A probability problem involving two cards (Part 3)

Probability, Tales from the Classroom 2 Minutes
In yesterday’s post, I gave two solutions that a student gave to the following probability problem. One of the solutions was correct, and one of the solutions was incorrect.
Two cards are dealt from a well-shuffled deck. Find the probability that the first is an ace or the second is a jack.

What follows is the incorrect (but, as we’ll see later, salvageable) solution.

Method #2 (incorrect):There are three ways that either the first or second card could be an a jack:
  1. The first card is an ace and the second card is a jack.
  2. The first card is an ace and the second card is not a jack.
  3. The first card is not an ace and the second card is a jack.
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 a jack, given first an ace) = \displaystyle \frac{4}{52} \cdot \frac{4}{51}
  2. P(first an ace) P(second not a jack, given first an ace) = \displaystyle \frac{4}{52} \cdot \frac{47}{51}
  3. P(first not an ace) P(second a jack, 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{4}{51} + \frac{4}{52} \cdot \frac{47}{51} + \frac{48}{52} \cdot \frac{4}{51}

= \displaystyle \frac{4 \times 4 + 4 \times 47 + 48 \times 4}{52 \times 51}

= \displaystyle \frac{4 + 47 + 48}{13 \times 51}

= \displaystyle \frac{99}{663}

When my student presented this to me, I must admit that it took me a couple of minutes before I found the hole in the student’s logic.

This answer is wrong because the second probability in Case 3 above was not calculated correctly. If we only know that the first card is not an ace, then we don’t have enough information to know how many of the remaining 51 cards are jacks. So the conditional probability \displaystyle \frac{4}{51} is incorrect in Step 3.

green lineEven though the above logic is slightly incorrect, it can be salvaged by splitting Case 3 above into two subcases.

Method #2:There are three ways that either the first or second card could be an a jack:
  1. The first card is an ace and the second card is a jack.
  2. The first card is an ace and the second card is not a jack.
  3. The first card is a jack, and the second card is a jack.
  4. The first card is neither an ace nor a jack, and the second card is a jack.
Each of these can be computed using the rule P(A \cap B) = P(A) P(B \mid A):
  1. P(first an ace) P(second a jack, given first an ace) = \displaystyle \frac{4}{52} \cdot \frac{4}{51}
  2. P(first an ace) P(second not a jack, given first an ace) = \displaystyle \frac{4}{52} \cdot \frac{47}{51}
  3. P(first a jack) P(second a jack, given first a jack) = \displaystyle \frac{4}{52} \cdot \frac{3}{51}
  4. P(first neither an ace or a jack) P(second a jack, given first neither an ace or a jack) = \displaystyle \frac{44}{52} \cdot \frac{4}{51}

By splitting the original Case 3 into the new Cases 3 and 4, there is no longer any ambiguity about how many jacks remain when the second card is chosen. Adding these together, we obtain the answer:

\displaystyle \frac{4}{52} \cdot \frac{4}{51} + \frac{4}{52} \cdot \frac{47}{51} + \frac{4}{52} \frac{3}{51} + \frac{44}{52} \cdot \frac{4}{51}

= \displaystyle \frac{4 \times 4 + 4 \times 47 + 4 \times 3 + 44 \times 4}{52 \times 51}

= \displaystyle \frac{4 + 47 + 3 + 44}{13 \times 51}

= \displaystyle \frac{98}{663}

Not surprisingly, this matches the answer obtained when the formula P(A \cup B) = P(A) + P(B) - P(A \cap B) was employed in yesterday’s post.

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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.

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One thought on “A probability problem involving two cards (Part 3)”

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

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