A probability problem involving two cards (Part 1)

Probability, Tales from the Classroom 2 Minutes
Here is a standard problem that could appear in an elementary probability class.
Two cards are dealt from a well-shuffled deck. Find the probability that the first is an ace or the second is a ace.
There are at least two legitimate ways to solve this problem:
Method #1: One law for probability states that
P(A \cup B) = P(A) + P(B) - P(A \cap B)
Another law of probability states that
P(A \cap B) = P(A) P(B \mid A)
Combining these, we find that
P(A \cup B) = P(A) + P(B) - P(A) P(B \mid A)
Written more colloquially,
P(first an ace or second an ace)
= P(first an ace) + P(second an ace) – P(first an ace AND second an ace)
=P(first an ace) + P(second an ace) – P(first an ace) P(second an ace, given first an ace)
Let’s look at these three probabilities on the last line separately.
  1. P(first an ace) is \displaystyle \frac{4}{52}.
  2. P(second an ace) is also \displaystyle \frac{4}{52}. No information about the first card appears between the two parentheses, and so this is similar to pulling a card out of the middle of a deck. Since no information is given about the preceding card(s), the answer is still \displaystyle \frac{4}{52}.
  3. P(second an ace, given first an ace) is different than the answer to #2 above. For this problem, the first card is known to be an ace, and the question is, given this knowledge, what is the probability that the second card is an ace? Since the first card is known to be an ace, there are only 3 aces left out of 51 possible cards. Therefore, the answer is \displaystyle \frac{3}{51}.

Putting these together, we find the final solution of

\displaystyle \frac{4}{52} + \frac{4}{52} - \frac{4}{52} \cdot \frac{3}{51}

= \displaystyle \frac{1}{13} + \frac{1}{13} - \frac{1}{13} \cdot \frac{1}{17}

= \displaystyle \frac{17+17-1}{13 \times 17}

= \displaystyle \frac{33}{221}

green lineHere’s a second legitimate solution, though it does take a little more work.

Method #2:There are three ways that either the first or second card could be an ace:
  1. The first card is an ace and the second card is an ace.
  2. The first card is an ace and the second card is not an ace.
  3. The first card is not an ace and the second card is an ace.
Each of these can be computed using the rule P(A \cap B) = P(A) P(B \mid A) in much the same way as above:
  1. P(first an ace) P(second an ace, given first an ace) = \displaystyle \frac{4}{52} \cdot \frac{3}{51}
  2. P(first an ace) P(second not an ace, given first an ace) = \displaystyle \frac{4}{52} \cdot \frac{48}{51}
  3. P(first not an ace) P(second an ace, given first not an ace) = \displaystyle \frac{48}{52} \cdot \frac{4}{51}

Adding these together, we obtain the answer:

\displaystyle \frac{4}{52} \cdot \frac{3}{51} + \frac{4}{52} \cdot \frac{48}{51} + \frac{48}{52} \cdot \frac{4}{51}

= \displaystyle \frac{4 \times 3 + 4 \times 48 + 48 \times 4}{52 \times 51}

= \displaystyle \frac{3 + 48 + 48}{13 \times 51}

= \displaystyle \frac{1 + 16 + 16}{13 \times 17}

\displaystyle \frac{33}{221}

Not surprisingly, the two answers are the same.

green line

In tomorrow’s post, I’ll describe the time that a student came to me with a similar-looking probability problem, but she obtained two different answers using these two different methods.

Unknown's avatar

Published by John Quintanilla

I'm a Professor of Mathematics and a University Distinguished Teaching Professor at the University of North Texas. For eight years, I was co-director of Teach North Texas, UNT's program for preparing secondary teachers of mathematics and science.

Published

One thought on “A probability problem involving two cards (Part 1)”

  1. Pingback: A probability problem involving two cards: Index | Mean Green Math

Leave a comment

This site uses Akismet to reduce spam. Learn how your comment data is processed.