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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?

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  • $\begingroup$ "I know how to calculate how many initial tickets I will need on average, I just need to calculate the expected value of each prize type." Please show what you mean with that with an example. At the moment it is not very obvious what do you mean. $\endgroup$ Jun 27 at 14:44

2 Answers 2

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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 $-\$\frac{40}9$

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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.

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