Question

You are interested in playing a game where youhave to flip three fair coins. To win, all three coinsmust show the same face (all heads or all tails), andyou will win four times your original wager.Otherwise, you lose your original bet. If you pay $1each time you play this game, and you play 20times, what is your expected net gain or loss.

Answer 1

First, let's calculate the probability of winning the game.

We win the game if we get three heads or three tails, so the probability is:

$\begin{gathered} P(3\text{ heads})=\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{8}\\ \\ P(3\text{ tails})=\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{8}\\ \\ P(win)=\frac{1}{8}+\frac{1}{8}=\frac{2}{8}=\frac{1}{4} \end{gathered}$

If the probability of winning is 1/4, the probability of losing is 3/4.

If the wager is $1, winning will return $4 plus the original bet, so a net earning of $4.

Losing will return nothing, so the net "earning" is -$1.

Calculating the expected value of one game, we have:

$\begin{gathered} E(x)=\sum x\cdot p(x)\\ \\ E(x)=\frac{1}{4}\cdot4+\frac{3}{4}\cdot(-1)\\ \\ E(x)=\frac{4}{4}-\frac{3}{4}\\ \\ E(x)=\frac{1}{4} \end{gathered}$

Playing 20 times will result in an earning of 20 * 1/4, that is, a gain of $5.

Correct option: fourth one.