4

Most likely you've had a situation, where you look down at your pocket cards, and they're exactly the same as the hand before. If they're junk, you suspect that the dealer is being lazy and didn't shuffle the deck. Yet, you saw him/her do it.

What are the odds that all the other players' cards,community cards, and remaining cards are exactly in the same order as the play before?

  • @Paparazzi That's not what the question asks. If I can improve the wording of it, please suggest how it can be done. – Roman Mik Apr 6 '17 at 18:12
  • I'm voting to close this question as off-topic because how many shuffles is not a poker question. Poker does not care about order and does not use the whole deck. – paparazzo Apr 6 '17 at 19:01
  • Poker AI developers will probably disagree with you :) – Roman Mik Apr 6 '17 at 19:22
  • I am a poker AI developer and I could care squat about how many unique shuffles. I will run hole cards against the 2,118,760 unique (order does not matter) boards. If I ran it against 50! shuffles it would never finish. – paparazzo Apr 6 '17 at 19:32
4

It turns out that each shuffled deck is in the order that may have never before existed in the history of the universe! :)

The odds of you getting two 52 card decks arranged in the exact same order are 52! ~= 8 x 10^67, which is waaay more than the number of atoms on Earth (~ 10^50).

For a detailed explanation, please check out a great video answer on TED.com How many ways can you arrange a deck of cards? - Yannay Khaikin

  • 1
    Great example of a self asked question & answer. Nice one. – Toby Booth Aug 16 '15 at 19:14
  • 1
    Does not need to be the exact shuffle for the same hand. KQ is same as QK. – paparazzo Apr 5 '17 at 11:56
  • @Paparazzi Not to mention, even in a hold-em game with 10 players that all see a showdown, you're only going to see 25 cards out of the deck. 25! is ~1.55 x 10^25 (and as you mentioned, it would be less than that still ignoring the order each player received their cards in) – user1934 Apr 5 '17 at 16:44
  • now am imagining a casino with a gimmicky promotion where every shuffle is tracked via RFID or something and if it matches any previous shuffle ever, a huge jackpot is paid out to everyone at the table... – user1934 Apr 6 '17 at 19:09
  • @Michael the subset of 28( 25 dealt and 3 burnt) first cards of the total set is still (52!). Granted, you don't see the 3 burnt card, but the total of possible combinations is not changed. You can check it yourself by arranging a deck of 3 cards (say AsKsQs) into all possible combinations, and then taking first two cards. You will see that out of 6 combinations, you still end up with 6 subsets of two cards. – Roman Mik Apr 6 '17 at 19:21
0

Depends on how many players are sitting at the table.

The number of the possibilities is:

52! / (47-(2*p))!

the p stands for how many players are currently sitting at the table.

  • That formula would give you the number of possible decks after each player got their two cards. – Roman Mik Aug 26 '15 at 17:27
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The odds of each player getting the same hand as before as well as the flop turn and river being the same as before are considerably better than the odds of the whole deck being the same unless you have a huge number of players. After all the cards are dealt the remaining cards don't exist.

0

So the other answers are correct that there are 52! possible deck combinations. From what I understand you're not asking about deck combinations, but rather the chance to get the exact same hand you received before.

For getting the exact same hand while paying respect to suit, for the first card there is a 2/52 chance to get one of the two specified cards out of the 52 card deck and for the second card, since the first card has been already picked and can't be picked again, there is a 1/51 chance to get a specific second card. For both of these to happen simultaneously you multiply them together for a (2/52) * (1/51) = .07% chance of both specific cards being picked. This is certainly unlikely but definitely possible if you play enough.

What is more likely is that you get the same two unpaired cards if you don't give respect to suit. For this there is a 8/52 chance to get one of the two cards, and a 4/51 chance to get the specific second card. Combined there is a (8/52) * (4/51) = 1.2% chance of this occurring which would definitely happen with some frequency over hundreds hands. The probability drops a lot if you are calculating the chance to get the same paired hand with there being a 4/52 chance for the first card and a 3/51 chance for the second (since one of the cards you need has already been used.) This gives us a total chance of (4/52) * (3/51) = .4% chance to get the same paired cards.

  • This is wrong. It is 2/52 * 1/51. The first card can be either. – paparazzo Apr 5 '17 at 17:59
  • Ahh true. I'll edit to reflect the correct answer. – TheSaint321 Apr 5 '17 at 19:16
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Order of the cards does not matter in poker and you don't play the whole deck

The odds of seeing the same board is 1 / 2,598,960
The odds of seeing the same hole cards is 1 / 1081
Same hole cards and same board 2,809,475,760

Second player same hole cards 1 / 990
Same board and two players same hole cards 1 / 2,781,381,002,400

  • The question was " how-many-ways-to-shuffle-a-standard-deck-of-52-cards?" – Roman Mik Apr 6 '17 at 18:25
  • @RomanMik And that is not a poker question – paparazzo Apr 6 '17 at 18:59

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