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# problem: counting posible value in blackjack

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Hi all, i need help in my blackjack game. in my blackjack, Ace card have 2 possible value 1 and 11. so if a player have a more then 1 ace. how many possible value for the cards? i assume there 2 ace card then it will be 3 posibilities 1. 1 1 = 2 2. 1 11 = 12 3. 11 11 = 22 if 3 ace card then it will be 4 posiblities?? 1. 1 1 1 = 3 2. 1 1 11 = 13 3. 1 11 11 = 23 4. 11 11 11 = 33 and etc etc, assuming we have infinite ace what the best way to count all the posibilities exist? thx.

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I've only just got out of bed, but isn't it the number of aces in the hand plus one, as you have shown above?

Let's have a think. A hand without aces has one possible score. Each ace you add to that hand creates two possible scores, so a hand with two aces in has three possible scores, a hand with three aces in has four and a hand with all four aces has five.

Think that is right and remains true regardless of the number of aces, since their position in the hand is not important.

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If you are doing standard blackjack, there are only 2 combinations for any number of aces: +10 and +0. This is because any score over 21 has already lost, and 2 aces would be 22 if you counted them as 11 each.

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thx for clearing up.

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Rutin
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JoeJ
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