# Why this strategy with options and zero risk is not possible?

I'm learning options and I've seen strategies with graphs like these: Why is not possible to combine a Butterfly with Straddle to get something like this (green is final result): I'm very new in options so sorry if I'm asking something obvious, but I think this example will help to understand it better.

EDIT I have found a nice site to create these graphs:
https://optioncreator.com
Playing with it I was able to create the graph from above, but it requires a big spread for the butterfly, is that a problem? The graph is cut, but the middle price is "100" and strategy becomes profitable if it ends below 45 or higher than 155. That means we need like 60% of change, this is not usual but is not impossible, and since is a non risk trade we could just use it in multiple stocks and wait one of them moves that amount? What I'm missing here?

• Have you considered giving this a try? This is probably the easiest way to find where the scheme fails - and, make no mistake, in a sufficiently efficient market it must fail. Jul 15 '18 at 16:34
• I cannot try it, I don't have a broker to try that, and will be too risky use real money just to learn. I have already explained this is a question to understand better how options work, I'm learning, that's the purpose of this site after all Jul 15 '18 at 19:44
• There's a redundancy in your trades. Line 2 is sell 2 \$100 calls and line 5 is buy one of the same. If you simplify the 5 trades by eliminating synthetically equivalent positions, your complex trade is equivalent to just buying the \$45p/155c strangle. Since these options are so far out of the money, they cost very little and therefore the risk graph shows almost no risk with a large possible gain if the underlying moves more than 55% up down. Jul 3 at 12:34

In terms of the risk graphs, your thought process is correct. If you combine a Short Butterfly with a Long Straddle, you end up with the risk graph of the green line. Unfortunately, because options cost money, there are no free lunches and that is the error in your assumption. Now I know that you're not going to accept that explanation on face value so let's try something more technical.

There are 6 basic synthetic positions relating to combinations of Puts, Calls and their underlying Stock. It's called 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

There are additional synthetic combinations. For example, a Bullish Vertical Spread is equal to Long Collared Stock. A bullish debit Vertical Spread is equivalent to a bullish credit Vertical Spread when options of the same strikes and series are used. Once you understand the Synthetic Triangle, you can simplify complex positions into positions with fewer legs. That has two benefits. First, it's often easier to visualize the simplified position's P&L and second, you incur less frictional costs when you transact with fewer legs.

A Butterfly Spread is comprised of a bullish and bearish Vertical Spread with a common central strike. It can be constructed several ways and they all have a similar R/R. Using the Synthetic Triangle you can verify that the following three positions are equivalent:

3) Buy one \$95 put, sell one \$100p, sell one \$100c, buy one \$105c

Now, let's take # 3 and add a long straddle at the center strike and simplify the equation. We then have:

• +1 \$95p - 1 \$100p -1 \$100c + 1 \$105c (Long Butterfly)

• +1 \$100p + 1 \$100c (Long Straddle)

• +1 \$95p +1 \$105c (Long Strangle)

The green line in your graph is the P&L of a Long Strangle. The problem is that you assumed that it would be free and you put the horizontal line at ZERO. Long Strangles aren't given out for free. They cost money. The base of your green line graph belongs in negative territory and that will always be the risk.

• I didn't understand why 3) is the same, you are mixing put and calls there, Also I don't understand what do you mean with "they cost money", do you mean the extra fee from the broker? because I thought the "cost" was already included on those graphs (so it is on mine when I just combined them). I mean the butterfly costs money, that's why we loose on the wings, the straddle costs money, that's why we loose on the center. But then combined the win and loose zones cancel each other and final cost is 0 there, and because Bfly is limited loss and straddle unlimited win we have a zone with profit Jul 15 '18 at 7:00
• 1) "I didn't understand why 3) is the same, you are mixing put and calls there." - The Synthetic Triangle provides you with the means to understand why all 3 Butterflies are equivalent. It's like an algebra equation. You need to study it and familiarize yourself with the substitutions. It's all right there in #'s 1-6. And yes, you're welcome for the detailed explanation. 2) "Also I don't understand what do you mean with "they cost money"." - If you place an order with your broker to buy a Strangle, does he give it to you for free? Jul 15 '18 at 11:04
• If you don't understand why all 3 Butterflies are the same then try Plan (B). Here are some made up quotes. XYZ is \$100 and the option prices are 95p (\$1), 100p (\$2), 95c (\$6), 100c (\$2) and 105c (1\$). Plug those quotes into all three Butterflies and determine what the debit cost of each Fly is. If done correctly, it will be the same for each. If the debit cost is the same then the potential profit is the same (vertical spread width minus debit cost) which then proves that all 3 Butterflies are equivalent (in this case, they all have the identical P&L). Jul 15 '18 at 11:05
• I will explain how I see it because I think I'm reading the graphs in thw wrong way, maybe you can point where is my error: I see the graph as the final net profit (or loss) when the options expires, and that includes already the cost to start the strategy. If that's true we can just sum graphs to get the final net profit for combined strategies. And if we sum the butterfly with the straddle we get the green graph. Jul 15 '18 at 12:10
• "I have edited my question with a simulated example. To generate it I needed to add a big spread to the butterfly, but I don't know if this is a problem, and if it is, why?" With the SPY at \$279.56, the narrowest Strangle that you can buy for \$2 cents (1 cent ask price per leg) expiring on 7/20 is the 243p + 291c. So while 2 cents approaches zero it's NOT zero and still costs money. The average delta for this Strangle is about .005 so there's about a 1 in 200 chance that this Strangle ends up ITM. If you think that's a good bet, have at it. Jul 15 '18 at 14:35

