At time $t = 0$, a stock sells for $75$. At time $t = 1$, the stock will sell for either $50$, $75$, or $100$. You can purchase (or sell) the following two options:
(1) An option to buy the stock at time $t = 1$ for $50$;
(2) An option to buy the stock at time $t = 1$ for $60$.
The price of the first option is $10$. The price of the second option is $5$.
Give an example of a set of buy/sell decisions for the two options that guarantees a profit. (Note: only buy/sell the options, not allowed to buy/sell the stock.)
My attempt: Assume we buy/sell $x$ options $1$ for $10x$, and buy/sell $y$ option $2$ for $5y$. The total value at t=1 of ALL two options is: $0$ if price $= 50$ at $t=1$, $25x+15y$ if price $= 75$ at $t = 1$, and $50x+ 40y$ if price $= 100$ at $t=1$. Now, we need to choose $x$ and $y$ so that $25x+15y = 50x+40y$, which implies $y = -x$.
Thus, the final profit $ = 10x - (10x+5y) = 5x$ if price $=75$ or $100$ at $t=1$, and $= -5x$ if price $=50.$ This means $x$ needs to be simultaneously positive and negative, which is impossible. So no arbitrage??
My question: For the case when we sell the option, do we need to worry about "having to buy it" in case the option holder exercises the option? Otherwise, just sell $2$ call options and we're done?
I suggest you sell $3$ options with a strike price of $50$ and buy $5$ options with a strike price of $60$, or some multiple of this
You will receive a net premium of $3\times 10-5 \times 5=5$
So whatever happens you both get the premium of $5$ and have options worth $0$ or $50$, for an overall profit of $5$ or $55$ (plus any interest on the premium)
You would lose money if there were no interest and the price at time $t=1$ were between $51.67$ and $72.50$, but the question suggests this is not possible
Your error seems to be in trying to have the same profit with final prices of $75$ and $100$; meanwhile just selling two call options will lose money if the final price is $75$ or $100$ as the options will be exercised