3.1 Expectancy, R-multiples and the System Quality Number

Motivation

A trader's results are a list of trades of different sizes in different instruments. To ask whether a method has an edge, and how much capital to put behind it, the results must be put on a common scale. Van Tharp's R-multiple does this by measuring every outcome in units of the risk taken.

Setup

A stop-loss (stop) is a pre-set price at which a losing position will be closed. For a trade entered at price $P_e$ with stop $P_s$ and quantity $Q$ , the initial risk is $1R = |P_e - P_s|\,Q$ in rupees: the planned loss if the stop is hit at its price. If the trade's realised profit after costs is $\Pi$ , its R-multiple is $R = \Pi/1R$ . A trade stopped out exactly at the stop has $R \approx -1$ ; a trade that gains three times the planned risk has $R = 3$ . For a trade list $R_1, \dots, R_N$ the expectancy is

E = \frac1N\sum_{i=1}^{N} R_i = p\,\bar W - (1-p)\,\bar L, \tag{3.1}

where $p$ is the win rate, $\bar W$ the average winning R and $\bar L$ the average losing R (as a positive number). Tharp's System Quality Number is

\text{SQN} = \sqrt{\min(N, 100)}\;\frac{\bar R}{s_R}, \tag{3.2}

with $s_R$ the sample standard deviation of the R-multiples.

Proposition 3.1SQN is a capped $t$ -statistic

If the $R_i$ are iid with mean $E$ and variance $\sigma_R^2$ , then for $N \le 100$ the SQN is the $t$ -statistic for $H_0: E = 0$ , asymptotically $N(\sqrt N E/\sigma_R, 1)$ . For $N > 100$ it is $\sqrt{100/N}$ times the $t$ -statistic.

Proof

The $t$ -statistic is $\bar R/(s_R/\sqrt N) = \sqrt N\bar R/s_R$ ; by the CLT and Slutsky's lemma it is asymptotically normal with mean $\sqrt N E/\sigma_R$ and unit variance. The cap replaces $\sqrt N$ with 10.∎

The cap is Tharp's device to compare systems with different trade counts: a system with 1,000 trades would otherwise score $\sqrt{10}$ times higher than the same system with 100. For $N \ge 100$ the SQN is therefore an estimate of the per-trade Sharpe ratio $E/\sigma_R$ times 10. Tharp's rough bands (1.6–1.9 poor, 2–2.5 average, 2.5–3 good, 3–5 excellent, above 7 suspect) come from his practice, not from statistics (reported by secondary summaries of his books; sources differ on the exact bands).

Worked example 3.1.1a breakout swing system

A swing-trading system of the kind described by O'Neil and Minervini (buy breakouts in uptrending stocks, stop 7–8% below entry; see M04) wins 40% of trades with an average win of $2.5R$ and loses $1R$ on the other 60%. Then $E = 0.4(2.5) - 0.6(1) = 0.4R$ per trade. The R-multiples take two values, so $s_R = 1.723$ (with the $N-1$ divisor over 100 trades), and $\text{SQN} = 10 \times 0.4/1.723 = 2.32$ : "average" in Tharp's bands, and a $t$ -statistic of 2.3 on 100 trades.

Costs in R units For a delivery equity trade in India (bought and sold on NSE through a zero-brokerage broker), statutory and exchange charges come to about 0.22% of the position value per round trip plus a ₹15.34 depository charge on the sale (STT 0.1% on each side, stamp duty 0.015% on the buy, exchange charges 0.00307% per side, GST; rates as of 6 Nov 2026 from the Finance Act, 2026 STT schedule and NSE circular NSE/FA/73061; M01 · 1.7 derives the cost algebra and gives the sources). With a stop 4% below entry, $1R = 4\%$ of position value, so costs are $0.22/4 = 0.055R$ per trade. With a 1% stop they are $0.22R$ , more than half the edge of the system above. Tight stops make costs large in R terms.

