Aces, Aces, Aces

matt21

Well-Known Member
#1
In a 6D game....

What's the probability of getting dealt A,A - splitting the aces and then getting another ace on top of each of the first aces?

I think I have the answer for the first hand of the shoe. But in all subsequent rounds, the probability would be driven by the number of cards dealt and the number of aces dealt prior to the round in play.

What's the best mathematical approach for handling this problem?

Look forward to any replies,
Matt21
 

iCountNTrack

Well-Known Member
#2
matt21 said:
In a 6D game....

What's the probability of getting dealt A,A - splitting the aces and then getting another ace on top of each of the first aces?

I think I have the answer for the first hand of the shoe. But in all subsequent rounds, the probability would be driven by the number of cards dealt and the number of aces dealt prior to the round in play.

What's the best mathematical approach for handling this problem?

Look forward to any replies,
Matt21
 

paddywhack

Well-Known Member
#3
matt21 said:
In a 6D game....

What's the probability of getting dealt A,A - splitting the aces and then getting another ace on top of each of the first aces?

I think I have the answer for the first hand of the shoe. But in all subsequent rounds, the probability would be driven by the number of cards dealt and the number of aces dealt prior to the round in play.

What's the best mathematical approach for handling this problem?

Look forward to any replies,
Matt21
Do you one better. Saw it once. Split aces each received an ace and all were the Ace of Diamonds in a 6 deck shoe.
 

matt21

Well-Known Member
#4
thanks for those replies.

getting ace of diamond 4 times - that would be 6/312 x 5/311 x 4/310 x 3/309 - very very unlikely. Multiply by four to get the probability of getting 4 suited aces in a row (any suit).
 

iCountNTrack

Well-Known Member
#5
matt21 said:
thanks for those replies.

getting ace of diamond 4 times - that would be 6/312 x 5/311 x 4/310 x 3/309 - very very unlikely. Multiply by four to get the probability of getting 4 suited aces in a row (any suit).
That would not be quite right that is because the between the first ace, second ace, and third ace would not be dealt sequentially in a round and it is possible that some aces were dealt in between.
 

Sucker

Well-Known Member
#6
matt21 said:
thanks for those replies.

getting ace of diamond 4 times - that would be 6/312 x 5/311 x 4/310 x 3/309 - very very unlikely. Multiply by four to get the probability of getting 4 suited aces in a row (any suit).
Actually; this is EXACTLY right.
 

Sucker

Well-Known Member
#8
As far as PROBABILITY is concerned, it does not MATTER in what order the cards are dealt. Suppose you're playing first base, and there are two other players besides the dealer. You will be dealt the 1st and the 5th card in the shoe. Your hit cards will be the 9th and the 10th card. The chance that THESE four cards are the ace of diamonds is EXACTLY the same as the chance that the FIRST four cards in the shoe are the ace of diamonds.
 

iCountNTrack

Well-Known Member
#9
Sucker said:
As far as PROBABILITY is concerned, it does not MATTER in what order the cards are dealt. Suppose you're playing first base, and there are two other players besides the dealer. You will be dealt the 1st and the 5th card in the shoe. Your hit cards will be the 9th and the 10th card. The chance that THESE four cards are the ace of diamonds is EXACTLY the same as the chance that the FIRST four cards in the shoe are the ace of diamonds.
I still don't understand your logic, the probability of drawing an ace at any moment is (number of aces left)/(total number of cards left ), if the player at first base receives 4 cards in a row, the overall probability would be 6/312*5/311*4/310*309

However in normal dealing procedure the player doesnt get all 4 cards at row as other players and dealer get cards, and the number of aces left and total number cards left would change.
 

RJT

Well-Known Member
#10
iCountNTrack said:
I still don't understand your logic, the probability of drawing an ace at any moment is (number of aces left)/(total number of cards left ), if the player at first base receives 4 cards in a row, the overall probability would be 6/312*5/311*4/310*309

However in normal dealing procedure the player doesnt get all 4 cards at row as other players and dealer get cards, and the number of aces left and total number cards left would change.
As no strategy changes or betsize changes can be made after the round is in progress, the extra information from the additional information makes no difference. You are really simply looking to calculate the probability of drawing the four suited aces from any random position in a starting composition shoe.

RJT.
 

Sonny

Well-Known Member
#11
Sucker said:
As far as PROBABILITY is concerned, it does not MATTER in what order the cards are dealt.
That is true. The probabilities before the cards are dealt are exactly the same. Even when the dealer follows standard dealing procedures and deals face down to 6 players, the probabilities after the deal (but before any cards are exposed) are exactly the same as before the deal.

iCountNTrack said:
However in normal dealing procedure the player doesnt get all 4 cards at row as other players and dealer get cards, and the number of aces left and total number cards left would change.
That is also true. Every time a card is exposed, the known probabilities change. Since the dealer is exposing cards during the dealing procedure (upcard, other players' hit, split, double down cards) we can update the probabilities as the hand is played. But what will they average out to in the long run? The same as the probability in the pre-deal situation. The probabilities of any particular situation will be different but they will converge.

Since the question seems to ask about the pre-deal probability (which is the same as the overall probability) I have to agree with the masses on this one.

-Sonny-
 
#12
Yes. You are going to receive 4 cards out of the shoe. The probability of cards 1,2,3,4 all being the Ace of Diamonds is the same as 13,14,15,16 being the Ace of Diamonds, or 19,101,54,207 all being the Ace of Diamonds.
 
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