How lottery odds are actually calculated — combinations explained from scratch

Every lottery odds figure you have ever seen — 1 in 292,201,338 for US Powerball, 1 in 8,145,060 for Australia's Saturday Lotto — comes from a single piece of school-level mathematics: counting combinations. No operator secret, no statistics degree, no simulation. This article builds the formula from nothing and then uses it to derive the jackpot odds of eleven major world games, every step shown.

Step 1: counting ordered picks

Suppose a machine will draw 6 balls from a barrel of 45, one at a time. How many different sequences of balls could come out?

• The first ball can be any of 45.
• The second can be any of the remaining 44.
• The third, any of 43 — and so on down to the sixth, which can be any of 40.

Multiply the choices together:

45 × 44 × 43 × 42 × 41 × 40 = 5,864,443,200

So there are about 5.86 billion possible ordered sequences. Mathematicians call these permutations. The general recipe for drawing k items from n, in order, is n × (n−1) × … × (n−k+1) — start at n and multiply k descending numbers.

Step 2: why order doesn't matter (and why we divide)

Here is the thing your ticket doesn't care about: the order the balls came out. If the winning numbers are 3, 17, 22, 30, 41, 45, you win whether the machine spat out the 22 first or last. The draw 17–3–45–22–41–30 and the draw 3–17–22–30–41–45 are the same result as far as prizes go.

How many orderings does one set of 6 numbers have? The first position could be any of the 6 numbers, the second any of the remaining 5, and so on:

6 × 5 × 4 × 3 × 2 × 1 = 720

That product is called a factorial, written 6! ("six factorial"). Every distinct set of 6 numbers appears 720 times in our list of 5.86 billion sequences — once per ordering. So the number of genuinely different sets is:

5,864,443,200 ÷ 720 = 8,145,060

That is where Saturday Lotto's famous figure comes from, and it matches the odds published by the operator, The Lott. Exactly one of those 8,145,060 sets wins Division 1, so one standard game has a 1 in 8,145,060 chance of the jackpot.

The general formula is called the binomial coefficient, written C(n,k) and read "n choose k":

C(n,k) = [n × (n−1) × … × (n−k+1)] ÷ k!

It counts the number of different k-number sets you can choose from n numbers when order is irrelevant. In the textbook's fully-factorial costume the same formula reads C(n,k) = n! ÷ (k! × (n−k)!) — for Saturday Lotto that's 45! ÷ (6! × 39!), where the enormous 39! on the bottom simply cancels the unused tail of the 45! on top, leaving exactly our 45 × 44 × 43 × 42 × 41 × 40 ÷ 720. Two sanity checks: C(45,1) = 45 (there are 45 ways to pick one ball), and C(45,45) = 1 (only one way to take the lot). Everything on this site — every odds table, every calculator — rests on this one formula.

One wording note before the worked examples. Strictly, "odds" and "probability" are different dialects: a probability of 1/8,145,060 corresponds to odds of 1 to 8,145,059 against. Lottery operators (and this site) use the looser everyday convention "1 in 8,145,060", meaning one favourable outcome per 8,145,060 equally likely outcomes. At lottery scale the distinction changes nothing you'd notice — one part in eight million — but it's worth knowing which convention a pay table is speaking before you compare figures across sources.

Worked example 1: US Powerball — the full 292,201,338

Powerball uses two separate machines: 5 white balls from a drum of 69, and 1 red Powerball from a different drum of 26. Because the drums are independent, we count each and multiply.

White balls — C(69,5):

69 × 68 × 67 × 66 × 65 = 1,348,621,560

5! = 120

1,348,621,560 ÷ 120 = 11,238,513

Red ball — C(26,1) = 26.

Total distinct tickets:

11,238,513 × 26 = 292,201,338

One of them wins the jackpot: 1 in 292,201,338, exactly as published on the official Powerball prize chart. Notice how much work the little red ball does — it multiplies the whole space by 26. There's a full anatomy of that trick in what a bonus ball does to your odds.

Worked example 2: EuroMillions — two mini-lotteries multiplied

EuroMillions asks for 5 main numbers from 50 and 2 Lucky Stars from 12. Same logic, two parts.

Main numbers — C(50,5):

50 × 49 × 48 × 47 × 46 = 254,251,200

254,251,200 ÷ 120 = 2,118,760

Lucky Stars — C(12,2):

12 × 11 = 132; 132 ÷ 2! = 66

Total:

2,118,760 × 66 = 139,838,160

1 in 139,838,160, matching the operator's published odds. Eurojackpot has used the identical 5-from-50 plus 2-from-12 format since March 2022, so its jackpot odds are the same 1 in 139,838,160 (Eurojackpot odds).

