Consider these two games:
Game $1$: Toss $4$ coins. If coins $1$ and $2$ are heads, you win $\$5$. If coins $3$ and $4$ are heads, you win an additional $\$5$.
Game $2$: Toss $3$ coins. If coins $1$ and $2$ are heads, you win $\$5$. If coins $2$ and $3$ are heads, you win an additional $\$5$.
My question is, which game will have the highest expected value for winnings? The case of game $1$ is simple. Receiving $\$5$ from tossing coins $1$ and $2$ and receiving $\$5$ from tossing coins $3$ and $4$ are independent events. Therefore, if the coins lands heads with probability $p$, the expected value of the winnings is$$
E(\text{Game} \; 1) = 5 p^2 + 5 p^2 = 10p^2.
$$
In game $2$, we still have the probability $p^2$ of winning the first $\$5$. However, the probability of receiving the additional $\$5$from coin $3$ is dependent upon having a heads in coin $2$. I figured so far that
$$E(\text{Game}\;2) = 5p^2 + x$$
I am not sure what $x$ is. Which result links the expected value of an event when conditional probabilities are involved?
EDIT: My solution approach is wrong, please see my comment below.
Best Answer
In game $2$ you win $10$ if all the first three come up heads, which is probability $p^3$. You win $5$ if the first three come up $HHT$ or $THH$, which is probability $2p^2(1-p)$. The total expectation is $10p^3+10p^2(1-p)=10p^3+10p^2-10p^3=10p^2$. The expectation is the same.
This is an example of the linearity of expectation. The expectation from each pair is $5p^2$. Each game has two pairs that matter. The linearity of expectation does not require independence.