 # Blackjack: Finding Expected Values of Games of Chance with Cards

In this video we learn about games of cards, and how to calculate probabilities. We look at the game of Blackjack and calculate the probability of getting certain hands. In addition we demonstrate how to calculate the expected value of a game that involves betting.

## Drawing Cards and Dependent Probabilities

Calculating probabilities with games of cards can be quite complicated depending on the game and the amount of players. In this video, we will simplify the games for ease of calculation.

Suppose you draw two cards from a standard deck of 52 cards, what is the probability they are both Aces?

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Just in case you were not sure, a standard deck of cards has 4 suits called diamonds, hearts, clubs and spades. Thus, there are 13 cards of each suit numbered from 2 to 10 and then the face cards of Jack, Queen and King. The final card, Ace, is also considered 1.

So, we are looking for two aces. Even though two cards are being dealt, to calculate the probability, they must be thought of as two cards being dealt one after the other. Let’s call them card 1 and card 2.

Since there are 4 aces in the pack and 52 cards, the probability that card 1 is an ace is 4 out of 52. Assuming we do not replace this card, there are now 51 cards left in the pack and only three Aces left. Thus, the probability that card 2 is an Ace is 3 out of 51. When the probabilities change like this from one event to another, we call them dependent events.

To calculate the final probability, we multiply the probabilities above since we want an Ace and an Ace. So the final probability of getting two Aces when you are dealt two cards from a standard deck of 52 is:

4 out of 52 multiplied by 3 out of 51, which equals 12 out of 2,652, which simplifies to 1 out of 221, or about 0.0045.

Thus, there is a 0.45% chance of getting two Aces.

Now let’s look at something a little more complicated, like the game of blackjack.

## Blackjack

James wants to learn how to win at the game of blackjack. Another name for this game is 21. The idea is to get as close to 21 points with the sum of the cards dealt to you from a standard deck of 52 cards.

In blackjack, or 21, each card is worth the same amount of points as the number shown on the face of the card. However, the face cards, Jack, Queen and King, are worth 10 points each, and Aces are worth either 1 or 11 points, depending on what you need to make the sum as close to 21 as possible.

The complete rules of blackjack in a casino are more complicated, but let’s examine a simple game to understand how the probabilities work.

James is dealt two cards from a standard deck of 52 cards. Let’s assume he is the only person playing this game. The rules are: He has to place a bet of \$5 to play a game. If he gets an Ace and a face card or an Ace and a Ten he will have a sum of 21, which is called blackjack! And for the blackjack combination, he wins \$10. But note his gain is only \$5 since he paid \$5 to play.

And, for anything else he loses his bet of \$5!

After every game, the cards are replaced and the pack reshuffled. If James continues to play this game, what can he expect to win or lose in the long run?

To calculate the expected value of this game, we first need to understand how cards work and find the probability of getting an Ace and a face card or Ten.

There are two scenarios to consider: Either James gets the Ace as card 1 and then he gets the face card or Ten as card 2, or he gets the face card or Ten first as card 1 and then the Ace as card 2.

Either one of those combinations gives James the winning hand. So what is the probability of getting this winning combination?

Let’s look at the scenario of getting the Ace first as card 1 and the face card or Ten as card 2. Since there are 4 suits, and thus four Aces in the pack to begin with, the probability of getting an Ace as the first card is 4/52.

The next card has to be a face card or a Ten. How many face cards are in the pack? Since there are 4 suits and 3 face cards in each suit, there are a total of 4 x 3 = 12 face cards in the pack. And there are four Tens in the pack. So we have 12 + 4 = 16 cards to choose from, out of how many left? The size of the pack has changed since we took card 1 out! There are now only 51 cards in the pack. And so the probability of getting a face card or a Ten as card 2 is 16 out of 51. These are called dependent events since the first outcome affected the second outcome. Dependent events are dependent on each other since the outcome of one affects the outcome of the other.

So the probability of getting an Ace as card 1 and a face card or a Ten as card 2 is:

4/52 multiplied by 16/51, which equals 16/663 when simplified.

Note, we multiplied the fractions because we want the probability of one event AND another event. In probability calculations, AND is a message to multiply.

So what about the probability of getting the face card or Ten first and then the Ace? Well, if we examine closely, this will be the same probability even though the numbers are reversed. This is shown in the completed table below.

The probability of getting the face card or Ten as card 1 is 16/52 and then the probability of getting the Ace as card 2 is 4/51. Multiplying them, we get 64/2652 which simplifies to 16/663.

We must consider both scenarios since both are possible. Since either one or the other will win us \$5 and OR in probability is an indication to add, the probability of getting an Ace and a face card or a Ten in any order when we are dealt two cards from a standard deck of 52 cards is 16/663 + 16/663, which equals 32/663, or about 0.048. In other words, when James is dealt two cards from a standard deck of 52 cards, assuming no other cards have been taken, he will get the combination of an Ace and a face card or a Ten about 4.8% of the time!

And this means that he will not get this winning combination 100 – 4.8% of the time which is about 95.2%! As a probability this is 0.952. Note, we switched from using fractions to decimals. This is perfectly fine as long as we do not round too much.

So what about the expected value of the game?

Let’s recall that James gains \$5 for the winning combination of Ace and a face card or Ten, but loses \$5 for anything else. To calculate the expected value, we multiply the values of each scenario by its probability. So, we have

\$5 * 0.048 = 0.24

Plus

-\$5 * 0.952 = -4.76.

So James can expect to have a gain of 0.24 minus 4.76, which equals negative \$4.52. In other words, James expects a loss of about \$4.52 every time he draws two cards! This is summarized in the table below.

Blackjack Value of X Probability of X P(X) X times P(X)
21 +\$5 0.048 0.24
Not 21 -\$5 0.952 -4.76
Expected Value -\$4.52

## Lesson Summary

In this video, we looked at probabilities involving games of cards. Since in most games cards are dealt and not replaced, the probabilities calculated were examples of dependent events, where the outcome of one event, in this case a card, affects the outcome of another event, or the next card.

To calculate these probabilities, we separated each card as an event. For example, when dealt two cards from a standard deck of 52 cards, the probability of getting two Aces was calculated as 4 out of 52 for the first Ace multiplied by 3 out of 51 for the second Ace. Using this method, we calculated the probability of getting 21 in the game of blackjack. We broke this down as getting an Ace first and then a face card or Ten OR getting a face card or Ten first and then the Ace. Since there were several ways to get this combination, we had to consider each option and then add the probabilities. Finally, we used the probability to calculate the expected value for a game in blackjack.

## Learning Outcomes

After this lesson is through you should be able to:

• Recall how to calculate the probability of dependent events
• Remember the rules and the winning combinations for blackjack
• Calculate the probability of winning a blackjack game
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