Sizu You are at the till and the scratch cards say “1 in 5 wins a prize”. You buy five. Surely one of them has to win? It feels that way, because 5 cards at 1 in 5 each seems to add up to a whole. It does not. The real chance of at least one win is 67.2%, so about one time in three you walk away with five losers.
(The 1-in-5 figure is invented for this example. Real cards print their own odds, and the method below works with any of them. We also assume each card is an independent 1-in-5 chance.)
Step 1: stop asking “at least one”
“At least one winner” is awkward, because it includes exactly one, exactly two, and so on up to five. There is a much shorter way to count it. The only way you do not get at least one winner is if every card loses. So find that, and subtract it from 100%.
Step 2: the chance one card loses
A card wins 1 time in 5, so it loses 4 times in 5. That is 4/5, or 0.8, or 80%.
Step 3: multiply across the cards
Each card is its own draw, so the chances that every card loses multiply. One card: 0.8. Two cards: 0.8 × 0.8 = 0.64. And so on:
Five cards: 0.8 × 0.8 × 0.8 × 0.8 × 0.8 = 0.328, which is 32.8%.
Step 4: flip it back
At least one win = 100% − 32.8% = 67.2%.
To check the result a second way, look at the other card counts. One card gives 20%. Two give 100% − 64% = 36%. Three give 48.8%. Four give 59.0%, and only then does “at least one win” tip past an even chance. Percentages here are rounded to one decimal place.
Why adding the chances goes wrong
Adding 20% five times gives 100%. The sum counts the lucky cases more than once: a run where cards 2 and 4 both win gets counted for card 2 and again for card 4. The more cards you add, the bigger that double counting gets.
Ten cards would “add up” to 200%, which cannot be a chance at all. The real figure is 1 − 0.8^10 = 89.3%. The same slip catches people with dice: six rolls “add up” to 6/6, yet the real chance of at least one six is 1 − (5/6)^6 = 66.5%. Even four rolls give only 51.8%, not 4/6, which is why a four-roll bet on a six is only just better than a coin toss.
How many cards for an even chance?
You can run the same sum in reverse. Each extra card multiplies the “all lose” chance by 0.8. After three cards it is 51.2%, still above half, so the chance of a win is 48.8%, just short of even. After four cards it drops to 41.0%, so a win is more likely than not (59.0%). To reach a 90% chance of at least one win you need 11 cards, because 10 cards give 89.3% and 11 give 91.4%. That is a lot of cards to buy, and it shows how slowly the odds creep up.
Two tickets, two shops
Say you can buy five cards at one shop, or three at one shop and two at another. It makes no difference to the sum, because every card is its own 1-in-5 draw. All that matters is the count of cards and the chance on each. This is also why comparing offers is easy: a “3 cards for the price of 2” deal changes how many cards you get per dollar, but it does not change the odds on any one card. Work out the chance of at least one win for each count, and set that beside the cost.
For example, with these 1-in-5 cards, going from 4 cards to 5 raises your chance from 59.0% to 67.2%, a gain of 8.2 points. Going from 9 to 10 raises it only from 86.6% to 89.3%, a gain of 2.7 points. Each extra card buys less than the last one did.
A note on “due” wins
Each card is a fresh 1-in-5 chance. If your first four cards all lose, the fifth is still 1 in 5, not “due”. The 67.2% figure is the chance before you start scratching. Once four have lost, you are only asking about the one left.
What this means when you buy
- More cards do raise the chance of at least one win, but with shrinking returns for each extra card.
- A win is never guaranteed, however many cards you buy, so only spend what you are happy to lose.
- The same idea covers any repeated risk: a 5% chance a delivery is late, ten times, makes at least one late delivery likely, about 40% (1 − 0.95^10 = 40.1%).
Try it with your own numbers
The free At Least One calculator takes the chance on one try and the number of tries. It shows the chance of at least one success, the chance of none, and how many tries you need for an even or a 90% chance. It comes from Chance Didn’t Have to Be This Hard, which covers “and”, “or” and “at least one”, lotteries and the long run.
Rule of thumb: for “at least once” in repeated tries, work out the chance it never happens, multiply it across the tries, and subtract from 100%. Never add the chances.
This is arithmetic for understanding odds, not gambling advice. Cards and games differ, so read the odds printed on the one you buy.