Module 3 Quiz: Uncertainty, Risk and Ruin

This module never once looked at a chart. It turned a method into two numbers, said what a run of them feels like, how long before the record can settle anything, how much of the account to put behind each one, where the stop belongs, what the tail costs, and which ten fields make any of it answerable. Six questions. The last takes the 45 per cent system the module has carried since lesson 17 and prices what trying to get the money back would cost.

Covers: Lessons 17 to 24, and the system winning 45 per cent of the time at a payoff of 2 that every table in the module was computed on.

Every question below hands you numbers and asks for a number back. Work all 6 with a calculator before you scroll to the answers; each answer shows the arithmetic, so a wrong result tells you which step to go back to rather than only that you were wrong.

The questions

1. An edge, read off six rows

Six trades from one record. The account stood at $50,000 throughout, and the stop column is the price as it was first placed.

Entry	Stop as placed	Size	Exit	Why out
84.00	82.50	400	87.60	target
85.20	83.80	430	83.80	stop
61.50	60.30	500	60.20	stop
86.40	85.00	425	85.55	manual
62.80	61.30	400	66.10	target
88.10	86.60	900	87.20	manual

Ask. What are p, b and the expectancy, and what payoff ratio would a broker statement have reported instead?

2. When the run is due

A method wins 38 per cent of the time, its winners pay 2.6 times its losers, and you take four trades a week. A run of seven losses would genuinely worry you.

Ask. How many trades until the first run of seven, roughly how long is that in months, and how much does the payoff of 2.6 change the answer?

3. How much record it would take

The same method: it wins 38 per cent of the time at a payoff of 2.6. Lesson 10’s charges come to s = 0.12 on this instrument.

Ask. What is the expectancy before and after costs, how much noise sits on one trade, and how many trades would it take to establish that the edge exists at all in each case?

4. Where the stop goes, and what it buys

You are long at $124.00. You entered because $121.50 held, so $121.50 is the price at which the reason is gone. The average true range is $0.90. The next level in your favour is $131.00. Your account is $40,000 and you risk 1.5 per cent a trade.

Ask. Put the stop one average bar beyond the level. How many shares, what is b, and what win rate does that need? Then do the same for a stop 2 per cent below the entry and say which one is wrong.

5. What the fraction does to the tail

A method wins 56 per cent of the time at even money, so every winner and every loser is the same size. Lesson 22’s classical result applies exactly: ruin is ((1 − p) ÷ p) raised to the power of how many bets your purse holds.

Ask. What is the risk of ruin at 2 per cent a trade and at 1 per cent, and what happens to both if the win rate turns out to be exactly 50 per cent?

6. The system this module carried, and the way out of the hole

Back to the system every table in this module was computed on: it wins 45 per cent of the time and its winners pay twice its losers. You are in an ordinary drawdown, the 9R median that half of all careers on this system meet. Lesson 24 simulated the next twenty trades: risking 1 per cent leaves the one-in-twenty career at 0.867 of its starting equity, and risking 5 per cent leaves it at 0.681.

Ask. What is the expectancy? What do one winner and two losers cost the account at 1 per cent and at 5 per cent, given that they come to exactly nothing in R? And what gain does each of those two one-in-twenty careers need to get back to where it started?

The answers

Each one is worked in full. Where a figure comes from a lesson rather than from this page, the lesson is named.

1. An edge, read off six rows

Each row carries its own R, because R is the distance from the entry to the stop as placed. Divide the move by that distance and the size drops out.

Risk a share	Money at risk	Result	In R
$1.50	$600.00	+$1,440.00	+2.40R
$1.40	$602.00	−$602.00	−1.00R
$1.20	$600.00	−$650.00	−1.08R
$1.40	$595.00	−$361.25	−0.61R
$1.50	$600.00	+$1,320.00	+2.20R
$1.50	$1,350.00	−$810.00	−0.60R

Two of six won, so p = 33.3 per cent. The winners average 2.30R and the losers average 0.82R, so b = 2.30 ÷ 0.82 = 2.80, and the expectancy is 0.333 × 2.30 − 0.667 × 0.82 = +0.22R a trade.

Lesson 17 wrote the same thing as p × b − (1 − p), which gives +0.27 here. That is not a contradiction, it is a different unit: lesson 17 measures in average losses and this measures in the R you planned, and the log says the average loss came to 0.82 of a planned R. Multiply 0.27 by 0.82 and you have 0.22 back. Only the record knows the two units differ.

Now the broker’s view of the same six trades. The winners average $1,380 and the losers average $605.81, so the payoff ratio is 2.28 rather than 2.80, wrong by about a fifth. The cause is the last row: five trades risked around $600 and one risked $1,350. Money mixes how good the trades were with how large the bets were. R separates them, and that separation is the only reason the stop column exists.

One thing the six rows cannot tell you is whether any of this is real. Lesson 19 puts the sample needed at hundreds. Six rows give you the method, not the answer.

Answer. p = 33.3 per cent, b = 2.80, and +0.22R a trade in planned R. The money view gives 2.28.

2. When the run is due

Lesson 18’s closed form is (1 − q^k) ÷ (p × q^k), with q the loss rate and k the run length. Here q = 0.62 and k = 7, so q^k = 0.0352.

That gives (1 − 0.0352) ÷ (0.38 × 0.0352) = 0.9648 ÷ 0.0134 = 72 trades. At four a week that is 18 weeks, a little over four months.

Now look at what never entered the calculation. Not b, not the expectancy, not a single thing about how much a winner pays. A run is a property of the win rate and of the length of the sample, and of nothing else, which is why the run you are living through carries no information about whether the method is any good.

And read the answer as a rate rather than an event. Not once in a career: once every 72 trades, for as long as you keep trading. Over five years at four a week you should expect about fourteen of them, and be surprised by none.

