combinatorics sum 2

Question

Player 1 and Player 2 both start with 100 rupees. Each round of a game consists of the following: Both players choose a number randomly and independently from 1 to 5. If both players choose the same number, then Player 1 gives rupees 10 to Player 2. Otherwise, Player 2 gives rupees 10 to Player 1. Then the expected amount of money Player 1 will be left with after playing 10 rounds of this game is A. 120 B. 100 C. 50 D. 160

My Approach: Required expectation for 1 round= (1/5*1*(-10)+1/5*1/4*(10))=-1.5 rupees Therefore Expectation for 10 rounds = -1.5*10=-15 rupees

So the player 1 will be left with 100-15= 85 rs.

But this does not match any of the answer options.

Answer 1

I think the term in your expectation for winning should be read $\dfrac 15 \cdot \dfrac 14 \cdot 10$, but I don't see how you got that. Player $1$'s chance for winning is $\dfrac 45$, so that term should be $\dfrac 45 \cdot 10$

Answer 2

The expected gain for player $1$ per turn is $\frac{1}{5} \cdot (-10) + \frac{4}{5} \cdot (10) = 6$. After $10$ turns, that's $60$ rupees gained. On top of the original $100$, that's $160$ rupees total.

One way to think of the probability is to first assume Player $2$ picks a number. Then the probability that player $1$ picks that number is $\frac{1}{5}$. The probability of not picking that number is $\frac{4}{5}$.