Imagine you have a gambling game with a house advantage of $1\%$. The game is very simple, you flip a coin and you win or lose. Odds of either is $50\%$. If you win, you get $1.98$ times your bet.
Now what would it do to the house advantage, if the player is allowed to flip the coin again if he loses? For example, a player bets $\$10$, flips a coin, loses, flips again (for free), and could win. What is the house advantage in this scenario if such rule would apply?
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The return when the player wins is $R(W)=-0.98$ (a loss for the house). The return when the player looses is $R(L) = +1.00$ (a win for the house).
House advantage is the expected return for the house. $\operatorname{E}_X[R(X)]$, where $X$ is the result of a game.
In the one flip game the expected return is: $$\begin{align}\operatorname{E}_X[R(X)] & = R(W) \Pr(W)+R(L) \Pr(L) \\ ~ & = (-0.98)(0.5) + (1.00)(0.5) \\ ~ & = +0.01\end{align}$$
In the two flip game the player looses if they loose two flips of the coin, the player wins otherwise. This gives: $$\begin{align}\operatorname{E}_Y[R(Y)] & = R(W) \Pr(W_1 \cup (L_1\cap W_2)) + R(L) \Pr(L_1\cap L_2)\end{align}$$ Here $W_n$ is the event of a win on flip $n$, $L_n$ is the event of a loss.
Hint: what is the chance the player wins? He need to win one flip of two.