Theoretical card deck question
May 5th, 2012 at 9:04:15 PM
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Hi! Me and my friends had a discussion about card shuffling and deck randomizing that we cannot reach a consensus.
Assume a regular 52 card deck, playing, say, whist with 4 players.
The theoretical discussion was:
1- there is a difference from
A) dealing packs of 13 cards TO a player, then another 13 cards TO the second player, then 13 cards TO the third player and the last 13 cards TO the forth player
OR
B) dealing one card at a time TO reach player in a clockwise fashion until everyone has 13 cards
They say that the game is not the same and that there is not a random effect in A) and that in B) there is. I say it doesn't Matter so as long as the dealer does not Have any information on the deck. Which one is truE? Is the game not random if we use the A) scenario?
2- we also discussed the following scenario: Assume a regular 52 card deck. I get the 4 aces, put all four together and place them in the deck. I do not know where they are in the deck, i just know that the 4 aces are together. I deal 4 cards in a row from the deck to each player. I can deal from anywhere in the deck but i Have to deal 4 cards in a row to each player.
They say the likelihood of dealing THE 4 ACES to one player is now HIGHER than if I dealt one card at each time TO each player in a clockwise.
I say that it doesn't Matter so as long as i don't know where in the deck are the 4 aces that i put together earlier.
Which is it?
3- a small variation. Suppose we just played a whist game. In the end, the cards are usually from the same suit since You Have TO follow suit if You Have it. So, at the end, if You don't shuffle the deck, there will be clumps of 4 cards of the same suit regularly. Assume You don't shuffle the deck for the next round (say, you are lazy!). You don't know where and how the cards are in the deck, You just know that most cards will be in clUmps of 4 from the same suit.
Again, they say there is a diference between dealing 13 cards in a row TO a player, then 13 in a row TO the second, then 13 in a row TO the third player and 13 in a row TO the fourth player
OR
dealing 1 card at a time TO each player in a clockwise.
I say it doesn't matter If You don't know where the cards are.
Can You help me figure this out?
Thanks!
Assume a regular 52 card deck, playing, say, whist with 4 players.
The theoretical discussion was:
1- there is a difference from
A) dealing packs of 13 cards TO a player, then another 13 cards TO the second player, then 13 cards TO the third player and the last 13 cards TO the forth player
OR
B) dealing one card at a time TO reach player in a clockwise fashion until everyone has 13 cards
They say that the game is not the same and that there is not a random effect in A) and that in B) there is. I say it doesn't Matter so as long as the dealer does not Have any information on the deck. Which one is truE? Is the game not random if we use the A) scenario?
2- we also discussed the following scenario: Assume a regular 52 card deck. I get the 4 aces, put all four together and place them in the deck. I do not know where they are in the deck, i just know that the 4 aces are together. I deal 4 cards in a row from the deck to each player. I can deal from anywhere in the deck but i Have to deal 4 cards in a row to each player.
They say the likelihood of dealing THE 4 ACES to one player is now HIGHER than if I dealt one card at each time TO each player in a clockwise.
I say that it doesn't Matter so as long as i don't know where in the deck are the 4 aces that i put together earlier.
Which is it?
3- a small variation. Suppose we just played a whist game. In the end, the cards are usually from the same suit since You Have TO follow suit if You Have it. So, at the end, if You don't shuffle the deck, there will be clumps of 4 cards of the same suit regularly. Assume You don't shuffle the deck for the next round (say, you are lazy!). You don't know where and how the cards are in the deck, You just know that most cards will be in clUmps of 4 from the same suit.
Again, they say there is a diference between dealing 13 cards in a row TO a player, then 13 in a row TO the second, then 13 in a row TO the third player and 13 in a row TO the fourth player
OR
dealing 1 card at a time TO each player in a clockwise.
I say it doesn't matter If You don't know where the cards are.
Can You help me figure this out?
Thanks!
May 5th, 2012 at 9:31:49 PM
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1. IF the cards are randomized in the deck, there's no difference between the methods.
2. It matters. If you deal 13/13/13/13, there's a chance that all four consecutive aces go to one player. If you deal 1/1/1/1/1/1/1/1/1/1/1...1, there's no way that 4 consecutive cards can all go to one player (no one person can get all 4 aces).
3. Pretty much the same as 2. Dealing 1 card at a time breaks up the clumps of cards, dealing 13 cards at a time keeps the clumps more-or-less together.
2. It matters. If you deal 13/13/13/13, there's a chance that all four consecutive aces go to one player. If you deal 1/1/1/1/1/1/1/1/1/1/1...1, there's no way that 4 consecutive cards can all go to one player (no one person can get all 4 aces).
3. Pretty much the same as 2. Dealing 1 card at a time breaks up the clumps of cards, dealing 13 cards at a time keeps the clumps more-or-less together.
"So as the clock ticked and the day passed, opportunity met preparation, and luck happened." - Maurice Clarett
May 6th, 2012 at 12:57:13 PM
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2. rdw answer is correct, although he wrote things as if you were still dealing 13 cards at a time as in scenario 1. In scenario 2, you say you are dealing 4 cards at a time. The likelihood is greater that a player will get all 4 cards, simply because if you were dealing 1 card at a time, and the 4 aces are clumped together, there is simply no way that a player will receive all 4 cards. If you're dealing 4 cards at a time, then there is a possibility of one player getting all 4 aces. I don't know the exact possibility, but basically: 1 card at a time equals 0%, 4 cards at a time equals something greater than 0%.