Answer. 72 trades, about four months, and the payoff changes nothing at all.

3. How much record it would take

The expectancy is 0.38 × 2.6 − 0.62 = +0.37R, and lesson 17’s breakeven of (1 + s) ÷ (b + 1) = 1.12 ÷ 3.6 = 31.1 per cent confirms that 38 per cent clears the bar. Subtract the charges and the edge is +0.25R a trade.

The noise comes from the same two inputs: σ² = p·b² + (1 − p) − E² = 0.38 × 6.76 + 0.62 − 0.37² = 3.053, so σ = 1.75R. One trade in a good method carries nearly five times as much noise as signal.

What you are establishing	The edge to catch	Trades	At four a week
There is an edge at all, before costs	0.37R	177	10 months
There is an edge at all, after costs	0.25R	390	1.9 years

Both come from n = 7.85 × σ² ÷ Δ², and the second row is the one worth keeping. Subtracting the same charge from every trade moves the average and leaves the spread exactly where it was, so the charges did not merely take a third of the edge. They more than doubled the record you need to prove you have one.

Answer. +0.37R before costs and +0.25R after. One trade carries 1.75R of noise. 177 trades before costs, 390 after.

4. Where the stop goes, and what it buys

One average bar beyond $121.50 puts the stop at $120.60, so the risk is $3.40 a share. One R is $40,000 × 1.5% = $600, and $600 ÷ $3.40 = 176 shares, a position worth $21,824. The target is $7.00 away, so b = 7.00 ÷ 3.40 = 2.06 and the breakeven win rate is 1 ÷ 3.06 = 32.7 per cent.

Now the percentage stop, which is where the trap is.

Where the stop goes	Distance	Shares at $600 risk	Reward to risk, b	Win rate to break even
2% below the entry, $121.52	$2.48	241	2.82	26.2%
One average bar beyond the level, $120.60	$3.40	176	2.06	32.7%

The first row wins every column. It buys 65 more shares, it shows the better reward to risk, and it needs six and a half fewer points of win rate. It is also the one that is certainly wrong: $121.52 sits two cents above $121.50, which is on the level rather than beyond it. The ordinary probe that tests the level takes you out, and it takes you out while the reason you entered is exactly as true as it was when you entered.

The extra 6.5 points of win rate is what a real stop costs. It is not an upgrade. It is a worse-looking trade that is an actual one.

Answer. 176 shares, b = 2.06, and a 32.7 per cent breakeven. The 2 per cent stop looks better in every column and is the wrong one.

5. What the fraction does to the tail

Risking 2 per cent means the purse holds 50 bets, so u = 50 and the base is 0.44 ÷ 0.56 = 0.7857. Raised to the 50th power that is 0.0000058, or 0.00058 per cent.

Halve the risk and u doubles to 100, which does not halve the answer. It squares it: 0.0000058² = 3.4 × 10⁻¹¹, about three chances in a hundred billion. Every halving of the fraction squares the odds of survival, which is why the distance between risking 2 per cent and risking 10 per cent is nothing like a factor of five.

Now set p to exactly 0.50. The base becomes 0.50 ÷ 0.50 = 1, and 1 raised to any power at all is 1. Ruin is certain however large the purse, and the size of the bet decides only how long it takes to arrive.

That is the sentence the whole module rests on. Sizing converts an edge into survival and cannot manufacture one. Without an edge you are not managing risk, you are choosing a pace.

Answer. 0.00058 per cent at 2 per cent risk and about three in a hundred billion at 1 per cent. At a 50 per cent win rate both are 100 per cent.

6. The system this module carried, and the way out of the hole

The expectancy is 0.45 × 2 − 0.55 = +0.35R a trade, which is the number the whole module has been spending. The sequence +2R, −1R, −1R sums to zero in R and does not sum to zero in the account.

Risk a trade	The three trades	Left with	Cost of standing still
1%	1.02 × 0.99 × 0.99	0.99970	0.030%
5%	1.10 × 0.95 × 0.95	0.99275	0.725%

Nothing about the trades changed and none of the three was a mistake. Only the fraction moved, and the cost of standing still grew 24 times, because it goes with the square of the fraction. Double what you risk and you very nearly quadruple what standing still costs you.

Then the way back. A career left at 0.867 needs 1 ÷ 0.867 − 1 = 15.3 per cent to reach its old high. One left at 0.681 needs 1 ÷ 0.681 − 1 = 46.8 per cent. The first is a good quarter and the second is a year.

Read the two together and the module closes on itself. The 5 per cent career was sizing up to get the money back, and it works: it reaches a new high inside twenty trades nine times in ten against fewer than half at 1 per cent. What it also does is turn a 13 per cent hole into a 32 per cent one in the one-in-twenty case, and 46.8 per cent is not a plan. The edge was identical in both. Everything that differs is a fraction you chose before any of it began.

Answer. +0.35R a trade. The three trades cost 0.030 per cent at 1 per cent risk and 0.725 per cent at 5, a ratio of 24. And the two careers need 15.3 per cent and 46.8 per cent to get back.

What this quiz was testing

Whether you can get a number out of your own record and then say what it is entitled to settle. Handed six rows, you produce an expectancy; handed a win rate, you produce the run and when to expect it; handed an edge and its noise, you produce the sample that would prove it; handed a level and a volatility, you produce a share count; handed a win rate and a fraction, you produce a probability of being finished. Nothing in the module asked what the market was doing, and nothing in it needed to.

Which is exactly the gap module 4 opens on. It starts with a stop placed a tenth of an average bar below the level: it shows the best reward to risk on the page at 9.2 to one, it needs a win rate of only 9.8 per cent, and it still loses money, because four trades in five never find out whether they were right.

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