See the chance that n tickets at least cover their cost
When you buy several scratch tickets, "how many must win to break even" and "how likely that is" are two separate calculations. Enter four values and the tool reports both.
The formulas used
The number of winning tickets needed is the total spent divided by the prize, rounded up: k0 = ceil(n x c / w). The chance of breaking even or better is the binomial tail sum P = sum over j >= k0 of C(n, j) pj (1-p)n-j. Binomial coefficients overflow quickly, so every term is computed in log space and then exponentiated before summing. Expected winning tickets is simply n x p.
Assumptions and what is out of scope
Each ticket is assumed independent with the same win probability. Real scratch games print a fixed number of tickets, so draws from one pack are not fully independent. The model also assumes a single prize tier, so it does not fit real games whose prizes are split across several tiers. Per-ticket expected value and expected profit or loss are not covered. Even a high break-even chance does not help when the prize multiplied by the win probability is below the ticket price: buying more tickets then raises expected loss. Reference material as of October 2026, and not an encouragement to buy.
Frequently asked questions
No. This calculation assumes a single prize amount, so it differs from the break-even odds of a real scratch game split across first, second and third tiers. Use it only as an approximation for one common small-prize tier.
No. The break-even chance only describes the possibility of clearing the cost line; it says nothing about how much you lose on average. When the prize multiplied by the win probability is below the ticket price, buying more tickets increases expected loss.
Only the N of 1-in-N odds is accepted. For 20% enter 5, and for 1% enter 100. Values of N at or below 1 are blocked.