Option Price

R

raf_bcn

Hi
An options newbie question.

Imagine this scenario
Underlying at 266 and 1 Dte.
The atm call is priced at 0.90 , What does it mean? Why the sellers are asking that price ? It seems that the sellers thinks the underlying will go as maximum to 266.90 until expiration.
If the volatility remains equal. Is that right ? otherwise they would not be selling.
_What is the probability of the underlying close above 266.90 ? Why are they satisfied with that probability?

I would like to know what is the logic behind that price.
thank you.
 
If:
Underlying at 266
1 day left
Strike price of 266? If it is the 266 and the put is also ~ 0.90, that makes the straddle ~1.80. That means "the market" is is pricing in a range of 1.80 for one day. That is around 0.676%. You are only looking at one side, the model implies you hedge. If you buy the call for $0.90, some one else will sell that call and hegde with buying around 50 shares per options.

I'll leave the probability up to you as to a close at or above 266.90. To get that, you would have to model the 266.90 call and look at the that delta.

Bob
 
Or, another way to put it, that the 90¢ pricing expects a 50|50 shot of $1.80 on the call side, alone, bringing the price to $267.80. That $1.80 movement becomes the target, the 50|50 odds bringing it to an instantaneous value of E(V) = $1.80*50% = $0.90....

A different way to say the same thing.
 
raf_bcn said:
_What is the probability of the underlying close above 266.90 ? Why are they satisfied with that probability?
More...

Look at the delta of the $267 call then fudge it down by 10%. It will be close enough for civil service work :)

If you prefer something more accurate, toss all of the option variables into an option pricing model, using a hypothetical $266.90 call
 
Hi


So, call 0.90 + put 0.90 = 1.80 The market is pricing in a range of +/- 1.80 , this is a 3.60 range. A 66 % probability, 1 sd, that the underlying will be between that two points at expiration with actual iv. All depends on the iv the market makers determine for that day.
One day I will ask how they arrive to that consensus.

The probability of the underlying closing above 1 sd, + 1.80 is (1-0.66)/2 = 17 % aprox.
But the probability of the underlying close above + 0.90 ,the atm call price, is higher , aprox. 34 % , looking at the delta. So the seller has a 66 % probability of winning.

And also If the market maker sells the two sides, call and put, they have a 0.66 probability of winning. That is what says a normal distribution, right?
Can anyone argue that the probability is 0.83 ? One time I almost thought this.

_Is that a market makers consensus , to price atm options with a win probability of 0.66 ? , Why not 0.50 or 0.75, why not at 1,5 sd or 2 sd ?

And talking about probabilities , this is probably a stupid question .

Thanks.
 
raf_bcn said:
So, call 0.90 + put 0.90 = 1.80 The market is pricing in a range of +/- 1.80 , this is a 3.60 range.
More...

No, it is pricing in a range of $1.80, not 3.60. And, I have no idea where you got that probability from.

You are also making this way too complicated. If you think market participants are wrong and the actual move will be more, then you should buy it. You don’t have to buy the straddle. If you are bullish and believe the stock will go up more than $.90 just buy it and take your chances knowing that you have to exit either way at the end of the following day.

That math is simple, down, up less than $0.90, held until near end of day, you lose. Up more than $0.90, you win. You need a process for a directional trade not calculate vol or Probability of the stock going up in any one day based on market prices.
 
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