Question

You and a friend play a game where you each toss a balanced coin. If the upper faces on the coins are both tails, you win $8; if the faces are both heads, you win $4; if the coins do not match (one shows a head, the other a tail), you lose $2 (win (−$2)). Calculate the mean and variance of Y, your winnings on a single play of the game. Note that E(Y) > 0.

Answer 1

Mean: $2

Variance: 18

Explanation:

There are 4 possible outcomes:

TT (both tails)

TH (tail, head)

HT (head, tail)

HH (head, head)

Find the probabilities:

`P(TT)=P(TH)=P(HT)=P(HH)=(1)/(2)\cdot (1)/(2)=(1)/(4)`

Complete the table

`\begin{array}{ccccc}&TT&TH&HT&HH\\ \\P&(1)/(4)&(1)/(4)&(1)/(4)&(1)/(4)\\ \\Win&\$8&-\$2&-\$2&\$4\end{array}`

The mean is

`(1)/(4)\cdot 8+(1)/(4)\cdot (-2)+(1)/(4)\cdot (-2)+(1)/(4)\cdot 4=2`

The variance is

`((8-2)^2+(-2-2)^2+(-2-2)^2+(4-2)^2)/(4)=(36+16+16+4)/(4)=18`