# Inverted strangle

I am researching inverted strangles and feel like I’m missing something. Suppose I buy 100 shares of XYZ @ \$50. I then sell one deep ITM call at \$45 for \$6, resulting in a \$1 profit. At the same time I sell a \$55 put for \$5, resulting in a cost basis of \$50 if the put is assigned. Assuming both the call and put are assigned, I receive \$1 premium profit and still get 100 shares of XYZ at the same price.

What am I missing? Thanks in advanced for any help.

Let's look at the three payoff zones:

If the stock goes above \$55, then you have to sell it to the call holder for \$45 for a gain of \$1 (\$5 loss on the stock + \$6 gain in premium).

If the stock is between \$55 and \$45, then you have to sell it to the call holder for \$45 for a net gain of \$1, but you also have to buy it from the put holder for \$55, for a net loss of (55 - price).

If the stock goes below \$45, then you are required to buy another 100 shares from the put holder for \$55, which doubles your losses.

So your maximum gain is only \$1, and your maximum loss is unlimited, and your exposure is doubled if the stock drops below the call price.

A more common strategy is a naked inverse straddle, where you have a net gain if the stock is between the strikes, and a net loss otherwise (modulo any premium you get in excess of the width of the straddle). There is still unlimited downside; owning the stock just removes the loss potential on a gain and doubles the exposure if the stock goes down.

• Great, thank you very much! Laid out very clearly.
– VDP
May 3, 2018 at 21:15
• It's hard to determine "safer" with two strategies with different R/R spectrums. One is bullish with an asymmetric payoff while the other is neutral, each with little margin for error on the short side. Apples and oranges. May 3, 2018 at 22:25
• @BobBaerker Good point - I've updated my answer to illustrate the differences. May 4, 2018 at 13:33
• Your math needs some assistance. Between \$45 and \$55 you failed to account for the \$5 in put premium received so the put's gain or loss in that range is (Price - \$50) not (\$55 - Price). 'Doubles your losses" could be clearer -> Below \$45, you lose \$2 for every \$1 the underlying drops. The maximum gain is \$6 (not \$1) because above \$55 you keep the \$1 of call premium and the \$5 of put premium. There is no unlimited loss with short puts since you are buying stock. Unlimited loss is from short calls becoming short stock. Sep 16, 2018 at 17:39

The problem is that you aren't dealing with an inverted strangle. You have sold two puts, one synthetic and one naked. Let's take a deep dive into equivalent positions.

There are 6 basic synthetic positions relating to combinations of put options, call options and their underlying stock (the Synthetic Triangle):

1. Synthetic Long Stock = Long Call + Short Put

2. Synthetic Short Stock = Short Call + Long Put

3. Synthetic Long Call = Long Stock + Long Put

4. Synthetic Short Call = Short Stock + Short Put

5. Synthetic Short Put = Long Stock + Short Call

6. Synthetic Long Put = Short Stock + Long Call

These are all variations of S + P - C = 0 which is the core of put/call parity.

Note that # 5 which shows that a short put equals a covered call (+ STK - Call). As applied to your example, you bought the stock and sold a \$45 call. This is equivalent to having sold a \$45 put. Then, you sold a \$55 put, ending up with one short \$45 put and one short \$55 put. This isn't an Inverted Short Strangle (often called a Guts Strangle). It's two naked/short puts. See the reply from D Stanley explaining the P&L of the position.

If you were executing the position all at once, it would be better to sell the equivalent OTM strangle because B/A spreads on OTM options tend to be much narrower and you'll avoid the possibility of early assignment (assuming these are American options).

+STK + \$45p = + \$45c

+STK + \$55p = + \$55c

Factoring both equations you end up with:

+STK + \$45p = + \$45c

+STK - \$55c = - \$55p

or

-STK - \$45p = - \$45c

+STK - \$55c = - \$55p

Selling the \$45c and the \$55p would be an Inverted Strangle. Add both sides:

-\$45c - \$55p = - STK - \$45p + STK - \$55c

Simplify:

-\$45c - \$55p = - \$45p - \$55c

(ITM \$45c/55p strangle = OTM \$45p/55c strangle)

Clear as mud?