# If an option's price is 100% made up of its intrinsic value, is there a way to guarantee a non-loss while having a chance at a profit?

Let's hypothetically say that a stock is trading at \$153, and I can buy its \$150 call option for \$3.

If I shorted 100 shares of the stock and bought 100 options, then I would be guaranteed to break even (let's just assume \$0 commissions). If the stock went up \$5, I'd gain \$500 on the options and lose \$500 on the short sell. If if went down \$3, the options would expire worthless (loss of \$300) but I'd gain \$300 on the short sell. And so on.

However, it seems like the option would be mispriced, because there of course is some chance that the stock will rise in value (while its \$3 price implies there is no time value component to it).

If this were the case, is there a strategy that would cap my loss at \$0 (as above) but give me some upside if the stock were to increase in price?

• Note that with the strategy you describe, your profit comes if the stock falls more than \$3. If it falls \$10, you gain \$1000 on the short sell, paying for the \$300 of options and leaving \$700 of profit. – Nate Eldredge Mar 5 '15 at 18:35
• Ah, that was really stupid of me not to realize. Thanks! So I guess the inverse (make money if the stock goes up but guarantee no loss if it goes down) is to buy out of the money puts (or maybe writing calls?) and going long the stock... – Jer Mar 5 '15 at 18:38
• By shorting the stock and buying a Call, you have created a synthetic Put. A Box spread locks P/L. – Optionparty Mar 6 '15 at 3:48