# How to Solve Basic Probability Problems Involving a Deck of Cards

#### 25 JUN 2018

#### CLASS

A common topic in introductory probability is solving problems involving a deck of standard playing cards. These can be handy if you are playing card games or just trying to understand how probability works. Following a few simple steps and principles will tell you how to solve probability exercises that involve decks of cards.

For all problems of this type, there are some important points that apply. First of all, the problem will probably refer to a standard deck of playing cards. This means that there are no tricks involved. The problem assumes a regular deck of cards, not "stacked," shuffled randomly, with cards drawn randomly.

Some students claim that questions of this nature are unfair, especially if they grew up in a culture that doesn't play games using what we call a standard deck of cards. While this may be the case, it is not hard to learn the facts about a deck of cards that you are expected to know.

A standard deck of cards contains 52 different cards. It contains cards of 13 different ranks, ranging from Ace (essentially 1) through 10, followed by Jack, Queen, King, which you could think of as 11, 12, and 13. In each rank there are cards of four suits: a heart, a club, a diamond, and spade. Hearts and diamonds are red, spades and clubs are black. There are 4 cards of each rank, and 13 cards of each suit. There are no jokers. That is all you need to know to answer any problem involving a deck of cards.

Here is a simple problem: "A person draws a card from a standard deck, and it is the Queen of Hearts. The card is replaced, and the deck is reshuffled. What are the chances of drawing the Queen of Spades on the next draw?" First of all, the word "replace" in this context means "put back."

This is actually a trick question. The fact that the Queen of Hearts was drawn on the first draw has nothing at all to do with the second draw, since it was returned to the deck, and the deck was reshuffled. The deck doesn't have a memory. It is incorrect to say that the Queen of Hearts is "on a roll" so is more likely to come up again, just as it is incorrect to say that the Queen of Hearts is less likely to come up again because other cards are "overdue." The answer to the question is simply 1/52.

Here is some other typical problems with some of the standard wording omitted here for brevity: "What are the chances of drawing a red card?" There are 2 red suits of 13 cards each, so the answer is 26/52 which we would probably reduce to 1/2. "What are the chances of drawing a seven?" There are four sevens out of 52, giving us 4/52 which we would probably reduce to 1/13. "What are the chances of drawing a club?" There are 13 clubs out of 52, giving us 13/52 which we would probably reduce to 1/13.

Be on the lookout for trick questions: "What are the chances of drawing a green card?" The answer is 0. There are none. "What are the chances of drawing either a red or a black card?" The answer is 52/52, which equals 1, or equivalently 100%. Every card in the deck is one or the other.

Here is a problem that is a bit trickier: "Two cards will be drawn from a standard deck without replacement. What are the chances of drawing the Nine of Clubs followed by a red card?" First of all, take note of the fact that we will not be putting the first card back in the deck after drawing it. The chances of drawing the Nine of Clubs on the first draw is 1/52. Now that card is gone, and we have 51 left. The chances of drawing a red card from the remaining cards are 26/51. There are still 26 red cards left, since the Nine of Clubs was not one of them.

This problem implies an "and" condition, and for those problems we multiply the individual probabilities. We must multiply 1/52 times 26/51, giving us 26/2652, which we would probably reduce to 1/102.

Here is another typical problem: "Two cards will be drawn from a standard deck with replacement, and with shuffling in between draws. What are the chances of drawing a Three on the first draw, and a diamond on the second draw?" Take note of the fact that we are dealing with a replacement scenario. The chances of drawing a Three on the first draw is 4/52. The chances of drawing a diamond on the second draw is 13/52. Each draw has nothing to do with the previous, since each draw started from a complete, shuffled deck. Multiply 4/52 times 13/52 to get 52/2704, which reduces to 1/52.

Here is a final typical problem which can be a bit tricky. "What are the chances of drawing a Five or a Diamond?" We are dealing with an "or" situation, which means we have to add (not multiply) the involved probabilities. The chances of drawing a Five are 4/52. The chances of drawing a diamond are 13/52. Many students just add those two fractions together to get 17/52, but that is actually wrong. The problem is that we counted the Five of Diamonds twice, once as a Five, and again as a Diamond. We have to subtract one of those times, so that we only count it once, so we end up with 16/52 which is the correct answer.

Another way to think about it is that there are 13 diamonds in the deck, and then we just have to count the three Fives that are not diamonds. That gives us 16 possible cards out of 52.

Students should make sure that they are comfortable working with the basic probability concepts discussed in this article since they come up quite frequently.