Solution: 2014 Fall Final - 22

Author: Michiel Smid

Question

I flip a fair coin, independently, 6 times, resulting in a sequence of heads ($H$) and tails ($T$). For each (consecutive) $HTH$ in this sequence, you win \$5. Define the random variable $X$ to be the amount of dollars that you win. For example, if the sequence is $$ THTHTH, $$ then $X = 10$. What is the expected value of $X$?

(a)

5/2

(b)

(c)

(d)

2/5

Solution

Probability of Getting $ HTH $:
The probability of getting a head followed by a tail followed by a head in three consecutive flips is:
$P(HTH) = P(H) \times P(T) \times P(H) = \left( \frac{1}{2}\right) \times \left( \frac{1}{2}\right) \times \left( \frac{1}{2}\right) = \frac{1}{8} $
Number of Possible $ HTH $ in 6 Flips:
Since we are looking for consecutive $ HTH $ patterns, we need to consider overlapping patterns in the sequence.
In a sequence of 6 coin flips, the positions where a $ HTH $ can start are:

1st position: $ {1, 2, 3} $
2nd position: $ {2, 3, 4} $
3rd position: $ {3, 4, 5} $
4th position: $ {4, 5, 6} $

Therefore, there are 4 possible positions for a $ HTH $ to occur.

Expected Value Contribution from Each Position:
For each position, the expected value contribution is:
$E(\text{one } HTH) = P(HTH) \times \text{payout} = \frac{1}{8} \times 5 = \frac{5}{8}$
Total Expected Value $ E(X) $:
Since there are 4 possible positions, the total expected value is the sum of the contributions from each position:
$E(X) = 4 \times \frac{5}{8} = \frac{20}{8} = \frac{5}{2}$
Therefore, the expected value of $ X $ is $ \boxed{ \frac{5}{2}} $.