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If a deck, being used for BlackJack, contains extra Aces, does the dealer have an advantage, or vice versa. Additionally, do less aces cause an effect on the dealer or players advantage?

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How many extra Aces are we talking here? Does the casino pay 3:2 on a Blackjack? Can you split, double down, double down after a split, etc. Are you playing a single deck, 4 decks, 8 decks, etc. –  user1873 Jul 10 '13 at 5:21
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This is a very complicated question. There are several factors to consider that make it impossible to answer without knowing a lot of details that you left out. Taken to the extreme, imagine a blackjack deck with 1 million extra Aces. In such a game, the player would have the advantage if the dealer was required to hold on 17. The player can hit until they have 18, confident that the dealer would most likely get 17 and be forced to stop. Here are some odds for adding a single extra Ace.

  • Decreased Chance of Player Blackjack

    • 52 card: ((4/52) × (16/51)) + ((16/52) × (4/51)) = 4.82655%
    • 53 card: ((5/53) × (16/52)) + ((16/53) × (5/52)) = 5.80552%

The odds of getting BJ are the odds of getting dealt an Ace then a 10/face-card plus the reverse order. Adding in an extra Ace increases your odds by almost one percent.

  • Increased Chance of Dealer Blackjack when player has BJ

    • 52 card: ((3/50) × (15/49)) + ((15/50) × (3/49)) = 3.67347%
    • 53 card: ((4/51) × (15/50)) + ((15/51) × (4/50)) = 4.70588%

The calculation here is much the same, except two cards are removed (the player BJ). Adding a single Ace hurts the player by over a whole percentage point.

Unless you give me a lot more detail (there are other things to consider) I have no idea who has the advantage. For example with 4 Aces, you have increased your odds of BJ by almost 3.5%, but the dealer has an increase of greater than 3.5% chance of having BJ when you have BJ. Who has the advantage would require extra information on what the payout is for player BJ.

  • 56 card player BJ: ((8/56) × (16/55)) + ((16/56) × (8/55)) = 8.31169%
  • 56 card dealer BJ after player: ((7/54) × (15/53)) + ((15/54) × (7/53)) = 7.33753%
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In your second formula, that should be 5/53, right? –  bwarner Jul 10 '13 at 15:03
    
@bwarner: Which changes it to 5.8%, a 1% increase in blackjack chance. Assuming dealer blackjack pushes, there is still a 0.88% increased blackjack for the player. OP is correct that blackjack payout is important (as is insurance payout) –  Guvante Jul 10 '13 at 18:02
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Extra aces (without extra other cards) increase the chances of blackjacks.

If we assume a normal deck, there are 4 aces, and 12 faces; this means a blackjack is (4/52)*(12/51)=48/2652, or about 12/663.

If we add 4 extra aces, it instead goes to (8/56)*(12/55) = 96/3080 = 12/385.

just over twice as many blackjacks, before accounting for splitting two aces.

If they pay even 6:5 on blackjacks, they have lost a good bit, because it's normally the highest paying hand, and blackjack has very low house odds favor. If they don't change the typical 3:2 payout for a blackjack, they've probably lost their house edge.

The house edge in blackjack is normally measured in hundredths of a percent. Double the frequency of blackjacks, and they lose money.

Heck, allow doubledown after a split in a single-deck blackjack table, and the house loses money, so doubling the number of blackjacks is a definite benefit to the player.

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Surely the house has the same increased odds of a blackjack? If a house blackjack beats any hand, this would work against the player. –  TimLymington Jul 10 '13 at 10:35
    
The way I am used to the rules, player blackjacks will pay before the dealer even reveals their other card –  Andrey Jul 10 '13 at 13:32
    
@TimLymington, you are absolutely right (and this answer is probably wrong (we need more detail in the question)). The increased odds that the deal BJ pushes your BJ lessens some of the increased advantage of getting a BJ. –  user1873 Jul 10 '13 at 14:13
    
Due to insurance and increased payout for blackjack, additional blackjacks are good for the player overall. –  Guvante Jul 10 '13 at 18:06
    
Shouldn't your calculation use 16 (not 12 cards of '10' pip value) since there is 10,J,Q K. Additionally, you can get Blackjack with an Ace dealt first or second with a complementary 10-value card, so you need to add IP both of those odds to get the total odds of getting BJ. –  user1873 Jul 11 '13 at 0:47
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