Every draw adds up to something

Where the middle of the curve comes from

Each draw takes 5 numbers from 56. Any one of them averages $(N+1)/2$, and expectations add regardless of whether the picks are independent, so:

$$E[S] = k \cdot \frac{N+1}{2}$$

For Mega Millions that is 142.5. For the classic 6/49 game it is exactly 150.

The spread needs more care, because the numbers are drawn without replacement — take a high one and the rest are slightly likelier to be low. That negative dependence tightens the distribution, and the finite-population correction accounts for it:

$$\mathrm{Var}[S] = \frac{k(N+1)(N-k)}{12}$$

Here $\sigma =$ 34.8. For 6/49 the variance is 1,075 and $\sigma \approx 32.79$. Treating the picks as independent would overstate the spread — the curve would be too wide, and the bars would look suspiciously tight against it.

The curve is exact, not fitted

The gold line is not a bell curve drawn through the data. It is the actual count of tickets at every total, computed from the format alone.

Think of the generating function $\prod_{i=1}^{N}(1 + x^i y)$: each number is either in a ticket or not, $y$ counts how many were taken and $x$ tracks the running total. The coefficient of is precisely the number of -ticket combinations summing to , and a small dynamic programme extracts it in milliseconds.

For 6/49 that gives 165,772 tickets at sum 150 — the single most common total, and still only about 1.185% of all 13,983,816 combinations. This game has 42,416 at its peak of 142, out of 3,819,816 possible tickets.

The dashed line is the normal approximation. It hugs the exact curve because a sum of many similar contributions tends toward a bell shape — the central limit theorem doing what it always does. It is an approximation of something we can compute precisely, which is why the exact curve is the solid one.

The category error

Here is the whole thing in one sentence: you bet on six numbers, not on their sum.

Every ticket has probability $1/\binom{N}{k}$ — for this game, 1 in 3,819,816. That figure does not consult the ticket's total. A sum of 142 is common because many tickets share it, not because any one of them is favoured, and picking one of those tickets gets you exactly one of them.

Conditioning on a sum range does not change any ticket's chances. It only counts how many tickets are in the range. Betting "a mid-range sum" is betting "a ticket", with extra steps.

How the fit badge is computed

The observed sums are binned, tail bins are merged until each expects at least five draws — otherwise a single draw in a thin bin produces an enormous $(O-E)^2/E$ term and the test reports drama that is purely an artefact of binning — and then:

$$\chi^2 = \sum_i \frac{(O_i - E_i)^2}{E_i}$$

against the exact expected counts, on one fewer degree of freedom than there are bins.

"The world is asking to be deceived"

Henk Tijms spent a career as a probabilist at the Vrije Universiteit Amsterdam, and in retirement he keeps returning to the same irritation: the industry built on selling people sum ranges.

His summary is the cleanest sentence anyone has written about this page:

You are betting on the six individual numbers that are going to be drawn, and not on what their sum will be.

That is the entire refutation. Everything else — the curve, the percentages, the software — sits downstream of it. He titled one piece Lotto Nonsense: The World is Asking to be Deceived, and the second half of the title is doing as much work as the first.

The most successful system ever sold

Gail Howard began publishing lottery systems in 1983 under the name Smart Luck. Books, software, a newsletter, television appearances; the Lottery Master Guide went through many printings. It was, commercially, the most successful lottery system in American history.

A substantial part of it was sum-range advice — the "Balanced Game". Keep your total in the middle band, avoid the extremes, and your ticket is "balanced".

It is worth being precise about what is wrong with it, because the claims themselves were generally true. It is genuinely the case that around 70% of winning combinations fall in the middle band. The trick is that this is not a fact about winners. It is a fact about tickets — and winners are drawn from tickets. The panel above computes both numbers side by side; they are the same number, because they were always the same number.

Survivorship framing is what makes it persuasive. Look only at winners, quote a property they share, and imply the property caused the winning. The same move works for any property that most tickets happen to have.

Why it feels right anyway

The pitch survives because it agrees with an instinct that is very hard to argue with: 1-2-3-4-5-6 feels like it cannot win.

Kahneman and Tversky's representativeness heuristic explains the feeling — we judge randomness by whether an outcome looks random, and a tidy sequence does not. The bell curve then arrives and appears to confirm it. The odds line refuses to.

In April 2008 an Ontario draw came up 40-41-42-43-44-45. Two hundred and thirty nine people had that ticket, having chosen it precisely because it looked impossible — and they split a second prize that would otherwise have been one person's. The sequence was never less likely. It was only ever more popular, and popularity is the one thing about a ticket that genuinely affects what you take home.

The most honest chart in the casino

The bell curve on this page is real. It is exact, it is beautiful, and it has been true of this game since before its first draw. It tells you a great deal about the structure of the ticket space.

It tells you nothing whatsoever about which ticket to buy. Those two facts have always sat comfortably together, and only one of them has ever been sold.