Sequence of Returns

The same returns, in a different order. With nothing taken out the order cannot matter. Take an income and it decides everything.

Worked example — Sequence of Returns

Final value, real order	$4,414,533
Final value, reversed	$9,790,543
Average annual return	10.31%
Compound annual growth	9.67%

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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The average return is not the rate your money grows at 🖖

The arithmetic mean of a return series is not the rate your money grows at, even before you withdraw a penny. Gain 50% then lose 50% and the mean is 0%, but a dollar is now 75 cents. The compound rate is always at or below the arithmetic mean, and the gap is close to half the variance — on the 1966 window at 60/40 the mean is 10.31% against a compound 9.67%, a gap of 0.63 points where half the variance is 0.69. That drag exists with no withdrawals at all, and it is why the average return in a brochure is the wrong number to plan with. Sequence risk is a second, larger penalty stacked on top of it.

The same withdrawal rule lasts thirty years, or six 🖖

Set the start year to 1982 and the draw to 12% and the pot is still standing after 30 years. Set the start year to 1929 and change nothing else and it is gone in 1934. Same span, same 60/40 blend, same 3% escalator, same rule; the only difference is which years arrived first. Sweep every 30-year start in the record and the largest draw that survives runs from 3.44% in 1929 to 12.10% in 1982 — a factor of three and a half. The famous 4% is not a property of markets. It is the number calibrated to the worst start this record happens to contain, which makes it wrong for every other one.

Surviving and being fine are different questions 🖖

Start in 1928 and draw 4% and the panel reports that the pot survived — with $240,302 left of a million. The draw has escalated at 3% for 29 years by then, so the last one was $40,000 × 1.03²⁹ = $94,263: two and a half years of spending, at the end of a plan that has just finished. The 1982 retiree under the identical rule ends with $14,116,071. The readout calls both of them the same outcome because ruin is the only failure it can define, and landing two years short of ruin is not the same result as ending with fourteen times what you started with.

Problems solved in full

$1,000,000 in a 60/40 blend across 1966 to 1995 with no withdrawals 5 steps
Show that the order of 30 years of returns cannot change the final pot, and then measure what the wobble inside them costs anyway. This is No withdrawals: $1,000,000 in a 60/40 blend across 1966 to 1995, nothing taken out, and the panel reporting the arithmetic mean of those years as 10.31%.
1. A year multiplies the pot by 1 + r and does nothing else to it. Run that 30 times and the whole retirement is a single product rather than a process.
  
  $V_{n} = V_{0} k = 1 \prod n (1 + r_{k})$
2. Multiplication commutes, so reversing the list writes the same 30 factors in a different order and cannot change what they multiply to. The two curves in the chart cross and recross and still land on the same point, and the tool has no choice in the matter.
  
  $k = 1 \prod n (1 + r_{k}) = k = 1 \prod n (1 + r_{n + 1 - k})$
3. Multiply the factors out. This window nearly 16-folds the pot.
  
  $V_{30} = 1 0^{6} \times 15.95399 = 15953990$
4. The compound growth rate is whatever single rate reproduces that product, so take the 30th root. It is not the average of the returns and it cannot be: one is built from a product, the other from a sum.
  
  $g = (15.95399)^{1/30} - 1 = 0.0967 = 9.67%$
5. Now imagine a pot that earned exactly the panel's 10.31% every single year, with the same average and none of the variation. Compound it for the same 30 years and subtract.
  
  $1 0^{6} \times (1.1031)^{30} = 18986571 ⟹ 18986571 - 15953990 = 3032581$
Answer

$15,953,990, from either end of the sequence. The steady-10.31% pot finishes at $18,986,571 instead, so varying returns cost $3,032,581 without a single year's number being changed — 19% of what the portfolio actually ends with. That is why 9.67% is the honest figure and 10.31% is not: an average you could only have collected by never deviating from it was never available. The penalty scales with the square of the spread, so a portfolio twice as jumpy around the same average gives up roughly 4 times as much to it. Order is free here. Volatility never was.

$\overset{r}{ˉ} - g = 10.31% - 9.67% = 0.64%$
Drawing 4% from $1,000,000 in a 60/40 blend across 1929 to 1958 6 steps
The same kind of sequence, but now $40,000 a year comes out of it. The window compounds at 6.58% and you are drawing 4%, which sounds like room to spare. This is Retiring into 1929: $1,000,000 in a 60/40 blend across 1929 to 1958, a first-year draw of 4% of the starting pot, rising 3% a year with prices.
1. A withdrawal rule fixes dollars, not a percentage. 4% of the starting million is $40,000, and every later year takes that same $40,000 raised by 3% compounding. The pot's own balance never enters the withdrawal again.
  
