Raffle Odds Calculator

Introduction to Raffle Odds and Ticket Coverage

A raffle is a drum holding $N$ numbered tickets. You hold $k$ of them, and the organiser reaches in and pulls $n$ winning stubs, one after another, never putting a drawn ticket back. Because each pull shrinks what is left in the drum, the chance attached to the next pull depends on the ones that came before it. That "draw without replacement from a finite pool" setup is exactly what the hypergeometric distribution was built to describe, and it is what this page computes. Enter your three numbers, press Compute Odds, and the results appear instantly — nothing leaves your browser.

The single most useful idea in raffle probability is coverage: the fraction $\frac{k}{N}$ of the drum that carries your name. Coverage is not the same thing as your chance of winning something, but it anchors it. With one prize drawn, your chance of winning is exactly your coverage. With several prizes drawn, your chance of winning at least once is a little less than coverage multiplied by the number of prizes, because the same ticket cannot win two different prizes and because the draws interact. Almost every raffle-odds mistake people make in conversation comes from treating $n\times \frac{k}{N}$ as if it were a probability. It is an expected count, and once it climbs past 1 it stops being interpretable as a chance at all.

Coverage also explains why buying more tickets has sharply diminishing returns in a big raffle and almost linear returns in a small one. Doubling from 5 tickets to 10 in a 10,000-ticket draw doubles a very small number and stays very small. Doubling from 5 to 10 in a 40-ticket draw moves you from a quarter of the drum to half of it, and the shape of the whole distribution changes with it. The table this calculator prints — the chance of exactly 0, 1, 2, … wins — is the honest way to see that shift rather than guessing at it.

How to Use the Raffle Odds Calculator

Three whole numbers drive everything. Total Tickets Sold ($N$) is the size of the whole pool once sales close, not the number printed. Your Tickets ($k$) is how many stubs carry your name; enter 0 if you are only exploring. Number of Prizes ($n$) is how many winning tickets get pulled from that one pool. The calculator returns the chance of at least one win as the headline figure, a row-by-row table of the chance of winning exactly 0, 1, 2, … prizes, the expected number of prizes, and the standard deviation of that count.

A good habit is to change one field at a time. Fix $N$ and $n$, then step $k$ up one ticket at a time and watch how much the headline moves; that difference is what one more ticket is actually buying. Then fix $k$ and raise $n$ to see how a longer prize list helps. The tool also prints the with-replacement comparison — what your odds would be if every prize were drawn from a freshly refilled drum — so you can see how far the two models diverge for your particular numbers.

Inputs are validated rather than coerced. Decimals, negatives, tickets owned above the pool size, and prize counts above the pool size are all refused with an explanation, because each of those describes a raffle that cannot exist. Zero prizes and zero tickets are legal and return a chance of exactly 0%, not a blank or an error. The Reset to defaults button restores the starting scenario, and Copy Result puts the whole readout on your clipboard.

The Raffle Odds Formula and the Hypergeometric Distribution

If $X$ counts how many of your tickets get drawn, the probability of winning exactly $x$ prizes is the number of ways to pick $x$ winners from your $k$ tickets and the remaining $n-x$ winners from everyone else's tickets, divided by all the ways to choose $n$ winners from the whole drum:

Formula: P(X = x) = (kCx × (N-k)C(n-x)) / NCn

$P(X=x)=\frac{\text{kCx}\times \text{(N-k)C(n-x)}}{\text{NCn}}$

That formula is only defined over the range $\max (0,k+n-N)\le x\le \min (k,n)$. The lower bound matters in small raffles: if you own 8 of 10 tickets and 4 prizes are drawn, you cannot possibly win fewer than 2, so $P(X=0)$ is exactly zero rather than merely small. The calculator prints the true zero in that case instead of a rounded stand-in.

The headline number most people want — the chance of winning anything at all — is quickest to get by turning the question around: instead of adding up every winning outcome, subtract the single losing outcome from 1.

Formula: P(X ≥ 1) = 1 - P(X = 0) = 1 - (N-k)Cn / NCn

$P(X\ge 1)=1-P(X=0)=1-\frac{\text{(N-k)Cn}}{\text{NCn}}$

Here $\text{aCb}$ is "$a$ choose $b$," the count of ways to pick $b$ items from $a$ without caring about order. Those counts explode fast — a 2,000-ticket raffle produces integers with hundreds of digits — so for ordinary raffle sizes the calculator does the combinatorics in exact BigInt integer arithmetic and only rounds to a percentage at the very end. For very large pools it switches to a numerically stable logarithmic form, $P(X=0)=\prod _{i=0}^{n-1}\frac{N-k-i}{N-i}$, evaluated with log1p and expm1 so that tiny probabilities keep their significant digits.

The expected number of prizes has a much simpler shape. It is your share of the drum multiplied by the number of draws, and it does not depend on whether tickets are replaced:

Formula: E[X] = n k / N

$E\left[X\right]=n\frac{k}{N}$

The spread around that average does depend on replacement. Sampling without replacement is less variable than the matching binomial, by the finite-population correction factor shown below, where $p=\frac{k}{N}$:

Formula: Var[X] = n p(1 − p) (N − n) / (N − 1)

$\mathrm{Var}\left[X\right]=np(1-p)\frac{N-n}{N-1}$

An expected value of 0.48 does not mean half a trophy. It means that if you ran this exact drawing thousands of times, your wins would average out to about 0.48 per drawing, and the standard deviation tells you how widely individual drawings scatter around that average.