The question was asked as to how one can place a Butterfly Spread by "mixing puts and calls".

The relationship between put and call prices involves an arbitrage position called a Conversion. This process dates back to the over the counter days when a dealer who owned a put was able to satisfy a potential call buyer by "converting" the put to a call. The formula is:

Stock - Strike Price + Put - Call + Carry Cost + Dividend = 0

To make this easier to follow, let's assume that the Strike Price is equal to the Stock Price, that there is no dividend and let's pretend that there is no Carry Cost. That leaves us with:

Stock + Put - Call = 0

There are 6 factored combinations of this equation (see the previously posted Synthetic Triangle info). For example, the following demonstrates that a Covered Call is synthetically equivalent to a Short Put.

Stock - Call = - Put

Now back to Butterflies. The first one listed in my previous answer was:

This is a pair of Vertical Spreads:

(B) = + 95p - 100p and (C) = - 100p + 105p

Let's take Vertical Spread (C) and do some Synthetic Triangle magic with it.

If S + P = C then S + 100p = + 100c and if factored with the signs changed it becomes:

(D) - 100p = S - 100c

Similarly, S + 105p = + 105c becomes:

(E) +105p = + 105c - S

Let's substitute (D) and (E) into (C)

(C) - 100p + 105p =

S - 100c + 105c - S =

(F) - 100c + 105c

which demonstrates that

(C) - 100p + 105p = (F) - 100c + 105c

Therefore, the vertical spreads (C) and (F) are equivalent. Substitute (F) for (C) in (A) and the result is:

(G) + 95p - 100p - 100c + 105c

which is buy one \$95p, sell one \$100p, sell one \$100c and buy one \$105c

and this Butterfly is equivalent to:

QED

• That's very interesting Bob, thanks for the detailed explanation! Jul 16 '18 at 7:21
• Anyway, maybe I didn't explain it better, but my question was about a strategy with 0 risk, I've started with butterfly + straddle just because I thought we could create the green graph in that way, but I didn't realize is just a strangle, and the problem is that we cannot alter the slope of the lines without increasing a lot the spreads and hence making the strategy almost useless because of the low probability. Jul 16 '18 at 7:24

OK this is my conclusion:

• The strategy is almost possible

• We don't need to combine Butterfly + Straddle, is just a Strangle

• The cost could be really low if we have a big spread in our strangle, but never 0

• With a big spread our chances to win are very low
• Because the chances to win are low, even when the cost is low too, is not 0, so we will end up losing money in most of the cases.

This is the closest i could get to a zero risk strategy. This strategy makes money when the price goes up or down. This consists of a put butterfly and a call butterfly.

Is it possible to make this strategy to have a positive theta value?

• They say that perfect is the enemy of the good. In this case, complex is the enemy of 'easily understood'. Your core position is a \$100 strangle. If you simplify using synthetic equivalents, you're converting your core \$100 straddle into a \$50/\$100/\$100/\$150 butterfly and adding two short verticals on either side to mildly reduce the cost of your core straddle. Selling deep ITM call verticals and deep ITM put verticals is problematic since their short legs are likely to be assigned early and therefore, synthetically equivalent OTM verticals would be a better choice. Jul 3 at 13:10