Worked example 3: a bonus-ball game — Saturday Lotto

Australia's Saturday Lotto draws 6 winning numbers plus 2 supplementaries from the same barrel of 45. Do the supplementaries change the jackpot odds? No — Division 1 requires matching the 6 winning numbers only, and we already counted that:

C(45,6) = 5,864,443,200 ÷ 720 = 8,145,060

The supplementaries exist to create extra prize divisions lower down (Division 2 is 5 winning numbers plus a supplementary, at 1 in 678,755 per The Lott). A bonus ball drawn from the same barrel never changes the jackpot; a bonus ball from a separate drum multiplies it. That single distinction explains most of the difference between the world's games, and gets its own full treatment in what a bonus ball does to your odds.

One level down: odds for partial matches

The same counting handles every lower division, not just jackpots. The pattern: to match exactly m of the 6 winning numbers, choose which m winners you hold — C(6,m) ways — and fill your remaining 6−m spots from the 39 losing numbers — C(39,6−m) ways. Multiply, then divide the total space by the result.

Worked example — Saturday Lotto Division 4 (match any 4 winning numbers):

C(6,4) = 15 ways to hold 4 of the 6 winners

C(39,2) = 39 × 38 ÷ 2 = 741 ways to fill your other 2 spots with losers

15 × 741 = 11,115 winning tickets out of 8,145,060

8,145,060 ÷ 11,115 = 732.8 → published as 1 in 733 by The Lott

Every division odds figure on every operator's site is this one move — mathematicians call the resulting distribution hypergeometric, but you never need the name, just the two C(…)s and a multiplication. Summing all the divisions' winning tickets gives the game's advertised "overall odds of winning", a number with its own marketing career examined in jackpot odds vs any-prize odds.

Jackpot odds for the major world games — every one derived

Each row below is the same C(n,k) machinery. "Formula" shows the counting; the last column is the derived number of combinations, which is also the "1 in …" jackpot odds for a single line.

Sources for the published figures: The Lott, OLG PlaySmart, Powerball, Mega Millions (matrix changed to 5/70 + 1/24 in April 2025), euro-millions.com, euro-jackpot.net, Caixa — Mega-Sena.

Three things jump out of the table:

1. Pool size beats pick size. SuperEnalotto's innocent-looking 6 of 90 (1 in 622.6 million) is more than twice as hard as US Powerball's 5 of 69 plus a red ball. Each extra ball in the pool inflates the numerator of C(n,k).
2. A separate drum is a multiplier. Australian Powerball's main game alone is 1 in 6,724,520; the 1-of-20 Powerball turns that into 1 in 134,490,400.
3. Same-barrel bonus balls are free. Saturday Lotto's supplementaries and UK Lotto's bonus ball change nothing at the jackpot level — they only add lower divisions.

A worked check you can do in your head

A useful approximation: C(n,k) grows roughly like n^k ÷ k!. For 6 of 49: 49^6 is about 13.8 trillion; divide by 720 and then knock a bit off for the "descending" numbers (48, 47, … instead of 49 each time) and you land near 14 million. If someone quotes you odds for a game, you can now smell-test the figure without a calculator app in sight.

Two misreadings to retire

"1 in 8,145,060" is per line, not per player. Buy 12 distinct lines and your jackpot probability is exactly 12 ÷ 8,145,060 — a clean 1 in 678,755. Chances in distinct lines scale perfectly linearly, which sounds encouraging until you price the linearity; that arithmetic gets its own cold shower in why two tickets change nothing.

Every combination is one combination. The derivation above never looked at which numbers you picked — only how many sets exist. The line 1, 2, 3, 4, 5, 6 is exactly one of the 8,145,060 sets, precisely as likely as last week's winners or your family's birthdays. What differs between lines is not the chance of winning but the expected crowd sharing a pari-mutuel prize if you do, since human-favourite patterns are massively over-picked — a genuinely useful edge you can explore with the number sharing risk tool.

What the number doesn't tell you

C(n,k) tells you the size of the haystack. It says nothing about the prize for finding the needle, how many other people will share it, or what a ticket costs per unit of chance — that is the territory of expected value and cost per chance. It also doesn't tell you what 292 million feels like, because human intuition gives up somewhere around a few thousand — for that, see picturing 1 in 292 million. And if the headline "overall odds 1 in 24.9" printed on a Powerball ticket looks like it contradicts everything above, it doesn't — it answers a much easier question, unpacked in jackpot odds vs any-prize odds.

The formula is the whole game. Everything else is presentation.