The Math Behind a Double-Six Domino Set (Why There Are 28 Tiles)

Anyone who's played with a standard domino set has probably had the number 28 drilled in without ever being told where it comes from. It isn't arbitrary — it falls straight out of a bit of combinatorics that's worth understanding once, because it explains every other set size too, from the small double-six sets used for two-player games up through the double-eighteen monsters used for very large tables.

What a domino tile actually is

A domino tile is an unordered pair of numbers, each between 0 and some maximum value n. "Unordered" is the key word: a 3-5 tile and a 5-3 tile are the same physical tile, not two different ones — there's no "first" or "second" half that matters. And unlike, say, a hand of playing cards, repeats are allowed: 4-4 is a perfectly normal tile (a "double"), not a forbidden combination.

So the question "how many tiles are in a set?" is really the question "how many unordered pairs, with repeats allowed, can you make from the numbers 0 through n?"

The formula

This is a classic combinatorics problem called combinations with repetition, and for pairs specifically it has a clean closed form:

Total tiles = (n + 1)(n + 2) / 2

Plug in n = 6 for a standard double-six set: (7 × 8) / 2 = 56 / 2 = 28. There's your 28, straight out of the formula rather than a number someone picked by hand.

Where the formula comes from

One way to see why this works: imagine listing every ordered pair (a, b) where a and b each range from 0 to n. That's (n+1) choices for a times (n+1) choices for b, or (n+1)² ordered pairs total. Among those, exactly n+1 of them are doubles (0-0, 1-1, 2-2, and so on up to n-n) — pairs where a and b happen to match.

The remaining (n+1)² − (n+1) pairs are non-doubles, and since each of those was counted twice in our ordered list (3-5 and 5-3 are the same tile, remember), we divide that portion by two and then add the doubles back in untouched:

[(n+1)² − (n+1)] / 2 + (n+1) = (n+1)(n+2) / 2

which is exactly the formula above, arrived at a slightly different way. Either derivation lands on the same number. There's a third, even quicker way to see it if you already know the "stars and bars" identity for combinations with repetition: choosing 2 values with repetition from n+1 options is C(n+2, 2), which expands to the same (n+1)(n+2)/2. All three routes — direct counting, the ordered-pairs argument above, and stars and bars — are just different lenses on the same underlying count.

Checking it against real sets

The formula holds for every common commercial set size:

• Double-six (n=6): (7 × 8)/2 = 28 tiles — the standard set for two to four players.
• Double-nine (n=9): (10 × 11)/2 = 55 tiles — common for larger groups.
• Double-twelve (n=12): (13 × 14)/2 = 91 tiles — the usual choice for Mexican Train and Chicken Foot.
• Double-fifteen (n=15): (16 × 17)/2 = 136 tiles.
• Double-eighteen (n=18): (19 × 20)/2 = 190 tiles — large enough for very big group games.

You can run any of these through the Domino Set & Combination Counter instead of doing the arithmetic by hand, including a breakdown of how many doubles are in the set and how many tiles carry any particular number, or see all five sizes side by side on the Domino Set Reference.

Two smaller facts that fall out for free

Once you have the main formula, two related counts come almost for free:

• Number of doubles. There's exactly one double for each value from 0 to n, so a set always contains n+1 doubles — 7 in a double-six set, 13 in a double-twelve set.
• Tiles containing a given number. Pick any single value, say 4. It appears on a tile paired with every value from 0 to n, including itself — that's n+1 tiles again. In a double-six set, the number 4 shows up on exactly 7 tiles: 4-0, 4-1, 4-2, 4-3, 4-4, 4-5, and 4-6.

Notice those two counts are always identical — the number of doubles and the number of tiles carrying any single value are both n+1, just arrived at from two different questions ("how many tiles have matching halves" versus "how many tiles include this one value at all"). That's not a coincidence: a double is simply the one tile, out of the n+1 tiles carrying a given value, whose other half also happens to be that same value.

