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🟠 Advanced • Lesson 61 of 85

The Number You Cannot Look Up

Reading time ~13 min • Module 7: The Other Side
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The adversarial frame is right and the arithmetic usually attached to it is wrong. Every version of the argument asks you to supply one number, the frequency with which a level is run through and then reclaimed, and that number is not a property of the market: on this course’s sixty closes it moves from 0.00 to 1.00 across ordinary choices of two definitions. So solve for the break-even instead of hunting the frequency. The two-state table everybody draws puts it at 25 per cent; size the two stops to the same risk and it drops to zero; put back the state the table leaves out, in which the level simply holds, and it becomes half the probability of a clean hold. Measured on this series, the level held on about three quarters of the approaches, so the wide stop needed a run-and-reclaim rate of at least 0.375 and never got past 0.250. On this series the whole genre’s recommendation is the wrong one, and the way to find that out was to add a row.

Prerequisites: Lesson 3, which established that the trades you get are chosen by somebody else, which is the whole of the adversarial frame, lesson 19, for what a sample this size can and cannot settle, and lesson 9, for the sizing arithmetic that turns two different stops into the same bet.

The number is not in the market, it is in your definition

Every article on this subject arrives at the same place. Stops cluster below obvious levels, because that is where every course puts them; somebody who needs to buy size wants to be filled where the supply is, which is exactly there; so the level gets traded through before it holds, and nobody had to coordinate anything for that to happen. All of that is sound, and this module has spent five lessons showing why the participants involved are acting under obligations rather than intentions.

Then the article tells you how often it happens. Seventy per cent, or eighty, or two thirds. That figure is the load-bearing beam of every conclusion drawn from it, and it is never sourced, because it cannot be: it depends on what you decided counts as a level and what you decided counts as a reclaim, and both of those are choices you make before you count.

Watch it move. Take the sixty closes this course has carried since lesson 38 and define a level as the lowest close of the previous few bars. Call it a run when a close falls below that level, and a reclaim when a later close gets back above it. Two knobs, both of them things a reasonable person would set differently, and both set here to values nobody would argue with.

Level taken asRuns throughReclaimed next barWithin threeWithin five
The lowest of the last five closes7477
The lowest of the last ten closes4244
The lowest of the last twenty closes2022

The frequency the argument wants runs from 0 out of 2 to 7 out of 7 across that grid. It is 0.00 with a twenty-bar level and a one-bar reclaim, and 1.00 with any level and three bars. Same sixty numbers, same market, no cherry-picking of dates: only two definitions moved, and the answer covered its entire possible range.

So the honest response to anybody quoting a percentage here is to ask what they defined a level as and how long a reclaim had to take. Without both, the number carries no information. And since you are not going to get a defensible one, the question has to change.

Solve for the break-even instead

Here is the trade the argument is about. You are long at 180 with support at 179. The tight stop goes just under the level at 178.50, which is what the textbook says. The wide stop goes at 177.00, which is a volatility buffer below it. Two states are usually drawn. In the first the level is run to 178.20, then reclaimed, and the price carries on to 183. In the second it genuinely breaks and the price goes to 175.

At a hundred shares each, the tight stop loses 150 dollars in both states, because 178.20 is below 178.50 either way. The wide stop survives the run and makes 300, or is stopped in the breakdown and loses 300. Write p for the chance of the first state and the expected values are minus 150 against 600p minus 300, which are equal at p of 0.25. That is where the familiar answer comes from, and it is the answer the whole genre repeats.

It is also an artefact. The two positions are not the same bet: one risks 150 dollars and the other risks 300. Lesson 9’s arithmetic says to compare them at the same risk, which means the wide stop takes fifty shares rather than a hundred. Do that and its payoffs halve to plus 150 and minus 150, its expectation becomes 300p minus 150, and it beats the tight stop for any p above zero at all.

The reason is worth saying slowly, because it generalises past this example. The tight stop’s expectation does not contain p. It is minus 150 in both columns, which means it is not a cheaper bet but a refusal to bet, and the refusal costs the entire upside. Once the two are sized to the same risk, anything with a positive payoff in any state beats something with a negative payoff in every state, and the sizes of the payoffs never enter the comparison.

The row the matrix leaves out

Which should be suspicious, because it proves too much. If the wide stop wins at any frequency above zero then it wins always, and a rule that is right no matter what the world does is usually a rule that has been asked the wrong question. The wrong question is in the table: there are only two columns.

