How Many Fish Are in the Ocean? Two Routes That Agree to a Power of Ten

One fish per ten thousand cubic metres times the whole ocean gives 130 trillion, and the arithmetic is exact. The error is that a density you can picture is a surface density, and confining it to the 200 metre sunlit layer drops the figure by a factor of 18. A second chain built from the annual catch, which touches no ocean geometry at all, lands in the same decade, and that agreement is the result rather than either set of digits.

How many fish are in the ocean? There is no scale on which this could be settled, and no census exists, so the question is really asking something else: can you build a chain of guesses whose answer is worth trusting to a power of ten, and can you tell which link is the weakest?

Two anchors are worth looking up rather than guessing. The world ocean covers about $3.6 \times 10^8$ square kilometres and holds about $1.3 \times 10^9$ cubic kilometres of water. Those two are consistent with each other, which is a check and not a coincidence: their ratio is a mean depth of 3.61 kilometres, the accepted figure. Everything else below is declared as a guess.

A density times the whole ocean

Guess one fish per $10^4$ cubic metres of water. That is a box a little over twenty metres on a side holding a single fish, which sounds sparse enough to be safe. A cubic kilometre is $10^9$ cubic metres, so the whole ocean holds $1.3 \times 10^{18}$ cubic metres and

N_{\text{naive}} \;=\; \frac{1.3\times 10^{18}\ \mathrm{m^3}}{10^4\ \mathrm{m^3\ per\ fish}} \;=\; 1.3 \times 10^{14}

(1)

A hundred and thirty trillion. The arithmetic is exact and the answer is off by more than an order of magnitude, because of where the density came from. A figure you can picture is a figure about water you have seen, and water anyone has seen is the top of the column. The guess is a surface density, and equation (1) applies it down 3.6 kilometres of ocean.

What the density actually describes

Sunlight is effectively gone by about 200 metres. Take that as the layer the guess belongs to. Its volume is the ocean's surface times that depth:

V_{\text{lit}} \;=\; 3.6\times 10^{8}\ \mathrm{km^2} \times 0.2\ \mathrm{km} \;=\; 7.2\times 10^{7}\ \mathrm{km^3} \;=\; 7.2\times 10^{16}\ \mathrm{m^3}

(2)

That is 5.5 percent of the ocean's water. At the declared density it holds

N_{\text{water}} \;=\; \frac{7.2\times 10^{16}}{10^4} \;=\; 7.2\times 10^{12}

(3)

The correction between equations (1) and (3) is a factor of 18.1, which is the size of the original mistake. Notice what this route does not claim. It assigns no fish at all below 200 metres, which is certainly false: the mesopelagic zone is thought to hold enormous numbers, and a genuine total would have to include them. Assigning nothing there makes equation (3) a floor rather than a census. The statement being defended is the narrow one: the density is a surface figure and cannot be carried down the column.

Fig. 1 — The guess belongs to the shaded strip. The trap consists of multiplying it by the whole column instead.

A flow instead of a stock

One chain is not an answer, because a chain that is linear in its weakest input gives no warning when that input is wrong. So build a second one that shares no factor with the first. The world's reported capture fisheries production runs near 90 million tonnes a year. At a kilogram per fish that is

90\times 10^6\ \mathrm{t} \times 10^3\ \mathrm{kg/t} \;=\; 9\times 10^{10}\ \text{fish landed per year}

(4)

Now the one piece of ecology this route needs: if a year's catch removes roughly one fish in thirty of the standing stock of that size, the stock is

N_{\text{catch}} \;=\; 30 \times 9\times 10^{10} \;=\; 2.7\times 10^{12}

(5)

This route uses no ocean geometry whatsoever. No area, no depth, no volume, no density. It converts a measured flow into a stock, where the first route measured a volume and filled it. The independence is testable rather than rhetorical: doubling the reported catch leaves equation (3) untouched, and doubling the ocean's volume leaves equation (5) untouched.

The agreement is the result

The two answers are $7.2\times 10^{12}$ and $2.7\times 10^{12}$ . They differ by a factor of 2.67 and both sit strictly between $10^{12}$ and $10^{13}$ . A few trillion, then. What is being claimed is the power of ten and the fact that two unrelated chains found it, rather than either set of digits.

Fermi chain — how the error compounds

Write the estimate as a product of declared factors, $N = \prod_i x_i$ . If each factor is trusted to within a multiplicative factor $f_i$ , meaning $x_i / f_i \le \hat{x}_i \le f_i x_i$ , then $N$ is trusted to within $\prod_i f_i$ . Relative errors add in $\log$ space, so the natural output of such a chain is an interval on a logarithmic scale, and the natural way to report it is a power of ten.

Sweeping every guess

The four declared guesses are the lit depth, the density, the average mass of a landed fish and the harvest fraction. Push each across its plausible range and the intervals are wide, exactly as equation-level linearity predicts. Taking the lit depth from 100 metres to 1 kilometre and the density across two orders of magnitude either side puts the water route anywhere in

3.6\times 10^{11} \;\le\; N_{\text{water}} \;\le\; 3.6\times 10^{14}

(6)

while taking the harvest fraction from one in ten to one in a hundred puts the catch route in $[9.0\times 10^{11},\, 9.0\times 10^{12}]$ , and taking the average landed fish from 0.1 kg to 10 kg puts it in $[2.7\times 10^{11},\, 2.7\times 10^{13}]$ . Three orders of magnitude of spread, from guesses that all sounded reasonable.

Fig. 2 — Two wide intervals whose point estimates land in the same decade. The dashed line is the answer from equation (1).

So why report anything at all? Because the useful question is not how wide each interval is, it is whether the two routes keep agreeing while the guesses move. They stay within a factor of ten of each other for every harvest fraction from one in ten to one in a hundred, and for every average landed mass from 0.1 kg to 3 kg. The stricter version, that both land inside the $10^{12}$ decade, survives average masses from 0.3 kg to 1 kg.

Where it breaks

Two places, both identified by sweeping rather than by intuition. Raise the average landed fish to 10 kg and the routes diverge by a factor of 25. Raise the harvest fraction to one fish in five and they diverge by 16. In each case the disagreement is the signal that one of the chains has been pushed outside the regime it was built for.

The average landed mass is the genuinely fragile input, since equation (4) is exactly inversely proportional to it and global landings are dominated by small pelagic species. One kilogram is a rough middle, not a safe number. It is worth recording that the first guess about which end would fail was wrong: the natural expectation is that the light end breaks the agreement, and the sweep shows the heavy end does. That is the argument for writing the sweep instead of reasoning about it.

More generally, the method stops working the moment a chain is asked for a digit. Equation (3) is linear in the density, so a density wrong by a factor of ten moves the answer by a factor of ten while every intermediate step still looks tidy. Nothing inside the chain detects this. Only a second chain does, which is why the cross-check is part of the method and not a flourish at the end of it.

Sources and further reading

The two anchors, and the mean depth that reconciles them: Ocean
Why 200 metres is the natural boundary, and what lies below it: Photic zone and Mesopelagic zone
The flow the second route rests on: Capture fisheries production
The style of reasoning: Fermi problem and Order of magnitude
For the mesopelagic caveat, the acoustic-survey literature revised estimates of that zone's biomass sharply upward: X. Irigoien et al., “Large mesopelagic fishes biomass and trophic efficiency in the open ocean”, Nature Communications 5 (2014), 3271.

Every chain above was recomputed in exact rational arithmetic and every sweep was run exhaustively rather than sampled. Nothing here asserts a true number of fish, and the deliverable is the decade the two routes share.