What are his expected winnings?

Question

Bob is playing a game of chance. It costs him $5 to play. He flips a coin 4 times and guesses its outcome on each flip. For each one he guesses correctly, he receives \$2. What are his expected winnings?

I tried :

Y = net winnings

X = number of correct guesses

The net winning equation I came up with is: $$Y = 2X-5$$

From what, I know that $E(Y)=-5+2E(X)$.

Since X~B(4,1/2), then E(X)=np=2

So in conclusion E(Y)=-1

Can a net winning expectation be - 1?

would that mean he will lose one 1$ on average?

I feel like he wins when he gets a correct bet but that can also be the case when he wins some money. I'm a bit confused and any help would be welcome.

Answer 1

Your answer is correct. Alternatively, you can make the probability distribution table: $$\begin{array}{c|c|c} X & P(X) & XP(X) \\ \hline -5 & \frac{1}{16} & -\frac{5}{16} \\ -3 & \frac{4}{16} & -\frac{12}{16} \\ -1 & \frac{6}{16} & -\frac{6}{16} \\ 1 & \frac{4}{16} & \frac{4}{16} \\ 3 & \frac{1}{16} & \frac{3}{16} \\ \hline & & -1 \end{array}$$

Answer 2

Since there are four flips, we just have to multiply the mathematical expectation for one flip to 4. For each flip, the mathematical expectation is (-3)* 1/2.

Therefore, the mathematical expectation for the whole flips is (-3)*(1/2)*4=-6

It is very unusual because in order for him to win some money, the money he receives should be higher than the price it takes him to participate in the game.

If I were him, I would never participate in that game! :)

( Sorry to say that I thought it cost him 5 dollars for each flips ......)

Answer 3

Your solution is perfectly correct.

Here's a slightly different (but not better) solution:

Let $X_i$ be the amount of money that is won on the $i$th toss. The expected value of $X_i$ is $(1/2)\cdot 2 + (1/2)\cdot 0 = 1$.

Let $X$ be Bob's net winnings, so that $X = X_1 + X_2 + X_3 + X_4 - 5$. Then \begin{align} E(X) &= E(X_1) + E(X_2) + E(X_3) + E(X_4) - E(5) \\ &= 1 + 1 + 1 + 1 - 5 \\ &= -1. \end{align}