  $V_{k} = V_{k - 1} (1 + r_{k}) - W_{k}, W_{k} = 40000 \times 1.0 3^{k - 1}$
2. 1929 returned -3.30% on the blend. Grow first, then subtract: the year's income comes out of what the year's market left behind.
  
  $V_{1} = 1 0^{6} (1 - 0.0330) - 40000 = 927000$
3. Unroll the recurrence and it separates into a product term and a sum term. The product is the saver's, untouched. In the sum, each withdrawal is multiplied only by the years that come after it — a 1929 withdrawal carries 29 subsequent factors, a 1958 one carries none. Reversing the sequence leaves the product alone and rebuilds the sum from scratch.
  
  $V_{n} = V_{0} k = 1 \prod n (1 + r_{k}) - k = 1 \sum n W_{k} j = k + 1 \prod n (1 + r_{j})$
4. You already have the product without touching the data. Compound growth is defined as its 30th root, so raise the panel's figure back up.
  
  $k = 1 \prod 30 (1 + r_{k}) = (1.0658)^{30} \approx 6.767$
5. Compute the sum both ways round. Identical factors, identical withdrawals; all that changed is which factors sit behind which withdrawal. Forward, the withdrawals carry a combined weight of 196.47. Reversed, 82.09.
  
  $S = k = 1 \sum 30 1.0 3^{k - 1} j = k + 1 \prod 30 (1 + r_{j}) ⟹ S_{\to} = 196.47, S_{\leftarrow} = 82.09$
6. The sum is proportional to the first year's draw, so setting the final value to zero and solving for that draw is a rearrangement, not a search. It gives the largest opening withdrawal each order can survive.
  
  $W_{1}^{*} = \frac{1 0 ^{6} \times 6.767}{196.47} = 34444 (3.44%), \frac{1 0 ^{6} \times 6.767}{82.09} = 82438 (8.24%)$
Answer

Forward the pot is exhausted in 1950; reversed, the identical 30 years finish at $3,484,161. The break-even opening draw is $34,444 one way round and $82,438 the other — 3.44% against 8.24%, a factor of 2.39 out of a reshuffle. That ratio is the quantity a retirement plan actually turns on, and neither headline percentage can supply it: the mean 7.63% and the compound growth 6.58% are both constructed to ignore order. It also disposes of the intuition this problem opened with. Drawing 4% from something growing at 6.58% is not a 2.58-point margin, because 4% is a percentage exactly once. After that it is $40,000 with an inflation escalator attached. By the fifth year that draw has grown to $45,020, and the forward pot it comes out of is $459,742 while the reversed pot at the very same point is $1,298,083 — the identical rule, biting at 9.79% in one order and 3.47% in the other.

$\frac{8.24%}{3.44%} = 2.39$

Learning path

Risk, measured

Leads to Rebalancing

References (4)

The withdrawal-rate question this tool is a demonstration of, tested against historical sequences rather than averages: P. L. Cooley, C. M. Hubbard and D. T. Walz, "Comparative Analysis of Retirement Portfolio Success Rates." Financial Services Review 12(2), 115–128, 2003.
Sequence-of-returns risk named and measured, which is the effect the two paths on this page differ by: A. Soni, "The Role of Annuities in Managing Sequence of Returns Risk Approaching and in Retirement." The Journal of Retirement 11(4), 26–39, 2023.
Ruin as the quantity of interest rather than a final balance — the reason this tool reports the year the pot is exhausted: D. Daraei and K. Sendova, "Determining Safe Withdrawal Rates for Post-Retirement via a Ruin-Theory Approach." Risks 12(4), 70, 2024.
And the standing objection to any single historical path, including the ones drawn here: M. E. Drew and A. Walk, "Just How Safe are ‘Safe Withdrawal Rates’ in Retirement?" Financial Planning Research Journal 1(1), 22–32, 2015.

By the 42Math team · Updated 2026-08-19

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

Retiring into 1929 - A 30-year retirement beginning at the 1929 peak, drawing 4% of the starting pot in the first year and raising it 3% a year from a 60/40 mix.
Retiring into 1966 - The 1966 start, whose real damage was inflation: the same 60/40 mix and the same withdrawal rule, over a window whose losses arrive early.
Retiring into 2000 - A 25-year window from 2000, opening with three consecutive losing years before anything recovers.
No withdrawals - The same 1966 window with the withdrawal set to zero, where both orders must end at the identical value.