### Brain Teasers

# Blackjack

Probability
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.

When playing a game of blackjack, what are the odds that you will be dealt blackjack on the first deal? Assume you are playing with one opponent and your opponent is the dealer.

Note: 10, jack, queen and king all are worth 10. Aces are worth 11. You have blackjack when the values of both your cards add up to 21. There is only one deck of cards being dealt out.

Note: 10, jack, queen and king all are worth 10. Aces are worth 11. You have blackjack when the values of both your cards add up to 21. There is only one deck of cards being dealt out.

### Answer

To receive a blackjack, either my first card is worth 10 and the second is an ace or vice versa, which makes the probability...(16/52)*(4/51) + (4/52)*(16/51) = 128/2652

...which is approximately 4.83%

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## Comments

Love probability.

The total number of two card hands is C(52,2) = 52!/ (50!* 2!)

Number of ways of getting a blackjack is : C(4,1) * C(16,1) = 4*16 = 64.

So, the desired probability is 64/1326 =32/663.

Good job!

The total number of two card hands is C(52,2) = 52!/ (50!* 2!)

Number of ways of getting a blackjack is : C(4,1) * C(16,1) = 4*16 = 64.

So, the desired probability is 64/1326 =32/663.

Good job!

I believe the answer may be slightly lower than this. If you consider that the player is simply given 2 cards from 52 then this number is correct. However, if the cards are dealt, with the first card going to the player and the second going to the dealer, then the player's 2nd card comes from a set of 50, not 51, and one of the useful cards may have already gone to the dealer.

Thank you, I see the mistake.

I'll have to keep this in mind. Great teaser

good teaser

well done

well done

Gizzer, no the cards the dealer deals to himself have no bearing on the probability unless they are known and then your getting into the area of card counting. In this case they are unknown , so the the cards could be tens aces or just as likely any card in the deck. Nor is the burn card relevant (after each shuffle the first card is placed directly into the discard shoe, also after each change of dealers). In fact even the number of players has no effect. There are web sites that contain every statistic imaginable in blackjack and none of these stats vary with number of players or number of burn cards. The only time this comes into play is when the cards are actually observed and assigned value for the purposes of card counting which is a vast and separate topic. Number of decks can change probability in many cases of blackjack play and basic strategy, but in the case of this teaser it does not.

Sorry, the last sentence in my previous post was false. There is a slight difference in the probability of getting a blackjack depending on number of decks. Everything below is copy/pasted from wizardofodds.com faq section:

"What is the probability of getting a blackjack in blackjack?

It depends on the number of decks; here are the probabilities for 1 to 8 decks:

1. 0.048265

2. 0.047797

3. 0.047643

4. 0.047566

5. 0.047520

6. 0.047489

7. 0.047468

8. 0.047451

How did you calculate those numbers?

The general formula is 2*(probability of 10)*(probability of ace). The reason for the 2 is that there are two ways to order the two cards; either the ace or 10 can come first. The probability that the first card will be a 10 is always 4/13, regardless of the number of decks, because there are 16 out of 52 tens in the deck and 16/52 = 4/13. The probability of an ace, given that a 10 has already been removed from the deck is the number of aces divided by the number of cards left. Let n be the number of decks. There are 4*n aces and 52*n-1 cards left. So for n decks the probability is 2*(4/13)*(4*n/(52*n-1)), which is conveniently about 1 in 21. For example for 6 decks the answer is 2*(4/13)*((4*6)/(52*6-1)) = 192/4043 = 0.047489."

"What is the probability of getting a blackjack in blackjack?

It depends on the number of decks; here are the probabilities for 1 to 8 decks:

1. 0.048265

2. 0.047797

3. 0.047643

4. 0.047566

5. 0.047520

6. 0.047489

7. 0.047468

8. 0.047451

How did you calculate those numbers?

The general formula is 2*(probability of 10)*(probability of ace). The reason for the 2 is that there are two ways to order the two cards; either the ace or 10 can come first. The probability that the first card will be a 10 is always 4/13, regardless of the number of decks, because there are 16 out of 52 tens in the deck and 16/52 = 4/13. The probability of an ace, given that a 10 has already been removed from the deck is the number of aces divided by the number of cards left. Let n be the number of decks. There are 4*n aces and 52*n-1 cards left. So for n decks the probability is 2*(4/13)*(4*n/(52*n-1)), which is conveniently about 1 in 21. For example for 6 decks the answer is 2*(4/13)*((4*6)/(52*6-1)) = 192/4043 = 0.047489."

More simply:

4/13 * 4/51 * 2!

Choose one of the 4 ranks for a 10, one of the 4 aces in the remaining 51 cards times the 2! = 2 ways to order the cards.

Doing the problem this way makes it trivial to adjust for the number of decks. Where d is the number of decks, the probability is:

P = 4/13 x 4d/(52d-1) x 2!

4/13 * 4/51 * 2!

Choose one of the 4 ranks for a 10, one of the 4 aces in the remaining 51 cards times the 2! = 2 ways to order the cards.

Doing the problem this way makes it trivial to adjust for the number of decks. Where d is the number of decks, the probability is:

P = 4/13 x 4d/(52d-1) x 2!

Good solution... but I think the question suggests that the dealer is dealing himself some cards in the meanwhile as well. Correct me if I'm wrong. And sorry if somebody has already mentioned this.

Also, is the first card not discarded immediately?

Sorry to be pedantic

Also, is the first card not discarded immediately?

Sorry to be pedantic

Lord Hawk:

It doesn't matter if the dealer first throws away half the deck. As long as you don't know which cards are discarded, the number of unknown cards is the same and the probability is the same.

It doesn't matter if the dealer first throws away half the deck. As long as you don't know which cards are discarded, the number of unknown cards is the same and the probability is the same.

Okay. Sorry, others have put this point up before and you have explained it before as well.

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