Averages and expected value

Goal: Compute expected values and use linearity: E[count] = n × (target cards ÷ 52).

Remember: Expected value is the long-run average. It need not be a value a single trial can produce.

1. 1.Five cards are drawn. What is the expected number of hearts?

   Answer: 1.25

   Working: 5 × 13/52

2. 2.Thirteen cards are drawn. What is the expected number of aces?

   Answer: 1

   Working: 13 × 4/52

3. 3.Classroom scoring exercise: a number card scores 0 points, a face card 1 point and an ace 3 points (the points are only a counting device). What is the expected score of one card?

   Answer: 24/52 ≈ 0.46

   Working: (36 × 0 + 12 × 1 + 4 × 3) ÷ 52

4. 4.Four cards are drawn. What is the expected number of face cards?

   Answer: 48/52 ≈ 0.92

   Working: 4 × 12/52

5. 5.Seven cards are drawn. What is the expected number of red cards?

   Answer: 3.5

   Working: 7 × 26/52

6. 6.A fair six-sided die is rolled. What is the expected value?

   Answer: 3.5

   Working: (1 + 2 + 3 + 4 + 5 + 6) ÷ 6

7. 7.True or false: the expected value must be a number that can actually occur on a single trial. Give an example.

   Answer: False. A die has expected value 3.5, which can never be rolled; 1.25 expected hearts is not a possible count either.