Question

Danielle may choose one of two options for the method in which she may be awarded a money prize. OPTION A: Spin a spinner twice. The spinner is divided into four equally-sized sectors numbered 1, 4, 4, and 5. If the sum of the two spins is greater than 6, Danielle is awarded $8. Otherwise, she must pay $2. OPTION B: Flip a coin three times. If heads appears once, Danielle is awarded $6. Otherwise, she must pay $1. Danielle chooses the option with the greater mathematical expectation. How much more money can Danielle expect to make by choosing this option over the other option?

Answer 1

The mathematical expectation is a weighted sum:

$E(X) = \displaystyle \sum_{i=1}^n x_ip(x_i)$

i.e. we multiply each outcome with its probability, and sum all these terms.

There are 16 possible outcomes for the spin, and here's table with wins/losses:

$\begin{array}{c|cccc}&1&4&4&5\\1&L&L&L&L\\4&L&W&W&W\\4&L&W&W&W\\5&L&W&W&W\end{array}$

So, there are 9 winning spins and 7 losing spins. Since all the spins have the same probability, the probablity of winning $8 is 9/16, and the probability of losing $2 is 7/16. This leads to a mathematical expectation of

$E(A) = 8\cdot \dfrac{9}{16}-2\dfrac{7}{16} = \dfrac{29}{8}$

In the case of the three coin flips, all triplets have the same probability of 1/8, and the eight triplets are

TTT, TTH, THT, HTT, THH, HTH, HHT, TTT

So, Danielle wins with 3 triplets, and loses with 5 triplets. The mathematical expectation is

$E(B) = 6\cdot \dfrac{3}{8}-1\dfrac{5}{8} = \dfrac{13}{8}$

So, the first method is better, and the difference is 29/8-13/8 = 2.