Mines Game Odds: Every Probability on the Board

Mines is a 5×5 grid with m hidden mines and 25 − m safe tiles. The probability of surviving k picks is a simple sequential product, because every tile you clear removes one safe tile from a shrinking pool:

P(survive k) = (25−m)/25 × (24−m)/24 × … × (25−m−k+1)/(25−k+1)

which is the same thing as C(25−m, k) / C(25, k). Every multiplier the game shows you is the reciprocal of that number, scaled down by the house edge. That is the entire game.

Working one out by hand

Three mines, five picks. Twenty-two of the 25 tiles are safe, and each successful pick removes one safe tile and one tile:

22/25 × 21/24 × 20/23 × 19/22 × 18/21
 = 0.8800 × 0.8750 × 0.8696 × 0.8636 × 0.8571
 = 57/115
 = 0.495652…  →  49.5652%

The fraction reduces exactly to 57/115, which is a useful sanity check that you have not slipped a decimal. The fair multiplier is 1 / 0.495652 = 2.0175×, and a table carrying a 1% edge shows 0.99 / 0.495652 = 1.9974×.

Notice the conditional probabilities falling as you go: 88.00%, 87.50%, 86.96%, 86.36%, 85.71%. Each pick is slightly more dangerous than the last, because the mines stay put while the pool of tiles shrinks. That is the sensation players describe as the board “tightening”, and it is real — it just does not mean anything is due.

The full odds tables

Every figure below is the product above, computed exactly and then converted to a fair multiplier and a 1% edge multiplier.

One mine (24 safe tiles)

The one-mine probabilities are unusually clean: with a single mine, P(survive k) = (25−k)/25. Clearing 24 tiles with one mine is the same 4% as picking the single safe tile with 24 mines, which is a nice structural symmetry — both amount to guessing which square is the odd one out.

Three mines (22 safe tiles)

The full clear with three mines is exactly 1 in 2,300 — which is C(25,3), the number of ways to place three mines on the board. Clearing everything means you happened to avoid one specific arrangement out of 2,300, and there is no cleverness available.

Five mines (20 safe tiles)

Three picks at five mines and five picks at three mines are the identical bet — both 49.5652%, both 1.9974× at a 1% edge. Different-looking configurations collapse onto the same probability all over this game, which is another way of seeing that the edge is a single constant applied to a payout table rather than anything to do with the layout.

Twenty-four mines

One safe tile, one pick, 1/25 = 4.00%. Fair payout 25×, and a 1% edge table shows 24.75×. This is the single best line to check on any operator’s table, because there is no ambiguity about what the correct number is.

Why cashing out later does not cost you anything

Assume the table is built as multiplier = (1 − e) / P(survive k). Then the expected return for stopping after k picks is:

EV(k) = P(survive k) × (1 − e) / P(survive k) = 1 − e

The P cancels for every k. Stopping after one tile returns 0.99 per unit staked; stopping after twenty returns 0.99 per unit staked. This is the same cancellation that makes every crash cash-out target equivalent and every dice slider position equivalent, and it is the signature of a properly constructed original.

Why tile choice and patterns do nothing

The mine layout is fixed before your first click, derived from the server seed, client seed and nonce — the same commitment mechanism behind every provably fair game. The implementation shuffles the 25 tile indices using hash-derived numbers and takes the first m as mines, which makes every one of the C(25,m) arrangements equally likely.

Two consequences:

• All 25 opening tiles are identical. There is no corner bias, no centre bias, no “mines cluster” tendency. P(tile is safe) = (25−m)/25 for every tile on the board.
• After any number of safe reveals, the remaining tiles stay exchangeable. Clearing four tiles with 3 mines tells you the mines are among the other 21, and nothing about which 21. Every unopened tile is safe with the same probability, 18/21, on the next pick.

The one thing your click order genuinely affects is nothing at all, which is why mines is a pure staking decision dressed as a game of skill.

Computing any configuration yourself

from fractions import Fraction

def survive(mines: int, picks: int) -> Fraction:
    p = Fraction(1)
    for i in range(picks):
        p *= Fraction(25 - mines - i, 25 - i)
    return p

for k in (1, 3, 5, 10):
    p = survive(3, k)
    print(k, p, float(p), "fair", float(1 / p), "1% edge", float(Fraction(99, 100) / p))

Using Fraction rather than floats matters more than it looks: the exact values reduce to small ratios like 57/115 and 91/460, which makes it obvious when a published multiplier has been rounded down rather than merely rounded.

Run it against the table your operator publishes, at three or four pick counts and two or three mine counts. If P × multiplier comes back at 0.99 everywhere, you know the edge and you know it is applied honestly. If it drifts, you have found a cost nobody advertised — and confirming the individual game results is a separate verification job that catches a different category of problem entirely.