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### Probability: Dealing with a Deck of Cards

There are a number of questions of probability that can be formed on a pack of cards. First, a little intro about a pack of cards:

### Standard Deck of 52 Cards

There are a total of 52 cards in a pack of cards, divided into two colors:
• Red (26 cards)
• Black (26 cards)
There are a total of 4 suits of cards, two each in red and black. They contain 13 cards each. They are:
• Spades (black) - 13 cards
• Clubs (black) - 13 cards
• Diamonds (red) - 13 cards
• Hears (red) - 13 cards
Each of these four suits, having 13 cards, contains the following cards:
• 1 King
• 1 Queen
• 1 Jack
• 1 Ace
• 2 to 10 numbered cards
Face cards: Each suit has a King, a Queen and a Jack. These three cards are known as the face cards of each suit. Since there are four suits, and three face cards in each suit, so there are a total of 4 * 3 = 12 face cards in a deck of 52 cards.

It would be logical to relate to the standard deck (or pack) of fifty two cards explained above whenever you encounter the phrase "the deck", or "the pack" in the discussion below.

### Probability of Getting a Particular Card

#### What do you think might be the probability of getting a particular card, say a King of Hearts, when you draw one card out of a standard deck of fifty two cards?

It is simple. There are a total of fifty two cards to draw from, so the total number of different possible outcomes is fifty two. There is just one King of Hearts in a standard deck of fifty two cards. Hence the total number of favorable outcomes is one. So, applying the formula for theoretical probability:
P("outcome") = "Number of favorable outcomes"/"Total number of possible outcomes"
P("King of Hearts") = 1/52
(If you didn't understand this one, you probably need to understand the basics of calculating theoretical probability. These are discussed in these posts:
So better read that before proceeding on.)

#### Another one: What is the probability of getting an Ace when drawing one card out of a standard deck of fifty two cards?

This one is a little different from the previous one. There are four aces in a standard deck of fifty two cards, one of each of the suits. We already know, from the previous question's solution, that there are a total of fifty two possible outcomes when you draw a single card out of a deck of fifty two cards. The number of favorable outcomes is, however, different in this question. There are four aces in a pack of cards, so if you draw any one of the four aces, you get a favorable outcome. Hence the number of favorable outcomes is four. Applying the formula for theoretical probability,
P("outcome")="Number of favorable outcomes"/"Total number of outcomes"
P("Ace")=4/52=1/13
Hence the probability of drawing an ace from a pack of 52 cards is 1/13, which means, that out of every thirteen cards you draw one by one, one of them is probably an ace.

#### What is the probability of getting a red Ace or a black Jack on drawing a card from a standard deck of fifty two cards?

Again the total number of possible events is fifty two, since there are fifty two cards in a standard deck of cards. However, the number of favorable events is different in this one. How many red aces and black jacks can you find in a pack of fifty two cards? Please refer to the section "Standard Deck of 52 Cards" above to try to get that yourself before reading on. If you need help, however, we state again that there are a total of four aces in a standard deck, each one of the four different suits. The four suits are spades and clubs (black) and hearts and diamonds (red). So there are two red aces and two black aces in a deck.

Similarly there are four Jacks in a deck of cards, each one of the four different suits. Likewise, there are two red Jacks and two black Jacks in a deck.

The question asks you to find the probability of getting a red ace or a black jack on drawing one card. Since there are two red aces and two black Jacks in a deck of cards and you can draw any one of them to get the favorable outcome, hence the number of favorable events is four (= 2 red aces + 2 black jacks).

Applying the formula for theoretical probability again, we get
P("outcome")="Number of favorable outcomes"/"Total number of outcomes"
P("red Ace or black Jack")=4/52 = 1/13
Till now you must have understood how to calculate the probability of getting a particular card or one of a particular combination of cards from a standard deck of fifty two cards. So let us advance to a few problems on probability that deal with drawing two cards from the deck.

Click here to see more solved problems on probability of selecting a particular type of card from a deck.

### Drawing Two Cards Without Replacing

This is a typical question with special regards to "without replacing" the first card. This means that you draw the second card without replacing the first card in the pack. So there are fifty one cards left to draw from after drawing the first card, and to draw the second card you have to choose from these fifty one cards. After drawing the second card, fifty cards are left in the deck.

#### What is the probability of drawing two red Aces one by one when drawing two cards out of a deck of fifty two cards?

Consider a deck of fifty two cards lying in front of you. There are a total of fifty two cards in the deck containing two red Aces. The first card that you draw has to be a red ace. On drawing the first card, there are a total number of fifty different possible outcomes possible and a total of two favorable outcomes from which one can occur. So, by the formula for theoretical probability, the probability of getting a red ace is:
P("first card - red ace") = 2/52 = 1/26
Suppose you did draw a red ace (against all odds). Then there is only one other red ace left in the deck of cards in front of you. The deck also now contains only fifty one cards, since you have drawn one card and did not replace it. Probability of getting a red ace in drawing the second card is:
P("second card - red ace") = "Number of red aces"/"Total number of cards"
P("second card - red ace") = 1/51
Consider: The probability of drawing a red ace in the first draw is 1/26 and that of drawing another red ace in the second draw is 1/51. What is the probability that both of these events can occur together? Here comes the concept of independent events, and their compound probability. The compound probability, that is, the probability of both of the above events happening together, is a product of their individual probabilities:
P("both cards - red ace")=1/26 * 1/51 = 1/1326
Hence the probability of getting a red ace in both the first and second draws is 1 in 1326 double draws. Quite a large number (or, equivalently, quite a small probability), isn't it?

#### What is the probability of getting two cards of the same color when drawing them one by one without replacing from a standard deck of fifty two cards?

There are cards of two colors in a deck: red and black. There are 26 red and 26 black cards in it. So you can either draw two red cards or two black cards from the pack.

First consider the probability of getting two red cards: Out of fifty two cards, 26 are red, so the probability of getting a red card when drawing your first card is 26/52 = 1/2. After drawing the first card, there are 51 cards left in the pack, out of which, 25 are red and 26 are black because the first card you drew was red in color. Thus, to draw a red card from the remaining set of cards, the probability of drawing a red card is 25/51.
The two events of drawing the cards one by one are two independent events and thus the probability of their occurring together is given by the product of their individual probabilities. Thus the probability of getting two red cards is 1/2 * 25/51 = 25/102

Now let us consider the probability of getting two black cards: Out of fifty two cards, 26 are black, so the probability of getting a black card when drawing your first card is 26/52 = 1/2. After drawing the first card, there are 51 cards left in the pack, out of which, 25 are black and 26 are red because the first card you drew was black in color. Thus, to draw a black card from the remaining set of cards, the probability of drawing a black card is 25/51.
The two events of drawing the two cards one by one are two independent events and thus the probability of their occurring together is given by the product of their probabilities. Thus the probability of getting two black cards is 1/2 * 25/51 = 25/102

P(both cards of same color) = P(both red or both black) = P(both red) + P(both black)
P(both cards of same color) = \frac{25}{102} + \frac{25}{102} = \frac{50}{102} = \frac{25}{51}

Thus the probability of selecting two cards of the same color from a deck of cards is 25/51.