Tournament Brackets: Byes, Seeds, and Why the Numbers Must Be Powers of Two

The bracket halves, so it has to start at a power of two

A knockout round pairs everyone off and sends half of them home. Start with 16 and you get 8, then 4, then 2, then 1: four rounds, no remainders, nobody idle. Start with 12 and the second round has 6, the third has 3, and now three players cannot be paired. The structure only closes when every round has an even number of survivors all the way down, and the only counts with that property are the powers of two. This is not a convention that tournament organisers chose; it is what repeated halving does, and it is why every published bracket you have ever seen has 8, 16, 32, 64 or 128 slots.

The number of rounds falls out of the same argument. If the bracket holds 2^r entrants, it takes r halvings to reach a single winner, so r = log2 of the bracket size, and for an arbitrary entry count n the bracket size is the smallest power of two not below n. Written out: rounds = ceil(log2 n), bracket size = 2^ceil(log2 n). Twelve entrants therefore play a 16 bracket over 4 rounds; 100 entrants play a 128 bracket over 7; 129 entrants play a 256 bracket over 8. That last jump is the whole problem in one line — the 129th entrant does not add a match to the schedule, they add an entire round to it.

Byes are exactly the gap to the next power of two

If the bracket has 2^ceil(log2 n) slots and only n of them are filled by real entrants, the remaining slots hold nobody. Every empty slot faces a real entrant who therefore advances without playing, and that is a bye. So the bye count is the subtraction and nothing more: byes = 2^ceil(log2 n) − n. Twenty-three entrants in a 32 bracket give 9 byes; 48 in a 64 bracket give 16; 100 in a 128 bracket give 28. The count is largest just above a power of two and zero exactly on one, which is why 129 entrants produce 127 byes — more byes than the number of people who actually play in round one.

The first round is smaller than half the bracket by exactly the same amount. If b entrants receive a bye, the other n − b play, and they play (n − b)/2 matches; substituting b = 2^r − n gives first-round matches = n − 2^(r − 1), which is n minus half the bracket. For 100 entrants that is 100 − 64 = 36 matches, and the check closes: 36 × 2 = 72 people play, 28 sit out, 72 + 28 = 100, and 36 + 28 = 64 survivors go into round two — exactly half the bracket, which is where a clean power-of-two structure resumes. From round two onward there are no byes at all, because the field is a power of two again by construction.

The seed order comes out of a recursion, not a memorised list

Start with a bracket of one: the seed order is [1]. To double it, take each seed s in a bracket that is about to have size m and replace it with the pair (s, m + 1 − s). One doubling gives [1, 2]. Two give [1, 4, 2, 3]. Three give [1, 8, 4, 5, 2, 7, 3, 6]. Four give 1, 16, 8, 9, 4, 13, 5, 12, 2, 15, 7, 10, 3, 14, 6, 11. Five give 1, 32, 16, 17, 8, 25, 9, 24, 4, 29, 13, 20, 5, 28, 12, 21, 2, 31, 15, 18, 7, 26, 10, 23, 3, 30, 14, 19, 6, 27, 11, 22. Nothing is memorised and nothing is looked up; the recursion is four lines of code and produces the standard order for any bracket size.

Read the sixteen-seed line as eight first-round matches and the pattern is immediate: 1 v 16, 8 v 9, 4 v 13, 5 v 12, 2 v 15, 7 v 10, 3 v 14, 6 v 11. Every pairing sums to 17, which is the bracket size plus one, because that is precisely what the substitution s to (s, m + 1 − s) enforces. It is the same rule at every level of the tree, so the strongest remaining seed in a block always faces the weakest remaining one, at every round, without anyone having to write a second rule down.

What the seeding actually guarantees, checked rather than claimed

Run the generated order through a few assertions and the design reveals itself. Split the sixteen-seed order into blocks and every block of the same size sums to the same number: 17 per pair, 34 per quarter, 68 per half, 136 for the whole bracket — always the block size times (n + 1) divided by 2. Take the best seed in each block and you recover the seed order for a bracket half the size: the blocks of two yield 1, 8, 4, 5, 2, 7, 3, 6, which is the eight-seed order; the blocks of four yield 1, 4, 2, 3. The bracket is self-similar, which is exactly what a doubling recursion should produce.

The guarantee people actually care about follows immediately. Seeds 1 and 2 sit in opposite halves, so if both keep winning they cannot meet before the final; seeds 1, 2, 3 and 4 sit in four different quarters, so no two of them can meet before the semi-finals. Tracing seed 1's earliest possible opponents in the sixteen bracket gives 16 in round one, 8 or 9 in round two, one of 4, 5, 12, 13 in the semi-final, and 2 or 3 only in the final. That is the point of seeding, stated precisely: it does not protect the favourite from strong opponents, it postpones them, so the tournament's best matches fall at the end rather than in the opening round.

The seeding also allocates the byes without a separate rule. Fill a 128 bracket with entrants 1 to 100 and phantom entrants 101 to 128, generate the seed order by the recursion, and read off which real entrants are paired against phantoms: they are exactly seeds 1 to 28, contiguous, in order. Nobody had to decide that the byes should go to the top seeds — the same substitution that pairs s with m + 1 − s puts the highest seeds opposite the highest-numbered slots, which are precisely the empty ones. That is worth checking in whatever tool you use, because a generator that hands byes to arbitrary entrants has broken the seeding, not just the byes.

