Mines probability: survival tables for 1, 3, 5 and 10 mines

The chance of surviving k picks on a 25-tile mines board with m mines is C(25−m, k) ÷ C(25, k): the number of ways to pick k tiles from the safe ones, divided by the number of ways to pick k tiles from all 25. With 3 mines, that gives 88.00% for one pick, 49.57% for five picks and 19.78% for ten.

The tables below give the exact numbers for the four most common mine counts. Every value was computed in Python and rechecked by simulation. If you want a value that isn't here, use the mines calculator.

The formula, in plain English

Think of your picks as a group of k tiles. You survive only if all k are safe.

• The number of possible groups of k tiles is C(25, k).
• The number of groups made only of safe tiles is C(25 − m, k).
• Every group is equally likely, so the chance is the ratio of the two.

C(n, r) is "n choose r". For example, C(25, 3) = 2,300 and C(22, 3) = 1,540, so with 3 mines the chance of three safe picks is 1,540 ÷ 2,300 = 66.96%.

The same answer, pick by pick

You can also multiply the chance of each pick in turn. With 3 mines and five picks:

pick 1: 22/25 = 0.8800
pick 2: 21/24 = 0.8750
pick 3: 20/23 = 0.8696
pick 4: 19/22 = 0.8636
pick 5: 18/21 = 0.8571

product = (22·21·20·19·18) / (25·24·23·22·21) = 0.4957  →  49.57%

Both methods give exactly the same result. The pick-by-pick version is easier to follow; the combination version is quicker for large k.

Survival table: 1 mine

With a single mine the numbers fall by exactly 4 percentage points per pick, because P = (25 − k) ÷ 25. Clearing all 24 safe tiles has a 4.00% chance.

Survival table: 3 mines

Five picks is roughly a coin flip. By ten picks, about four rounds in five end on a mine.

Survival table: 5 mines

Three picks with 5 mines (49.57%) is exactly as likely as five picks with 3 mines. That's not a coincidence: C(20,3)/C(25,3) and C(22,5)/C(25,5) are equal.

Survival table: 10 mines

With 10 mines, odds collapse quickly. Fifteen safe picks would mean clearing every safe tile on the board, which happens about once in 3.3 million rounds.

Worked example (fictional)

Fictional example for illustration. Alex sets 5 mines and wants to know the chance of getting four safe tiles.

1. Safe tiles: 25 − 5 = 20.
2. C(20, 4) = 4,845. C(25, 4) = 12,650.
3. 4,845 ÷ 12,650 = 0.3830 → 38.30%.

That matches the 5-mine table. Alex should expect four safe picks in a row in a bit under two rounds out of five.

Check it yourself

from math import comb
def p_survive(m, k):
    return comb(25 - m, k) / comb(25, k)

for m in (1, 3, 5, 10):
    print(m, [round(p_survive(m, k) * 100, 2) for k in (1, 2, 3, 5, 10)])
# 1 [96.0, 92.0, 88.0, 80.0, 60.0]
# 3 [88.0, 77.0, 66.96, 49.57, 19.78]
# 5 [80.0, 63.33, 49.57, 29.18, 5.65]
# 10 [60.0, 35.0, 19.78, 5.65, 0.09]

We also ran 200,000 simulated rounds of several combinations; the share of rounds that survived landed within a few tenths of a percentage point of these values each time (for example 76.91% observed vs 77.00% expected for 3 mines, 2 picks).

Patterns worth noticing

A few things stand out once the four tables sit side by side.

• Adding mines hurts more than adding picks, at first. Going from 3 to 5 mines cuts one-pick survival from 88% to 80%; going from 1 to 2 picks with 3 mines cuts it from 88% to 77%. Later on, extra picks bite harder, because the board is running out of safe tiles.
• Equal products, equal odds. Some combinations share a probability exactly: 3 mines with 5 picks and 5 mines with 3 picks are both 49.57%, and 3 mines with 10 picks equals 10 mines with 3 picks at 19.78%. The formula is symmetric in m and k.
• The drop accelerates. With 3 mines, the chance of the next pick being safe falls from 88.00% (first pick) to 85.71% (fifth pick). Small on each step, but it compounds.
• Long runs are rare. Even with 1 mine, clearing 15 safe tiles happens only 40% of the time.

Turning percentages into something you can feel

Percentages are easy to misread. Converting them to "rounds out of 100" helps:

Those are long-run averages. Any short stretch can be much better or much worse.

What the tables don't tell you

A probability is not a prediction about the next round. A 77% chance means roughly 77 rounds in 100 over a long run, with streaks of losses mixed in. It also says nothing about value: to know whether a multiplier is fairly priced, you need to compare it with 1 ÷ P. That's covered in fair vs house-edge multipliers. For the rules of the game itself, start with how the mines game works.

Quick reading checklist

• Find your mine count's table.
• Read across to the number of picks you plan.
• Turn the percentage into "rounds in 100" to make it concrete.
• Compare the offered multiplier with 1 ÷ P to see the edge.