What probability distribution do TCG raffles use?

TCG raffles and random spot breaks follow the hypergeometric distribution because each spot is drawn without replacement from a finite pool. Binomial approximations overstate your odds when the pool is small.

Is a TCG raffle mathematically fair?

A raffle is mathematically fair when the total cost of all spots equals the expected value of all prizes. The COLLEX raffle calculator surfaces this directly so you can spot raffles priced above expected value.

How many spots should I buy in a TCG break?

Buying more spots increases your probability of hitting at least one prize, but the marginal gain decreases. Use the outcome distribution to see exactly how each additional spot moves your odds.

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Raffle Odds Calculator

Real odds for TCG raffles and spot breaks — calculated from the actual math of drawing without replacement, not a guess.

1Enter spots and prizes remaining
2Add cost per spot and prize value
3Pick how many spots you'd buy
4Read your odds and verdict

Odds of landing at least 1 hit

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Expected hits: —

Buying 5 of 100 spots · 10 hits in pool

Is it worth it?

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Outcome distribution

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Probability curve by spots bought

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Value check

Top-prize multiplier
what the top prize is worth vs. one spot —

Fair multiplier for this raffle
= spots ÷ top-tier hits —

Expected value per spot
sum of (hits ÷ spots) × value across all tiers —

Net EV for your play
(EV − cost) × spots bought —

These odds assume every unsold spot has an equal chance of being yours. If the raffle assigns spots by number or lets buyers pick specific slots, the pure probability can differ. Always verify the raffle's draw rules.

How the math works

All outputs come from the hypergeometric distribution — the exact probability model for drawing items from a finite pool without replacement. Here are the formulas the calculator runs, with every variable defined.

Variables

N = total spots remaining in the raffle
K = total hits (prizes) still in the pool
n = spots you buy
k = number of hits you end up with
C(a, b) = "a choose b", the binomial coefficient

1 \cdot Odds of at least one hit P(\geq1 hit) = 1 - C(N - K, n) C(N, n) Computed in product form for numerical stability: P(0 hits) = (N-K)/N \cdot (N-K-1)/(N-1) \cdot \dots \cdot (N-K-n+1)/(N-n+1)

2 \cdot Full outcome distribution P(exactly k hits) = C(K, k) \cdot C(N - K, n - k) C(N, n) Summing these from k = 0 up to k = min(K, n) always equals 1. That's how the outcome-distribution bars are calculated.

3 \cdot Expected hits and spread μ (expected hits) = n \cdot K N σ (standard deviation) = \sqrt ( n \cdot (K/N) \cdot ((N-K)/N) \cdot ((N-n)/(N-1)) )

4 \cdot Fair-prize multiplier A raffle is mathematically fair when the expected value of a spot \geq the cost of a spot. For a single-tier raffle, the fair multiplier for the top prize is: fair multiplier = N K The "10\times the spot cost" rule you'll see online only holds when K/N = 1/10 — that is, a 10%-hit-rate raffle. For a 1-in-500 chase, the fair multiplier is 500\times . The calculator computes the real multiplier from your inputs.

5 \cdot Expected value per spot (multi-tier) EV per spot = Σ i hits i N \cdot value i Summed over every prize tier. If EV per spot \geq cost per spot, the raffle is fair on average.

Worked example · 100 spots, 10 hits, buy 10

P(0 hits) = 90/100 · 89/99 · 88/98 · 87/97 · 86/96
          · 85/95 · 84/94 · 83/93 · 82/92 · 81/91
        = 0.3305

P(≥1 hit) = 1 − 0.3305 = 0.6695  =  66.95%
μ = 10 · 10/100 = 1.00 hit (average)
σ = √(10 · 0.1 · 0.9 · 90/99) = 0.905

Naive "10 spots × 10% each = 100%" is wrong. The with-replacement approximation (1 − 0.9¹⁰) gives 65.13% — close, but off by about 1.8 points because each pick really does change what's left.

Things to keep in mind

Why the naive math is wrong

Buying 10 spots in a 10%-hit raffle is not a "100% chance" — and it's not "10%" either. Each pick changes the pool, which the hypergeometric distribution captures exactly. Simple multiplication over-counts at low buy-ins and under-counts at high ones.

Why the 10× rule is wrong too

"10× the spot cost" is only fair when the chase is a 10%-hit raffle. A 1-in-500 raffle needs a 500× prize to be fair. The calculator shows the true fair multiplier for your exact setup.

Expected value is an average

Positive EV doesn't mean you'll win. It means that on average, over many raffles, a player in this spot comes out ahead. Any one raffle can still end at zero.

Set a budget first

Probability is not a promise. 66% odds on 10 spots means one in three plays still hits nothing. Decide what you're comfortable losing before you buy in.