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