Lottery Multi-Ticket Calculator

Defaults: odds 1 in 13,983,816 (classic 6/49 jackpot, no bonus drum), tickets 10, draws 5. You are buying ten lines a night for a week of five draws, or ten lines for five independent drawings — the math is the same if each drawing is independent and you do not carry combinations across correlated games. Enter the 1-in-N jackpot (or any single-tier) odds, how many lines per drawing, and how many drawings.

Calculate. Total Entries = tickets × draws = 50. At Least One Jackpot = 1 − (1 − p)^50 with p = 1/13,983,816 ≈ 3.5756×10^−6, about 0.0003576 percent, or 1 in 279,677. Miss Probability = 1 minus that, 99.999642 percent. The highlighted chance looks like a rounding error next to 100 percent because it is.

Scale it. 100 tickets × 52 weekly draws = 5,200 entries: 1 − (1−p)^5200 ≈ 5,200/13,983,816 ≈ 0.0372 percent, about 1 in 2,689. A lifetime of 10,000 tickets: about 0.0715 percent, 1 in 1,398. To push the at-least-once jackpot chance near 1 percent on 6/49 you need on the order of 0.01 × 13,983,816 ≈ 140,000 independent entries. At $2 that is $280,000 of stake for a 1 percent shot at the top prize, ignoring splits and tax.

The birthday-style formula assumes independent trials with replacement of luck — each ticket is a Bernoulli(p) against a fresh draw, or distinct lines on one draw approximated as independent when tickets ≪ C(n,k). If you buy every combination, p_session = 1 and the formula is the wrong tool (use a covering argument). If your ten tickets include duplicates, you have fewer than ten trials.

Do not confuse this with expected value. Fifty tickets at negative EV are fifty times the loss, not a clever way to “get closer.” The multi-ticket calculator answers a probability question: how often does at least one jackpot occur in a bundle of independent entries. It does not add lower-tier hits, which occur far more often (match 3 on 6/49 is about 1 in 57 per ticket). If you want those, price them with the odds calculator and a binomial, not this page.

People are terrible at compounding small probabilities. Ten tickets feels ten times luckier; five draws feels like a campaign. The product tickets × draws is the right exposure, and 1−(1−p)^n is the right at-least-once probability, and the result is still tiny for jackpot p. Diaconis and Mosteller’s “law of truly large numbers” (1989) is the other direction: in a large enough population, rare events happen to someone. Both statements are true and they do not contradict. Someone hits 6/49; it is almost never you.

This is the same math as the birthday problem, except the birthday problem has a surprisingly high collision probability because you compare every pair. Here you are not looking for a collision among tickets; you are looking for a hit against a fixed draw with success probability p ≈ 7.15×10^−8. Linear approximation n p is excellent until n is a non-trivial fraction of 1/p. For n = 50, n p ≈ 3.576×10^−6, and the exact 1−(1−p)^n differs only at the n(n−1)p^2/2 ≈ 10^−12 level.

Syndicate marketing often quotes “we hold 10,000 tickets, so our odds are 10,000 times better.” Relative odds yes: 10,000/13,983,816 versus 1/13,983,816. Absolute probability remains 0.0715 percent per draw for a 6/49 jackpot. A syndicate can still be a rational way to buy a wheel or to chase a theoretically +EV rollover; it is not a way to make a 1-in-14-million event common. Office pools that buy 20 tickets and talk as if they “have a real chance this week” are using language the formula does not support.

Independent draws: five consecutive 6/49 drawings are independent if the machines and balls are fair. Buying the same ten combinations each night does not create negative dependence that helps you; it also does not hurt jackpot probability relative to ten fresh random combinations, for a fair draw. What repeated combinations do change is the correlation of lower-tier outcomes and the boredom of the slip.

SorteCalc’s defaults (10 tickets, 5 draws, 6/49) are a realistic recreational bundle, not a recommendation. Run your actual habit: two tickets twice a week for a year is 208 entries, still 1 − (1−p)^208 ≈ 1 in 67,230. If that number feels colder than the habit, the calculator did its job. Pair with jackpot EV to price the bundle in dollars: 50 × (−$1.66) is the wrong EV for 6/49 at unknown jackpot, but the method is the same — probability first, then money.

People are terrible at compounding tiny probabilities. Ten tickets feels ten times luckier; five drawings feel like a campaign. The right exposure is the product of tickets and drawings, and the right at-least-once probability is one minus (1 minus p) to the power of that product, the complement of a total miss. Persi Diaconis and Frederick Mosteller’s law of truly large numbers is the other direction: in a large enough population, rare events happen to someone. Both statements are true. Someone hits six-from-forty-nine; it is almost never you.

The default odds are one in 13,983,816, the classic six-from-forty-nine jackpot with no bonus drum, ten tickets, five drawings. Total entries n equal 50. Single-trial p equals 1 / 13,983,816, about 7.151 times ten to the minus eight. The at-least-once probability is 1 minus (1 minus p) to the 50, about 3.576 times ten to the minus six, or 0.0003576 percent, roughly one in 279,677. Miss probability is the complement, 99.999642 percent. The highlighted chance looks like a rounding error next to 100 percent because it is.

Linear approximation is excellent while n p is much less than one. Here n p equals 50 / 13,983,816, the same 3.576 times ten to the minus six, and 1 minus exp(minus n p) agrees with n p to an absolute error near 10 to the minus 12. Reporting 1 in 279,677 from the exact complement or from 13,983,816 / 50 is the same headline. Birthday-collision intuition does not apply: you are matching a pre-specified combination, not matching other tickets. Pairwise collisions among tickets are a shared-jackpot problem, not an at-least-one-win problem.

Independence fails in named ways. Duplicate tickets on one drawing are one trial, not two. Exhaustive coverage of every combination on one drawing gives probability one if you truly hold the matrix. A bonus ball already folded into N must not be multiplied again; if you wanted six-from-forty-nine plus one-from-ten, the 1-in-N is 139,838,160, not 13,983,816. Distinct lines on one drawing are slightly negatively dependent, sampling without replacement from the combination set. For n much smaller than 13,983,816 the correction is negligible: the exact chance of at least one jackpot with 50 distinct lines on one drawing is exactly 50 / 13,983,816.

Inversion of the complement is the planning identity. For a target at-least-once probability P, n is about minus ln(1 minus P) / p. Half a chance on six-from-forty-nine needs about 0.693 / p, near 9.69 million tickets — half a matrix. A 1 percent shot needs on the order of 140,000 independent entries; at two dollars that is 280,000 dollars of stake before splits and tax. Fifty recreational entries do not live on that scale.

Assumptions: independent Bernoulli trials at p, or distinct lines with n much less than the matrix. This page does not add match-three consolation, which is about one in 57 per six-from-forty-nine ticket and is a different binomial. Expected value of fifty negative-expectation tickets is fifty times the loss, not a clever way to get closer. Recalculate with your real 1-in-N, ticket count, and draw count. The complement stays tiny for jackpot p.