Bingo, priced exactly: every line count from zero to twelve and the return behind 38 balls

One card, 38 balls

The game is 75-ball bingo played alone. The engine deals a 5×5 card — five numbers from 1–15 in the B column, 16–30 in I, and so on — marks the centre cell free, then draws 38 of the 75 balls. Every row, every column and both diagonals is a line, twelve in all, and the payout depends only on how many of the twelve are complete. Designing Bingo explained why 38 balls and why those seven pay steps; it quoted the line-count distribution from a million simulated rounds. This article computes it exactly.

One fact makes the exact calculation tractable. The card holds 24 real numbers, and which of them get marked depends only on which 38 balls are drawn: a specific set of k card numbers being marked, and the other 24 − k not, has probability C(51, 38 − k) ÷ C(75, 38), the same for every card layout. So the distribution of line counts is the sum over all 2²⁴ = 16,777,216 mark patterns of that weight — a few seconds of computation, and no sampling error.

Every line count, priced

Share of return = chance × payout, as a percentage of the stake. The design notes’ simulated figures (62.9 / 28.6 / 7.05 / 1.27 / 0.20 / 0.03%, return ≈ 96.5%, hit 37.1%) sit within sampling error of these; the exact house edge is 3.34%.

Two things stand out in the table. First, the money is at the bottom of the paytable: the single-line ×1.5, hit on two cards in seven, carries 43% of the return, and one and two lines together carry 71%. The ×100 and above rungs that give the game its top are worth 4% of the return between them. Second, the steps are steep on purpose: each extra line is roughly five to seven times rarer than the last, and the payout roughly triples, so the return per rung falls with rarity rather than staying flat. That is a design choice — Keno makes the same one — and it is what keeps the whole card at 96.66% while the rare counts stay memorable.

Eleven is impossible because every cell sits on one row and one column, so a single unmarked number always breaks at least two lines: line counts are lost two at a time, never one. Ten lines is one unmarked number off both diagonals; twelve means all 24 numbers drawn, C(51, 14) ÷ C(75, 38).

The free star and the average card

A ball lands on any given card number with probability 38 ÷ 75 = 50.67%, but a line needs all of its numbers, and the draws are without replacement. A line that avoids the centre needs five specific balls among the 38: 38 ÷ 75 × 37 ÷ 74 × 36 ÷ 73 × 35 ÷ 72 × 34 ÷ 71 = 2.908%, one in 34.4. A line through the free star needs only four: 6.073%, one in 16.5. Four of the twelve lines pass through the star — the middle row, the middle column and both diagonals — so the expected number of completed lines on a card is 8 × 2.908% + 4 × 6.073% = 0.476.

Hypergeometric: C(24, k) × C(51, 38 − k) ÷ C(75, 38). The mean is 12.16 marks; the card looks half full on an average round, which is why one line is common and three is not.

The marked-number table is the honest picture of a typical round: about twelve of the twenty-four numbers get a mark, scattered across the grid, and the question is only whether five of them happen to sit in a row. Fourteen or more marks happen one round in four and still usually complete at most a line or two; the three-line card, one in 78, is nearly always a card with an unusually lucky arrangement rather than an unusually full one.

Because the line counts and the payouts are fixed, the per-round swing is fixed too: the standard deviation of one round’s result is 3.26 stakes. Over 100 one-unit rounds the expected loss is 3.3 units and the swing about 33 units, so a session that ends 30 units up or down is ordinary. Risk of ruin turns that into how long a given bankroll usually lasts.

Checking a card

1. From the round’s fairness panel take the revealed server seed, the client seed and the nonce.
2. For cursors 0 to 24 compute HMAC-SHA256 keyed by the server seed over “clientSeed-nonce-cursor”, take the first 8 bytes as a fraction of 2⁶⁴, and walk the card column by column: each draw picks index floor(u × remaining) from that column’s fifteen numbers, swapping the last number into the hole. Cell 12 is then the free star.
3. For cursors 25 to 62 do the same pool hop over 1–75 to get the 38 balls, mark the card, count the complete lines among the twelve, and read the payout from the table above.

The open-source verifier’s bingo command does the three steps in one call. The layout and the balls come from one committed seed pair, so a card cannot be dealt after the balls are known, and the balls cannot be drawn to fit the card — which is what the exact distribution above assumes, and what the commitment lets you check.