I understand your explanation. However, you made assumptions about delta. If this was to be priced by a bank would they make the same assumptions? Is there any specific reason for choosing 7 delta, why not 9 or 5?
The number 7 was just an example to show how the calculation would work, in the same way that you used a delta of 20 in your original post to refer to the 110 call and 90 put - in reality with the stock at 100, the 110C and 90P would have deltas much more numerically bigger. Lets take a real example with real data : On 18-Mar, the stock LMND was priced at 100.03 at 12:30EST. At that time the 110 call and 90 put expiring on 16-Apr (29 DTE's) had deltas of 40.0 and -29.4 respectively. Now, the 80 strike put had a delta of -16.2. Why are we looking at the 80 strike, which is 20 points away from the current stock price? Soon to be clarified. The probability at 12:30pm on 18-Mar that LMND would hit the 110 strike within the next 29 days is = 2 * delta = 2*0.40 = 0.80 (or 80%). Assume that the stock moves to 110 in that next instant, so no time decay, no changes in volatility, and all other things are constant except the delta. Now, the 90 strike put will have a delta close to -16.2 (because it is 20 points away from the now ATM price of 110). Therefore, the probability of the stock going down to 90 = 2 * 0.162 = 0.32 (32.4%). Thus, probability that LMND will touch 110 and then 90 is 0.80 * 0.32 = 0.2592 (roughly 26%). Banks have their complex, more finely-tuned, higher-level mathematics based computations, but for me as a retail trader, the above, back-of-an-envelope calculation, is sufficient. Of course, it is simplistic and makes some assumptions. Hope that helps.
But what do you think of @Same Lazy Element explanation on this thread? He arrived at a different answer. But the logic behind it makes sense, or you don't think so?
Since options are priced at their discounted expectation, an undiscounted (set r=0) binary can almost always be interpreted as a probability. The normalized DTKO combo in my answer actually pays out the complement of the binary, so you have to subtract the normalized price from one to get the probability. So a slight mistake in my answer. The problem with all the other analyses on this thread so far (besides delta*2 not being a very good estimate of probability of touch), is that the first barrier touched is not likely to be touched instantly, so that the time left to hit the second barrier is less than thirty days. For, e.g. B), you need to integrate the FPTD (first passage time density) of the upper barrier times the probability of touch (at that instant time) over the interval 0 to 30/365. For GBM (Geometric Brownian Motion), there is an exact solution for the FPTD and (in log price) it is the Inverse Gaussian Distribution. For other SDE's you usually have to solve for the FPTD numerically. My intuition is that you could take the mean (expected) first passage time in that interval times the probability of upper barrier touch times the probability of lower barrier touch in the time from mean first passage to expiry, and you wouldn't be far off. But my intuitons are often wrong, so you'd have to test this. Edit: to get A from B) doubling B won't work. Solve for U then L as above, then for L then U, and add them. Assume sticky strike. ;
LOL, my answer is completely wrong and the only excuse is that I was very high at the time. P(u) * P(l) is just the maximum upper bound for the probability. The double-touch probability is path-dependent, so you can't just multiply the probability of each touch to get the joint probability. I.e. if one of the barriers is touched, the probability of touching the other decreases severely because it's pretty far now. PS. You'd need a proper model to price this, I can't think of any back of the envelope way of doing. PPS. I would recon that the probability of a double touch would decrease exponentially with decrease in delta.
Yes, it helped. Just curious what the probability would be for the price to first touch 110 and then touch 90 and still touch 110 again? I don't think it would be 2*0.20*2*0.7*2*0.7 Right? Assuming no time decay.