Practitioner notes

R-multiples measure outcomes relative to plan. A trade that gaps through its stop and closes at $-2.5R$ must be recorded as $-2.5R$ , not $-1R$ . The tail of the R distribution is set by gaps (3.9).
Expectancy is a mean. Its standard error is $s_R/\sqrt N$ . From M02 · 2.4, 100 trades with SQN 2.3 give a 95% confidence interval for $E$ of roughly $0.4 \pm 0.34R$ : the edge could be almost zero.
Expectancy per trade is not expectancy per unit time. A system with $E = 0.2R$ and 200 trades a year beats one with $E = 0.6R$ and 30 trades a year, if both are real.

3.1.1How many trades distinguish edge from noise#

Proposition 3.1 turns the question "does this setup have an edge?" into a sample-size calculation. It is placed here, before any trading, because every journal review, paper-trading graduation test (M14 · 14.10) and playbook post-mortem (M15) uses it.

Setup

Trades produce i.i.d. R-multiples with mean $E$ (expectancy) and standard deviation $s$ . After $N$ trades the sample mean is $\bar R$ and the $t$ -statistic is $t = \bar R\sqrt N/\hat s$ (the SQN of (3.2) without the cap).

Corollary 3.1.1Trades needed for $t \ge 2$ in expectation

If the true expectancy is $E > 0$ , the expected value of the numerator, $E\sqrt N/s$ , reaches 2 when

N \ge \left(\frac{2s}{E}\right)^2. \tag{3.2a}

Proof

$E\sqrt N/s \ge 2 \iff \sqrt N \ge 2s/E$ .∎

This $N$ only gives a 50% chance of clearing $t = 2$ : half the time noise pulls the statistic below its expectation.

Corollary 3.1.2Trades needed for power $1-\beta$

Under the normal approximation of Proposition 3.1, $t \approx N(E\sqrt N/s, 1)$ , so $P(t \ge 2) = 1 - \Phi(2 - E\sqrt N/s)$ , and $P(t \ge 2) \ge 1 - \beta$ requires

N \ge \left(\frac{(2 + z_{1-\beta})\,s}{E}\right)^2, \tag{3.2b}

with $z_{0.8} = 0.8416$ .

Proof

$1 - \Phi(2 - E\sqrt N/s) \ge 1 - \beta \iff 2 - E\sqrt N/s \le -z_{1-\beta} \iff \sqrt N \ge (2 + z_{1-\beta})s/E$ .∎

Numbers

Setup	$E$ (R)	$s$ (R)	$N$ for $t = 2$ in expectation	$N$ for 80% power
Typical swing setup	0.20	1.50	225	455
Breakout system: 35% winners at +2.16R, 65% losers at −0.70R	0.30	1.36	83	166
Thin intraday edge	0.10	1.00	400	808

At three trades a week, 225 trades take 75 weeks. Paper trading alone therefore cannot establish a swing edge in a few months. The edge has to be established on historical data with the safeguards of M04 (out-of-sample tests, walk-forward, deflated Sharpe ratio); paper and small live trading then only check that live results are consistent with the backtest distribution, which needs far fewer trades (M14 · 14.10.3).

False positives Under the null $E = 0$ with normal R-multiples, $P(t \ge 2) = 1 - F_{t,N-1}(2)$ : 0.0241 for $N = 100$ and 0.0275 for $N = 30$ (normal limit 0.0228). The simulation in m14_graduation.py reproduces 0.0249 for $N = 100$ . For a skewed breakout-type distribution (two-point, mean zero) the rate at $N = 30$ is 0.012, because positive skew correlates the sample mean with the sample standard deviation; at $N = 100$ it is back to 0.024.

Selection The real danger is choosing among setups. If you paper-trade 10 setups with no edge and keep the best, the probability that at least one passes $t \ge 2$ is $1 - (1 - 0.0228)^{10} = 0.21$ ; with 20 setups it is 0.37. M04 · 4.6 gives the correction (the expected maximum Sharpe ratio of $N$ noise trials); apply it to setups, not only to backtests.