The missing column is the one where nothing happens. The price approaches the level, the level holds, no run occurs, and the rally to 183 arrives with neither stop touched. In that state the tight stop is not stopped out. It is long a hundred shares into a three-point rally and makes 300, while the wide stop is long fifty and makes 150. It is the state in which being tight pays, and it is left out of every payoff table in the literature because the literature is about the state where you get hunted.

OutcomeLevel holdsRun, then reclaimedReal breakdown
Tight stop, 100 shares+300-150-150
Wide stop, 50 shares+150+150-150

Call q the chance of a clean hold and p the chance of a run and reclaim. The tight stop is worth 450q minus 150; the wide stop is worth 300q plus 300p minus 150. Subtract and the difference is 300p minus 150q, so the wide stop wins when p is greater than q over 2. Not 25 per cent, not zero, but half the probability that the level simply holds.

That is a better answer than either of the others because it is a comparison rather than a threshold, and because both of its terms are countable on your own instrument. It also says something the two-column version cannot: if levels in your market usually hold, the wide stop needs a high run-and-reclaim rate to be worth it, and if they rarely hold, it barely needs any.

What the series actually recommends

Both terms are measurable, so measure them. Count an approach whenever a close comes within one bar’s standard deviation of the level, which on this series is 1.54, and count runs and reclaims as before. That gives q and p at once, from the same passes over the same sixty numbers, under each of the definitions the first table used.

With a five-bar level and a three-bar reclaim there are 29 approaches: 22 held, 7 ran through, and all 7 were reclaimed. That is q of 0.759 and p of 0.241, so the threshold is 0.379 and the wide stop loses. With a ten-bar level: 16 approaches, 12 held, 4 ran, 4 reclaimed, giving 0.750 and 0.250 against a threshold of 0.375. With a twenty-bar level: 10 approaches, 8 held, 2 ran, 2 reclaimed, giving 0.800 and 0.200 against 0.400. Tighten the reclaim window to one bar and p falls to 0.129, 0.111 and 0.000 while q barely moves.

Six definitions, and the tight stop wins under all six. On this series, at these settings, the advice the entire adversarial literature gives is the wrong advice, and it is wrong for a reason that has nothing to do with whether stop hunting is real. Levels here held about three quarters of the time they were approached, so the state in which the wide stop earns its extra room simply did not come up often enough to pay for the size it costs.

Notice what did not decide it. Not the frequency of stop runs, which is the number the argument fixates on and which was measured here at everything from 0.000 to 0.250. The threshold moved with q, and q was the stable quantity: it sat between 0.75 and 0.80 under every definition tried. The variable everybody argues about turned out to be the one that mattered less.

So the module closes on a method rather than a conclusion. Everything in the five lessons before this one was about what the other side is obliged to do: a quoter with inventory, a desk with a schedule, a dealer with a hedge that is a function of price. None of that tells you what will happen next, and this lesson is the reason that is not a disappointment. You are not trying to predict the other side. You are trying to build a position whose expectation you can write down in terms of quantities you can count, and then counting them.

What this does not settle

That 29 approaches is a sample. Lesson 19’s arithmetic settles this quickly: 7 runs out of 29 gives a p of 0.241 with an interval running from 0.086 to 0.397, and the threshold it has to beat is 0.379. The interval straddles the threshold, so the count leans one way without deciding anything, and the smallest cell, 2 runs out of 10, has an interval from 0 to 0.448 and decides less than that. What the exercise establishes is not that the tight stop is better here. It is that the comparison is decidable by counting at all, and that as far as the count goes it does not go the way the literature says.

That an approach is a well-defined event. It is not, and the definition used here — a close within one standard deviation of the level — is exactly the same kind of arbitrary choice the first section complained about. Widen it and q rises, because more of the near misses get counted as holds; narrow it and q falls. The threshold q over 2 is exact; the q you feed it is as constructed as the p it is being compared with, and this page does not get to exempt its own inputs from its own complaint.

That the payoffs are the right payoffs. The three-point rally, the run to 178.20 and the breakdown to 175 are the legacy example’s numbers and they are illustrative. The algebra survives changing them — the result p greater than q over 2 holds whenever the wide stop’s win in the run state equals the tight stop’s win in the hold state per unit of risk, which is what equal sizing produces — but the specific 0.379 and 0.375 do not.

