# Do I get blackjack when splitting aces?

If I am dealt two aces, face up, split them, and receive two face-cards, does that give me two blackjacks, or just two 21's ??

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• 4 years ago

Dear Robert W:

Here is how it works:

If a player is dealt an ace and any ten-value card (10, J, Q, K) on the first two cards, that is called a “natural” or “blackjack”. If the player holds a natural, and the dealer does not, the player wins immediately, and is typically paid 3:2 (that is, a \$10 bet wins \$15). If the player holds a natural, and the dealer does also, that is a tie (called a “push” or “standoff”). The player wins nothing and simply takes back his original bet. If the dealer holds a natural, then any player who does not also hold a natural loses immediately. He will not get a chance to ask for a “hit” (additional card) in an effort to get a value of twenty-one. This hand is over right then and there.

If a player is dealt a pair, and the dealer does not hold a natural, the player has the option to “split” the pair, lay down an amount equal to his original bet, and play two separate hands. If a player is dealt a pair of aces, and chooses to split them (side note: always split aces; always), he lays down an amount equal to his original bet, and plays two separate hands. The dealer will then deal a second card to each ace. If each of those cards is a ten-value card – sorry. The player does not have two naturals, because they did not occur on the first two cards. He merely has two hands, each with a value of 21. Play continues. If the dealer’s hand has a value of 17 (soft or hard) or higher, the dealer must “stand” (draw no more cards). If the dealer’s hand is less than 17, he must “draw” (take another card) until his count becomes 17 or higher, whereupon he draws no more cards. If, in the end, the player beats the dealer (that is, the dealer’s hand has a value between 17 and 20) or the dealer’s hand “busts” (exceeds 21), the player wins on both hands, and each hand is paid 1:1 (that is, a \$10 bet wins \$10 - twice). However, if the dealer ends up with a value of 21, it is a push, and neither player hand wins. The player simply takes back both his original bet and the split bet.

In brief: If the player’s first two cards are an ace and a ten-value card, that‘s a natural (“blackjack”). If the dealer does not also have a natural, the player is paid 3:2.

If the player otherwise ends up with a hand of value 21, and the dealer does not also have 21, that is merely a winning hand, and pays 1:1.

Flash Kellam

• 4 years ago

21s

• Anonymous
4 years ago

I think by the rules of the game

a Natural Blackjack is on the 1st 2 cards only.

Your situation would result in just a pair of 21 total hands

and both could push with the Dealer's hand

• Jay P
Lv 7
4 years ago

No. You only end up with two 21's.

Blackjack can only be made with the first 2 cards dealt.