Worked Example: 2,000 Tickets, 3 Prizes and 25 Stubs

A school fundraiser sells 2,000 tickets and will draw 3 prizes; you bought 25. Your coverage is 25 / 2,000 = 1.25%. The chance of winning nothing is $\frac{\text{1975C3}}{\text{2000C3}}$ = 96.2948%, so the chance of at least one win is 3.7052%, or roughly 1 in 27. Broken out by outcome, the chance of exactly one prize is 3.6605%, exactly two is 0.0445%, and all three is 0.00017%. Expected prizes are 3 × 25 / 2,000 = 0.0375. The naive shortcut $3\times 1.25\%=3.75\%$ overshoots the true 3.7052% because it double-counts the outcomes where you win twice.

Now scale everything down so the arithmetic is checkable by hand: $N$ = 50 tickets, $k$ = 6 of them yours, $n$ = 4 prizes drawn. Then $\text{44C4}=135751$ and $\text{50C4}=230300$, so $P(X=0)$ = 135751 / 230300 = 58.9453% and the chance of at least one win is 41.0547%. The full row set is 58.9453%, 34.5046%, 6.1615%, 0.3821% and 0.0065% for zero through four wins, and those five numbers add to exactly 1. Expected prizes are 4 × 6 / 50 = 0.48; buy two more stubs and it climbs to 4 × 8 / 50 = 0.64, a concrete way to price what those extra tickets bought.

Comparison Table: Sampling With and Without Replacement

The table below fixes a 1% share — 10 tickets out of 1,000 — and varies only the number of prizes drawn. The middle column is the with-replacement answer $1-{(1-p)}^n$, which applies when every prize is drawn from a refilled drum. The right column is the without-replacement answer this calculator reports. Notice that the without-replacement figure is always the larger of the two, and that both fall further and further behind the naive product as the prize list grows.

The gap in the last row is worth sitting with. Fifty prizes drawn from 1,000 tickets is half a percent of the drum removed per draw, and the naive product claims a coin flip while the true answer is about 40%. The with-replacement model gets much closer but still understates the real chance by 0.77 percentage points, because it never lets a losing ticket leave the drum.

Limitations and Assumptions Behind These Raffle Odds

Everything here rests on one clean picture, and the assumptions are worth stating plainly. Every ticket is equally likely to be drawn; a drawn ticket never returns to the drum; owning several tickets means you can win several prizes, one per stub; and the totals you enter are the final totals after sales close. Displayed percentages are rounded, so a genuinely tiny chance is shown as "<0.01%" rather than a misleading 0.00%, while a true zero is shown as "0%". The tool weighs probability only, never whether the ticket price is a fair deal; most fundraising raffles are deliberately designed to return less than you pay, because the point is the cause.

Real raffles bend those assumptions in a few predictable ways, and it is worth matching the fine print before trusting a number. If the event caps each person at one prize no matter how many tickets they hold, your chance of two or more wins collapses to zero and the redraw rule changes the arithmetic, because the pool of still-eligible tickets shrinks with every winner. If prizes are handed out in separate drawings by tier — a grand prize, then several runners-up — and tickets are returned between rounds, each round behaves like its own fresh raffle, and the with-replacement column of the table above is the right model rather than this calculator's default. If organisers set aside house or sponsor tickets, guarantee a winner from a subgroup, or run a 50/50 draw where the prize depends on sales, some stubs no longer share the same odds; shrink the effective pool for that prize or treat the output as a ballpark.

Two further limitations apply to the arithmetic rather than the raffle. Extremely large pools are evaluated in floating point rather than exact integers, which keeps roughly fifteen significant digits — far more than any raffle needs, but not literally exact. And the model says nothing about fairness of the draw itself: if the drum is not properly mixed, or tickets differ in size or stiffness, the equal-likelihood assumption fails and no formula can rescue it.

Sources Used to Check the Raffle Odds Formula

The probability mass function, its support, and the mean used on this page were checked against the following references. The multiple-prize case is the hypergeometric distribution; the refilled-drum case reduces to the binomial complement.

• Encyclopedia of Mathematics (EMS Press / Springer), Hypergeometric distribution — gives the mass function as choose(M, m) · choose(N − M, n − m) / choose(N, n) over max(0, M + n − N) ≤ m ≤ min(n, M), the mean np with p = M/N, and the variance npq(N − n)/(N − 1).
• Wolfram MathWorld, Hypergeometric Distribution — states the same mass function in the urn form and the mean μ = Np, confirming the expected-prizes formula used here.
• NIST/SEMATECH e-Handbook of Statistical Methods, Binomial Distribution — the with-replacement comparison column uses the binomial complement 1 − (1 − p)ⁿ and the mean np given there.

Questions People Ask About Raffle Odds

Privacy

Every calculation runs locally in your browser. Your inputs are never sent to a server, and no ticket numbers, totals or results are stored anywhere.