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Thread: The number of ways a six deck shoe can be ordered is 2.1 X 10**644.

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    Senior Member BigJer's Avatar
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    The number of ways a six deck shoe can be ordered is 2.1 X 10**644.

    The number of atoms in the known universe is on the order of 10**80.

    Was just thinking about that.....
    Last edited by BigJer; 02-22-2015 at 07:57 AM. Reason: Edited for correctness.
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    Out of curiosity, does 2.1e644 include or omit equivalent orders, and if omit, which equivalent orders does it omit?
    May the cards fall in your favor.

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    Senior Member BigJer's Avatar
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    Quote Originally Posted by Dieter View Post
    Out of curiosity, does 2.1e644 include or omit equivalent orders, and if omit, which equivalent orders does it omit?
    It includes equivalent orders. Let me see about that calculation. BRB.

    ETA:

    The order of things is important so this calculation is correct.
    Last edited by BigJer; 02-22-2015 at 09:03 AM.
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    Senior Member BigJer's Avatar
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    According to http://gizmodo.com/there-are-more-wa...ato-1553612843 there are more ways to shuffle one deck than there are atoms on earth.
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    Senior Member Aslan's Avatar
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    Quote Originally Posted by BigJer View Post
    According to http://gizmodo.com/there-are-more-wa...ato-1553612843 there are more ways to shuffle one deck than there are atoms on earth.
    The first commenter stated: "I just threw up in my brain." That about sums it up. laugh1.gif
    Last edited by Aslan; 02-22-2015 at 10:57 AM.

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    Quote Originally Posted by BigJer View Post
    It includes equivalent orders.
    Thanks.
    May the cards fall in your favor.

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    For BJ suits don't matter and there are only 10 possible card values. The calculation was done with 52 distinct cards, 52 factorial.

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    Quote Originally Posted by Tthree View Post
    For BJ suits don't matter and there are only 10 possible card values. The calculation was done with 52 distinct cards, 52 factorial.
    "Equivalent for blackjack" is different than what I was thinking.

    In a 6 pack deck, there will be 6 Queens of Hearts. Ordinarily, I don't care which of those is at position 6 or 79, or 87, or 112, or 221, or 278, but some calculations would count all the orderings where it's just which QH lands in which of those spots separately (all other cards in exactly the same place/order), and some of the calculations would lump all those arrangements together as 1.

    At least blackjack is simpler than Tarot reading, where you're concerned about orientation of a particular card, as well as position in the deck. 78! is a big number, and that's before counting reversals.
    Last edited by Dieter; 02-22-2015 at 08:40 PM.
    May the cards fall in your favor.

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    Senior Member BigJer's Avatar
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    Anyways it's a huge number.
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    Quote Originally Posted by BigJer View Post
    Anyways it's a huge number.
    I got to take my shoes off and pull down my pants to count that high.

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    Quote Originally Posted by Dieter View Post
    "Equivalent for blackjack" is different than what I was thinking.

    In a 6 pack deck, there will be 6 Queens of Hearts. Ordinarily, I don't care which of those is at position 6 or 79, or 87, or 112, or 221, or 278, but some calculations would count all the orderings where it's just which QH lands in which of those spots separately (all other cards in exactly the same place/order), and some of the calculations would lump all those arrangements together as 1.

    At least blackjack is simpler than Tarot reading, where you're concerned about orientation of a particular card, as well as position in the deck. 78! is a big number, and that's before counting reversals.
    Tthree is right, there are at least two much more reasonable interpretations of "distinct" arrangements of cards in a 6-deck shoe, neither of which yields 312! = 2.1x10^644 different orders. If we buy 6 identical standard packs of 52 cards, and consider all cards to be distinct, *except* for the 6 identical copies of each of the 52 cards in a single standard deck, then there are 312!/(6!)^52 = 5.5x10^495 possible arrangements.

    If instead we consider *blackjack* shoes, where neither suits nor 10-valued face cards are distinguishable, then there are 312!/(24!)^9/96! = 1.6x10^280 possible arrangements.

    Both of which are still incomprehensibly large numbers. Thoroughly shuffle a 6-deck shoe, and it is effectively certain that no one anywhere, ever, has encountered that particular arrangement of cards (even from the more relaxed "blackjack" perspective).

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