The formula and why it looks backwards
To find the chance of at least one hit, it is easiest to compute the chance of zero hits and subtract from one. If each pack hits with probability p, each pack misses with probability 1 minus p. Since packs are independent, n packs all missing happens with probability (1 minus p) to the power n. So the chance of at least one hit is 1 minus (1 minus p) to the power n.
The backwards route exists because at least one is a messy event to count directly, covering exactly one hit, exactly two, and so on, while zero hits is a single clean case. Probability, like this website, is often easiest to approach through its most negative outcome.
The worked example everyone gets wrong
Suppose a card's estimated rate is 1 in 100 packs, so p is 0.01. Opening 100 packs does not guarantee the card. The chance of at least one hit is 1 minus 0.99 to the power 100, which is about 0.63, so roughly 63 percent, assuming independent odds. In other words, about one in three people who open 100 packs at those odds find nothing, and they did nothing wrong, and no candle would have helped, because candles do not do that.
This 63 percent figure is not a coincidence of the numbers chosen. Whenever you open exactly the number of packs suggested by the odds, meaning n equals 1 over p, the chance of at least one hit lands near 63 percent for small p. The number of expected hits is 1, but expectation spreads itself unevenly, giving some people two hits and a third of people none.
How many packs for the confidence you actually want
Rearranging the formula tells you the pack count for a target probability. At a 1 in 100 rate, reaching a 50 percent chance of at least one hit takes about 69 packs. Reaching 90 percent takes about 230 packs. Reaching 95 percent takes about 299 packs. Certainty is never available at any pack count, which is an underrated argument for the singles market, where certainty retails at market price.
Notice how the costs pile up at the high confidence end. Going from 63 percent to 95 percent roughly triples the packs required. Randomness charges heavily for reassurance, and it does not take coupons.
Expected count versus at least one
Two different questions hide in every chase. How many hits should I expect on average, which is simply n times p, and what is the chance I get at least one, which is the formula above. At 1 in 100 odds across 100 packs, the expected count is exactly 1, while the chance of at least one is about 63 percent. Both are true at once, because the average includes lucky openers with multiple hits balancing the third of openers with zero.
For rates like a card appearing 1 in 20 packs, the same machinery applies. Twenty packs gives 1 minus 0.95 to the power 20, about 64 percent. The pattern holds all the way down the rarity ladder.
The fine print on the whole calculation
Everything above assumes the rate p is correct and packs are independent. In reality p is usually a community estimate with error bars of its own, and product structure can bend independence slightly, for instance within a single box where collation follows patterns. Treat these calculations as good approximations for planning, not physics.
And the standing disclaimers, sung as a round. The math describes averages, not your box. No blessing, curse, or candle on this site changes p by any amount, including amounts too small to measure, because the amount is zero. If a specific card is the goal, the singles market solves in one step what this entire page solves in exponents.