A related formula: total pip value

The tile-count formula answers "how many tiles," but a related question — "how many pips, total, across the whole set" — has its own clean formula, useful for understanding just how much a blocked round in a bigger set can swing a score by:

Total pip value = n(n + 1)(n + 2) / 2

For a double-six set, that's 6 × 7 × 8 / 2 = 336 / 2 = 168 total pips spread across all 28 tiles. Notice the resemblance to the tile-count formula — it's exactly n/2 times as many, since total pip value and tile count share the same (n+1)(n+2) core and differ only by that factor of n. For a double-twelve set, the same formula gives 12 × 13 × 14 / 2 = 1,092 pips — more than six times the double-six total, not merely twice as much as the roughly 3× increase in tile count, which is exactly why blocked rounds in a bigger set can swing scores so much harder than the tile count alone would suggest.

Deriving the pip-value formula

The derivation follows the same doubles-plus-non-doubles split used for the tile count above, just weighted by pip value instead of by tile. The n+1 doubles each carry pip value 2a for their value a, so they sum to 2(0 + 1 + … + n) = n(n+1), using the well-known formula for summing consecutive integers. The non-double tiles are trickier: summing (a+b) over every ordered pair with a ≠ b gives n(n+1)² minus the diagonal, and since every non-double is double-counted in that ordered sum (once as (a,b), once as (b,a)), halving it gives their true contribution: n²(n+1)/2. Add the two parts together — n(n+1) + n²(n+1)/2 — and factor out n(n+1) to get n(n+1)[1 + n/2], which simplifies to the formula above: n(n+1)(n+2)/2.

Run across every common set size, the same formula gives 495 pips for double-nine, 2,040 for double-fifteen, and 3,420 for double-eighteen — all five sizes, with their tile counts and doubles alongside, are on the Domino Set Reference page.

Both formulas are triangular numbers in disguise

There's a tidy way to see both formulas at once using triangular numbers — the sequence 1, 3, 6, 10, 15… where the k-th triangular number T(k) = k(k+1)/2 counts how many items fit in a triangle k rows deep. The tile-count formula (n+1)(n+2)/2 is exactly T(n+1): a double-six set's 28 tiles is the 7th triangular number, a double-twelve set's 91 tiles is the 13th. The pip-value formula is just n times that same triangular number — n(n+1)(n+2)/2 = n × T(n+1) — which is a clean way to remember both formulas as one idea (a triangular tile count) rather than two formulas to memorize separately.

Double-checking a formula like this yourself

A closed-form formula is convincing once you've seen the derivation, but it's worth knowing how to sanity-check one independently, without trusting the algebra alone. The brute-force method is simple: list every tile a ≤ b for a and b from 0 to n, count them, and sum their pips separately. For n=6 that's a short enough list to do by hand — 28 tiles, pips ranging from 0 (the 0-0 tile) up to 12 (the 6-6 double) — and adding them all up lands on exactly 168, matching the formula. For bigger sets, the same brute-force count is easy to run as a short script rather than by hand, and it's exactly how the figures on this page and on the Domino Set Reference were checked before publishing: an independent, tile-by-tile count confirming the closed-form formula, not just the algebra trusted on its own.

This habit — deriving a formula, then checking it against a slower but unambiguous brute-force method — is worth applying to any "neat" combinatorics claim you come across, domino sets included. A formula that only works for one example you happened to pick isn't actually the formula; one that holds up against several independently-counted set sizes almost certainly is.

Why this matters beyond trivia

Knowing the exact tile count isn't just a fun fact — it's the baseline for anything that depends on set completeness. If you're checking whether a domino box is missing a piece (see the Tile Inventory Tracker) or working out the odds of drawing a double into your opening hand (see the Draw Probability Calculator), it all starts from this same formula: (n+1)(n+2)/2 tiles, no more, no less. For a practical look at how these numbers should actually inform which set to buy or bring to a game night, see our guide to choosing a domino set size.