Total matches: n − 1, for every n, in one line

Every match in a single-elimination tournament eliminates exactly one entrant — that is what single elimination means. At the end, exactly one entrant has not been eliminated, so exactly n − 1 have been. One elimination per match therefore means exactly n − 1 matches, whatever the bracket size, whatever the byes, whatever the seeding. Twenty-three entrants play 22 matches; 100 entrants play 99; 129 entrants play 128. You never need to add up the rounds, and the byes do not enter the calculation at all, because a bye is not a match and eliminates nobody.

The same counting argument prices the other formats. In double elimination everyone except the champion must lose twice, so 2(n − 1) losses have to be produced, and since each match produces exactly one loss the schedule needs 2n − 2 matches. If the finalist coming out of the losers' bracket wins the grand final, they have handed the previously unbeaten player a first loss, and a reset match is played to give a second — 2n − 1 matches in that case. Both figures are exact, and which one you get is decided on the day. Round robin is a different creature: every pair meets once, so the count is C(n,2) = n(n − 1)/2, which grows quadratically. At 100 entrants that is 4,950 matches against 99 for a knockout, a factor of exactly 50.

Choosing a format from the numbers

The three formats trade matches against information. Single elimination is the cheapest possible tournament — n − 1 matches, ceil(log2 n) rounds — and it produces exactly one reliable fact, the identity of the winner. Everything below first place is an artefact of the draw: the losing semi-finalists are not ranked against each other, and a strong entrant who meets the champion in round two finishes indistinguishably from a weak one. Double elimination buys a second chance for roughly twice the matches and roughly one extra round, and it removes the single worst failure mode, which is a good entrant knocked out by one bad day.

Round robin gives a full ranking and charges quadratically for it. Twelve entrants play 66 matches instead of 11; 23 play 253 instead of 22; 48 play 1,128 instead of 47. It also needs n − 1 rounds when n is even and n rounds when n is odd, because with an odd field somebody sits out each round. The practical compromise most large events use is a group stage followed by a bracket: round robin inside small groups produces a defensible ranking cheaply, and the knockout then costs one match per eliminated qualifier. Whichever you pick, compute the match count before you book the venue — the difference between 99 and 4,950 is not a scheduling detail.

Frequently asked questions

How many byes does a tournament with 23 entrants need?

Nine. The bracket size is the smallest power of two not below 23, which is 32, and the bye count is that minus the entry count: 32 − 23 = 9. The formula is general — byes = 2^ceil(log2 n) − n — and it is a subtraction, not a rule of thumb. The first round then holds n minus half the bracket, that is 23 − 16 = 7 matches, and the arithmetic closes: 7 × 2 = 14 people play, 9 receive a bye, 14 + 9 = 23, and 7 + 9 = 16 survivors enter the second round, which is exactly half the bracket. From that point on there are no more byes, because the field is a power of two again. The whole tournament runs 5 rounds and 22 matches.

What is the standard seed order for a 16-player bracket?

1, 16, 8, 9, 4, 13, 5, 12, 2, 15, 7, 10, 3, 14, 6, 11 — read as eight first-round matches, that is 1 v 16, 8 v 9, 4 v 13, 5 v 12, 2 v 15, 7 v 10, 3 v 14, 6 v 11. Rather than memorising it, generate it: start from the list [1] and repeatedly replace every seed s by the pair (s, m + 1 − s), where m is the size the list is about to reach. Four rounds of that doubling give the order above; five give the 32-seed order, and so on. The result is verifiable rather than trusted — every first-round pairing sums to 17, every quarter of the list sums to 34 and every half to 68, seeds 1 and 2 land in opposite halves, and seeds 1 to 4 land in four different quarters.

Why are the top seeds paired against the bottom ones?

To push the meetings between strong entrants as late as possible. Pairing 1 against 16 and 2 against 15 is not about giving the favourites an easy opening; it is the only way to place them so that they cannot meet early. The recursion that produces the order puts seeds 1 and 2 in opposite halves and seeds 1, 2, 3, 4 in four different quarters, so seed 1 can meet 2 or 3 only in the final and cannot meet any of 4, 5, 12 or 13 before the semi-final. Trace seed 1's earliest possible opponents in a 16 bracket and you get 16 in round one, 8 or 9 in round two, one of 4, 5, 12, 13 in the semi and 2 or 3 in the final. The design goal is a tournament whose best matches happen at the end, and it produces that automatically from a single substitution rule.

How many matches will my tournament need?

For single elimination, n − 1, and there is nothing to look up. Each match eliminates exactly one entrant, at the end exactly one entrant remains, so exactly n − 1 eliminations and therefore n − 1 matches have taken place. Byes change nothing, because a bye is not a match. Double elimination needs everyone but the champion to lose twice, which is 2(n − 1) losses and therefore 2n − 2 matches; if the entrant coming from the losers' bracket wins the grand final, a reset match is played and the total is 2n − 1. Round robin plays every pair once, so C(n,2) = n(n − 1)/2 matches, over n − 1 rounds if n is even and n rounds if n is odd. For 100 entrants the three formats cost 99, 198 or 199, and 4,950 matches respectively.

Should the byes go to the top seeds?

They already do, if you build the bracket properly — you never have to decide it separately. Fill the bracket up to its power-of-two size with phantom entrants numbered above the real ones, generate the seed order by the doubling recursion, and read off who faces a phantom. For 100 entrants in a 128 bracket that turns out to be exactly seeds 1 to 28, contiguous and in order, because the same substitution that pairs s with m + 1 − s puts the highest seeds opposite the highest-numbered slots, and those are precisely the empty ones. Two consequences follow. First, the number of byes and the identity of the entrants who get them come from one construction, not two rules that can disagree. Second, a generator that scatters byes among arbitrary entrants has broken the seeding as well, and is worth replacing.