What's the probability of being dealt a blackjack?

What is a blackjack? Blackjack (or pontoon, twenty-one or – if you're French - vingt-et-un), is a card game. If you've never heard of it, you can find more info and rules here. The important rules as far as this question goes are:

• Aces can be worth either 1 or 11 points (as the player prefers).
• Picture cards (Jack, Queen and King) are worth 10 points.
• All other cards are worth their number value.
• The goal is to get as close as you can to a score of 21 with the cards you're dealt without going over.
• Each player is dealt two cards to start the game.

To be 'dealt a blackjack' means that the first two cards that you're given add up to 21, and you're done without needing any more cards. If you're dealt this hand, it's impossible for you to lose (although you could tie). So what's the probability of receiving this unbeatable hand? To answer this question, for simplicity's sake we'll assume that the cards are coming from a full, standard pack. And we'll look at two equivalent ways of doing it: with logic and with a tree diagram. The way with logic: how can you get 21 with only two cards? This is fairly simple – the only way you can get a score of 21 using just two cards is to get an ace (11 points) and either a 10 or a picture card (worth 10 points each). Obviously, it doesn't matter which order you draw these in, because 10 + 11 = 11 + 10. We're only interested in events when we get a card that's worth either 10 (10, J, Q, K) or 11 (Ace), so we can ignore everything else. So there are only two possible ways we can get 21:

1. We're dealt an 11, and then a 10. (11 + 10 = 21)
2. We're dealt a 10, and then an 11. (10 + 11 = 21)

Option 1:

• First event (getting an 11 card): There are 4 aces in a pack, and we've got a full pack (52 cards), so the probability of getting one of the aces first is 4 out of 52 (4/52).
• Second event (getting a 10 card): There are 16 cards worth 10 in a pack (4 of each of 10, J, Q, K), but we've already got one of the cards from the back, so the probability of getting a 10 now is 16 out of 51 (16/51).

We want event 1 AND event 2 to happen, so we have to multiply the probabilities: 4/52 x 16/51 = 64/2652. Option 2:

• First event (getting a 10 card): There are 16 cards worth 10, as before, but this time we've got a full pack, so the probability is 16/52.
• Second event (getting an 11 card): There are 4 aces, as before, but this time we've got rid of one card, so the probability is 4/52.

Again, we want event 1 AND event 2 to happen: 16/52 x 4/51 = 64/2652* Now, we want either option 1 OR option 2 to happen (because either way will give us a blackjack). There's an OR there, so we've got to add the two probabilities together: 64/2652 + 64/2652 = 128/2652. The probability bit's finished. All that's left is to tidy up, and make the numbers easier to read:

• Simplify the fraction: 128/2652 = 32/663

The probability tree method We start off by thinking about what can happen with the first card – it can be an ace (which is great), a face card (which is good) or something else (which is useless, as far as blackjacks go). We draw a branch for each of these possibilities starting on the left – upwards to A for ace, left to F for face card, and downwards to O for other. There are four aces in a pack of 52, so the probability on the A branch is 4/52. There are sixteen face cards, so the F branch is 16/52; and thirty-two other cards, so the O branch has a probability of 32/52. Now, if we started with an ace, if we're only interested in a blackjack, the next card is either a card worth 10 (yay!) or something else (boo!). We draw two branches from the A: F for a face card or ten, and A or O for everything else. Sixteen out of the 51 cards left in the pack are worth ten, so the branch to the F in the top right has a probability of 16/51 - one way of getting a blackjack has a probability of (4/52) x (16/51) = (64/2652), which cancels down to 16/663. If instead we had a face card first, we want an ace next - there are four of those left out of 51, so the probability of that branch (to the A in the middle on the right) is (16/52) x (4/51) = (64/2652) again, or 16/663. Adding the two probabilities together gives 32/663 – the same number as the other way. What on earth is 32/663? That fraction doesn't really tell us much – it's a very strange number. Let's try turning it into a decimal or a percentage:

• 32 / 663 = 0.04826546... (about 0.048, or about 4.8%).

Another way of looking at it is to turn it upside-down to find the odds: 663 ÷ 32 is (aptly) about 21 – so if you played 21 games you'd only expect to get one blackjack. Something to think about How many games would you have to play so that the probability of getting at least one blackjack was greater than 0.5?