Ok. Let's assume the original shuffling creates those hands/flops. Is the variation of those "identical" hands included in our 52!/4! shuffle combinations?

[TomCollins]

Sure.

[Siegmund]

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I know the number of possible shuffles of a 52 card deck is 52!. How many of these shuffles are, for all intents and purposes, basically the same?

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Define what your intents and purposes are and what "basically the same" is.

In stuffy mathematicsese: your notion of "basically the same" defines an equivalence relation. That equivalence relation breaks up the set of 52! permutations into some number of equivalence classes.

In holdem, we don't care about the 4! permutations of suits, which of two hole cards is dealt first to each player, what order the three flop cards are in, or what order the stub cards are in. In a 10-handed game, it looks like there are 52!/(2^10 * 3! * 1 * 1 * 27! * 4!) ~ 50234623599558030217607550528000000 unique deals, but you could probably argue there are somewhat fewer than that e.g. by tracking down the situations where flushes are completely impossible.

In stud it's harder - we sometimes care about the suits (if we have to choose which of two equal cards brings it in) and sometimes don't - so the sets of equivalent deals are not equal in size.

In bridge it's easier - we always care about suits, but never care about what order each player receives his 13 cards.

[kiddj]

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Define what your intents and purposes are and what "basically the same" is.

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The intent is to quantify deals in hold'em that are equivalent. My example is a shuffle of the deck that deals out the same and plays exactly the same, like if you are dealt two offsuit connectors and the flop consists of rainbow undercards. Now that I've thought about it, correcting for the different suits by dividing by 4! makes sense. This still gives us many combinations.

I'm sure there are a lot of deals that may be played the same, even though they aren't the exact same cards, but this would be difficult, and therefore, not worth calculating. IMO

[shday]

How about calculate the number of permutations based on the number of cards used. All the randomness left in the unused part of the deck wouldn’t matter.

The size would be further reduced because the order of cards doesn’t matter for each players hand and for the 3 cards of the flop. Then the equivalency of the suits would cut it down more.

For heads-up the total number of effectively unique shuffles would be close to (in case I missed something):

P(52,4)/4*C(48,3)*47*46/4!

or 2 530 869 413 800

I think this is very close to what you get using Seigmunds calculation from above (I didn't have a way to accurately calculate 52! at hand).

[shday]

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For heads-up the total number of effectively unique shuffles would be close to (in case I missed something):


P(52,4)/4*C(48,3)*47*46/4!

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oops, try this instead:

C(52,2)*C(50,2)*C(48,3)*45*44/4!

=2 317 817 502 000