# 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. • You don't need to do it so complicated. Just need to multiply the money he get if he wins in one flip, and multiply it by 4 and subtract the payoff! The money he gets for one flip this case is 1/2 times 2+1/2 times 0=1. Multiply by 4 to get 4, and subtract 5! – user547075 Apr 8 '18 at 4:15 ## 3 Answers Your answer is correct. Alternatively, you can make the probability distribution table: $$\begin{array}{c|c|c} X & P(X) & XP(X) \\ \hline -5 & \frac{1}{16} & -\frac{5}{16} \\ -3 & \frac{4}{16} & -\frac{12}{16} \\ -1 & \frac{6}{16} & -\frac{6}{16} \\ 1 & \frac{4}{16} & \frac{4}{16} \\ 3 & \frac{1}{16} & \frac{3}{16} \\ \hline & & -1 \end{array}$$ 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}

• I got the same answer but I was a bit confused by the question as it asks the expected winnings. Does that mean the number of time he wins or the amount he gets? – Mahamad A. Kanouté Apr 8 '18 at 3:56
• probably the amount – user547075 Apr 8 '18 at 3:56
• oh and sorry I thought it cost him 5dollars for each flip. lol – user547075 Apr 8 '18 at 3:56

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 ......)

• How do you know that the outcome is "1/2*(-5)+1/2*(-3)=-4 ". What does -4 means? why did you write it this particular way? – Mahamad A. Kanouté Apr 8 '18 at 3:41
• Do you have a way to do what you did using the Expectation function as I did? – Mahamad A. Kanouté Apr 8 '18 at 3:42
• This way is called the mathematical expectation – user547075 Apr 8 '18 at 3:47
• multiplying the probability of something to the price that a person eventually gets becomes the mathematical expectation. And, if there are several of them, you add them. For example, If the price for a gamble is 5 dollars and the money you get when you win is 35 dollars and the probabilty to win is 1/15, then the mathematical expectation is (1/15 )*30. – user547075 Apr 8 '18 at 3:49
• Sorry I made a mistake in the answer . – user547075 Apr 8 '18 at 3:53