Calculating odds with an extra draw as a possible outcome

Question

_{I rephrased the whole problem to be simpler and more generic.}

Consider a game where playing costs $\$10$. You have a $80\%$ chance to lose, and $20\%$ chance to win $\$25$. The expected profit can be calculated as:

$$E[X]=-\$10 \cdot 0.8 + \$15 \cdot 0.2 = -\$5$$

Now consider a second game, where you have to buy a ticket for $\$10$ to play. Then you have a $75\%$ chance to lose, $20\%$ chance to win 25 dollars, and $5\%$ chance to win 2 tickets, that you can only use to play again. If you get the 2 tickets, you can have another shot at the game like you would normally, plus one "free" game. Another shot at the 25 dollars but also an opportunity to get 2 tickets once again.

How do you calculate the expected profit for this game?

Answer 1

Let $Y$ be the expected winnings from using a single ticket (that you already own).

$$Y=\$25\cdot 0.2+0.05\cdot2Y$$

Then, using basic algebra, you solve and get $Y=\$\frac{50}{9}$.

Now, considering that the cost of a ticket is $\$10$, your expected winning is $-\$\frac{40}9$

Answer 2

To get a cleaner and more complete answer:

Considering a more general case, where in a game there is a $p_1$ chance to win $x_1$, a $p_2$ chance to win $x_2$, etc. ($0<p_1+\cdots+p_k\leq1$). The expected gain from a game is:

$$E = x_1 p_1 + x_2 p_2 + \cdots + x_k p_k$$

Now if there is a chance $p_t$ to win two tickets for the game ($0<p_1+⋯+p_k+p_t≤1$), the expected gain $E'$ is:

$$E' = x_1 p_1 + x_2 p_2 + \cdots + x_k p_k + p_t \cdot 2E'$$

That with some algebra can be rewritten as:

$$E' = \frac{x_1 p_1 + x_2 p_2 + \cdots + x_k p_k}{1-2p_t}$$

$$E' = \frac{1}{1-2p_t} E$$

So adding to the game a chance $pt$ to get 2 more tickets multiplies the expected value by $\frac{1}{1-2p_t}$, at least if $p_t < \frac{1}{2}$. I can't make sense of the results of $p_t=0.5$ or if $p_t>0.5$ however.