Two players play the following game with a machine:
1) Each player has 20 tickets. 2) Player One takes 1.23 Euro per ticket from the machine and also has 5% chance that the machine will return his ticket and he can use it again for another 1.23 Euro 3)Player Two takes 1 Euro per ticket from the machine and also has 36% chance that the machine will return his ticket and he can use it again for another 1 Euro
Which one of the two players has higher expected earnings after he uses all the tickets and what is the expected margin in percents between them?
My logic: Player A: 20 turns multiplied by 1.05 for the 5 percents give 21 usages which leads to 21*1.23 = 25.83 euro Player B: 20 turns multiplied by 1.36 for the 36 percents give 27 usages which leads to 27*1 = 27 euro Expected margin - 27/25.83 = 4.53 percents
Is this correct?
Your reasoning is incorrect. Notice that for any player, a ticket that is returned can be used once again, and it can be returned some more times afterwards. Actually, to model how much is gained in the game can be calculated by considering each ticket as a geometric random variable. Let $X_i$ be the random variable that counts how many times ticket $i$ was used, for player $1$. Notice that $X_i$ are Geometric with parameter $0.95$ of success (here, success means that the ticket did not return). Then, the amount that player $1$ gained is $$ 1.23 \cdot (X_1 + \ldots + X_{20}) $$ And its expected value is given by $$ 1.23 \cdot 20 \cdot \mathbb{E}[X_1] = \frac{1.23 \cdot 20}{0.95} \approx 25.8947, $$ since the expected value of a Geometric of parameter $p$ is $\frac1p$. Analogously, for player $2$ the expected gain is given by $$ \frac{1 \cdot 20}{0.64} = 31.25 $$