That being stopped means getting the stop price. It does not. A stop is a market order with a trigger, as lesson 58 said, and it fires when the book in front of it is thinnest. Both stops here are assumed to fill exactly where they sit, and in the state that matters most for the tight stop — a fast run through the level — that is the assumption least likely to hold. Correcting it makes the tight stop worse, which pushes the threshold in the wide stop’s favour, and this page has not tried to price it.

That the stop is the decision. The whole comparison holds the entry fixed and varies only where the loss is cut, which is how the literature frames it and is not how a position is actually built. A trader who waits for the run rather than buying the level is choosing a different entry, not a different stop, and none of the arithmetic here evaluates that. The table has two rows because the argument it answers has two rows.

And the concession that costs this lesson most: it has replaced a wrong number with a correct method and made the method sound easier than it is. Counting q and p on your own instrument means defining an approach, a level, a run and a reclaim in advance, then not adjusting them when the answer comes out uncomfortable. Every quoted frequency in this field was produced by somebody who adjusted them. The arithmetic on this page is the easy part, and the discipline it requires is the part that fails. Lesson 62 is where that discipline gets written down as a sentence, and it opens by judging one ordinary observation twice: 6 of 9 beats a coin and is adopted, and the same 6 of 9 against a 54.24 per cent baseline settles nothing at all.

Problems

  1. Watch your own number move. On two hundred bars of one instrument, count runs and reclaims at three level lengths and two reclaim windows, the way the first table does. Write down the six frequencies. An hour, and the spread between the largest and the smallest is the honest measure of how much any single quoted figure is worth.
  2. Count the state nobody counts. On the same two hundred bars, count approaches that held cleanly as well as approaches that ran. That gives you q, and half of it is your threshold. Compare it with the p you measured in the first problem. Two hours, and unlike every article on this subject you will have an answer that came from your instrument rather than from somebody’s recollection.
  3. Re-derive the threshold with your own payoffs. Replace the three-point rally and the five-point breakdown with the numbers your instrument actually produces, keep the two stops at equal dollar risk, and solve the three-state table again. Half an hour, and you will find out whether q over 2 survives your payoffs or whether the equal-risk assumption was doing more work than you thought.

Sources. John von Neumann and Oskar Morgenstern, Theory of Games and Economic Behavior (Princeton University Press, 1944), for the payoff table as the object you reason with, and for the discipline of writing every state down before evaluating any of them. John Nash, “Non-Cooperative Games” (Annals of Mathematics, 1951), for equilibrium in mixed strategies, which is why unpredictable behaviour can be stable rather than clever. Carol Osler, “Currency Orders and Exchange Rate Dynamics” (The Journal of Finance, 2003), for stop and limit clustering measured on real order books rather than asserted, and for how much of the round-number effect it accounts for. Daniel Kahneman and Amos Tversky, “Prospect Theory” (Econometrica, 1979), for why a stop that is certain to be small feels safer than one that is usually zero, which is the preference the first table exploits.

The adversarial frame is right and the number attached to it is not available. On this course’s sixty closes the run-and-reclaim frequency moved from 0.00 to 1.00 across two ordinary definitional choices, so it cannot be looked up and should not be quoted. Solve for the break-even instead, and be careful which table you solve: the two-state version says 25 per cent, equal sizing sends that to zero, and putting back the state where the level simply holds gives half the probability of a clean hold. Measured here, levels held on about three quarters of approaches under every definition tried, so the threshold sat between 0.375 and 0.400 while the measured run-and-reclaim rate never exceeded 0.250, and the tight stop won six times out of six. That is one series and it settles nothing about yours. What it settles is that the question has an arithmetic answer, which is the only thing five lessons about other people’s obligations were ever going to leave you with.

Related Lessons
Lesson 3

What a Fill Actually Is

The trades you get are chosen by somebody else, which is the whole adversarial frame.

Read Lesson →
Lesson 19

How Long Until You Know

What twenty-nine approaches and seven runs can and cannot settle.

Read Lesson →
Lesson 9

Who Else Is Here

The sizing arithmetic that turns two different stops into the same bet.

Read Lesson →
Educational only. Trading involves substantial risk of loss. Not financial advice. Past performance does not guarantee future results.

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