Calculator · Mines

Mines odds calculator

Exact probability that every tile you pick is safe — and what a fair payout would be. Any board size, any number of mines. No sign-up, no real money.

✓ Math checked · Updated · How we calculate
Many Mines games use about 1%. Use 0 for a perfectly fair game.
Chance all safe
49.57%
1 in 2.02
Fair multiplier
2.0175×
1 ÷ chance
Pays with edge
1.9974×
1% house edge
Next tile safe
85.00%
after these picks
Chance and multiplier after each safe pick
Safe picksChanceFair ×Pays ×
188.00%1.13641.1250
277.00%1.29871.2857
366.96%1.49351.4786
457.83%1.72931.7120
549.57%2.01751.9974
642.13%2.37362.3498
735.48%2.81862.7904
829.57%3.38243.3485
924.35%4.10714.0661
1019.78%5.05495.0044
1115.83%6.31876.2555
1212.43%8.04207.9615

Exact rational arithmetic (no floating-point rounding) · methodology

How Mines odds work

In Mines, a fixed number of mines is hidden at random on a grid. You reveal tiles one at a time; every safe tile raises the multiplier, and one mine ends the round. Because the mines do not move and are placed uniformly, the chance of surviving k picks depends only on three numbers: the board size, the number of mines and how many tiles you reveal.

On the default 5×5 board with 3 mines, the first pick is safe 88.00% of the time (22 of 25 tiles). After that, one safe tile and one tile are gone, so the second pick is safe 21 of 24 times — and so on. Multiply the fractions and you get the chance that all five picks are safe:

P(5 safe) = 22/25 × 21/24 × 20/23 × 19/22 × 18/21 = 49.57%

The general formula is the hypergeometric probability C(safe, k) ÷ C(tiles, k), where C is "n choose k". Our calculator evaluates it with exact fractions, so the result is not affected by floating-point rounding.

Fair multiplier vs. what the game pays

A fair bet pays exactly the inverse of its probability: if something happens 49.57% of the time, a fair payout is 2.0175× your stake (including the stake). Real games subtract a house edge. With a 1% edge the game pays 1.9974× — and that 1% is taken from every multiplier on the board, no matter how many tiles you reveal.

Push to 10 safe picks with 3 mines and your chance drops to 19.78%. The multiplier grows, but in the no-rounding model the expected return per dollar staked is still 99.00% at this 1% house edge — the same share the edge leaves at any pick count, including stopping after one tile.

Extreme settings

  • 1 mine, 1 pick: 96.00% chance, pays 1.0313×. Almost always safe, almost nothing won.
  • 5 mines, 5 picks: 29.18% chance (1 in 3.43), fair 3.4269×.
  • 24 mines, 1 pick: 4.00% chance (1 in 25), fair 25.0000×.

Different settings trade a small, frequent win for a large, rare one. None of them changes the long-run return, which is fixed by the house edge.

Why Mines "strategies" don't work

Patterns, corners, "cold" tiles and predictor apps all rely on the idea that some tiles are safer than others. With uniform hidden placement they are not: every unrevealed tile has the same probability of hiding a mine. A provably fair game fixes the mine positions from a server seed before you click, and in our reference scheme you can check the placement afterwards with our provably fair verifier.

FAQ

What are the odds of revealing 5 safe tiles with 3 mines?

On a 5×5 board with 3 mines, the chance that 5 picks in a row are all safe is 49.57%. A fair game would pay 2.0175× your stake; with a 1% house edge the payout is 1.9974×.

Is there a Mines strategy or pattern that works?

No. If mines are placed uniformly at random and hidden, every unrevealed tile has the same chance of being a mine. Corners, diagonals, patterns or "hot" tiles do not change the odds. Given the board size, only the number of mines and the number of picks matter.

Why does the multiplier grow so fast?

Each extra safe pick multiplies the risk: the chance of surviving k picks is a product of shrinking fractions. The fair multiplier is simply 1 divided by that chance, so it grows as the chance falls. The house edge then shaves a fixed percentage off every multiplier.

Does the order in which I pick tiles matter?

No. The probability that a set of k tiles contains no mines is the same whichever order you reveal them in: C(safe, k) ÷ C(tiles, k).

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