Last updated: 10 September 2026
Quick Answer: Mines game odds are hypergeometric: with three mines on 25 tiles, five safe picks in a row land 49.57% of the time. Each safe pick removes a tile, so the next pick is fractionally harder than the one before. The multiplier Power.win offers comes from those falling probabilities, less the operator’s margin.
Every Mines round shows you a multiplier. What it never shows you is the multiplier a fair game would pay at that same depth. Both numbers can be worked out, and the house edge is what sits between them: one minus the offered multiplier divided by the fair one. That calculation is what this article is for — the probability behind each pick, the multiplier those probabilities imply, and how to run the check against any operator, including this one. The step-by-step Mines rules guide covers how a round is played.
What are the odds of each pick in Mines?
Your first pick against three mines is safe 88.00% of the time, and every pick after that is slightly worse, because a revealed safe tile leaves the same three mines in a smaller grid.
The grid holds 25 tiles. Three of them are mines, which leaves 22 safe, so your first pick is a 22/25 draw. Reveal a gem and that tile leaves the pool while all three mines stay in it, which makes the next draw 21/24, then 20/23. You’re sampling without replacement, not flipping the same coin over and over, and sampling without replacement is the hypergeometric distribution. That is why the figure worth knowing is the cumulative one. Clearing five tiles means five successful draws in a row, each conditioned on the one before it.
Per-pick and cumulative tables for one, three, five and ten mines are already published in the rules guide linked above, so they are not repeated here. What follows is the layer underneath them: where those figures come from, what the multiplier does with them, and how you check an operator’s margin without taking anybody’s word for it.
How is the Mines multiplier derived from those odds?
The fair multiplier is one divided by your chance of reaching that depth, whereas the multiplier the game actually pays is that number reduced by the house edge.
Take the five-pick run. The probability of drawing a combination of five tiles that avoids all three mines is simply the number of combinations of five tiles free of mines divided by the total number of five-tile combinations on the board. In combination terms, this can be expressed as C(22,5) divided by C(25,5), which results in 26,334 divided by 53,130 or 49.5652%.
Both counts are worth doing by hand once. C(25,5) counts every way five tiles can be chosen from twenty-five: 25 × 24 × 23 × 22 × 21 ÷ 120 = 53,130. C(22,5) counts the same choice restricted to the twenty-two safe tiles: 22 × 21 × 20 × 19 × 18 ÷ 120 = 26,334. Dividing one by the other asks a single question: what share of all possible five-tile selections happens to miss all three mines. Nothing in that question refers to the order you clicked in, and that is the formal reason why click order never changes anything.
Invert that fraction and you have the break-even payout: 53,130 ÷ 26,334 = 2.0175×. At 2.0175×, over the long run the game hands back exactly what it takes in and keeps nothing. Run the same inversion at every depth and you get the ladder a fair game would pay:
| Safe picks (3 mines) | Fair multiplier (1 ÷ chance of reaching that depth) |
|---|---|
| 1 | 1.1364× |
| 2 | 1.2987× |
| 3 | 1.4935× |
| 4 | 1.7293× |
| 5 | 2.0175× |
| 6 | 2.3736× |
| 7 | 2.8186× |
| 8 | 3.3824× |
| 9 | 4.1071× |
| 10 | 5.0549× |
| 11 | 6.3187× |
| 12 | 8.0420× |
Three mines on a 25-tile grid. Each figure is one divided by the exact cumulative probability, rounded to four decimals.
A fair multiplier is a frequency written backwards. The 4.1071× on the nine-pick row is another way of writing “about once every four rounds,” and every other row reads the same way. The four decimals are not decoration. At this depth a one percent margin is worth about 0.02 on the multiplier, and rounding to two decimals can move a figure by half of that on its own. Round both sides and you compare 2.02 against 2.00 instead of 2.0175 against 1.99737 — which, as the worked example below shows, misstates the margin by more than a tenth of itself. The example below shows exactly how much that costs you.
A worked example: 2.00× at five picks
Say your round offers 2.00× after five safe picks, against the fair value of 2.0175×. Wager one unit a hundred times and you win 49.5652 of those rounds on average, paying 2.00 units each:
- Returned: 49.5652 × 2.00 = 99.13 units
- Staked: 100 units
- House edge: 1 − (2.00 ÷ 2.0175) = 0.87%
Now look at that 0.87% skeptically, because Power.win publishes 99% RTP and a 1% house edge for Mines. The two figures disagree, and the reason is the rounding this section just warned about. The exact fair value at five picks is 2.017543859…, and 99% of that is 1.99737×, which a two-decimal display shows as 2.00×. The 2.00× listed in the rules guide is therefore a rounded 1.99737×, not a genuinely better offer. Divide the exact numbers and you get 1.00% on the nose. Divide the rounded ones and you get 0.87%. The missing 0.13 points is the display, not the operator — which is exactly why the ladder above runs to four decimals.
The method itself is sound, and it works against anyone: divide the multiplier you are offered by the fair multiplier for that mine count and depth, then subtract from one. Three things keep it honest. First, match both settings and not just the depth — the fair figure at five picks against three mines is 2.0175×, but at five picks against ten mines it is 17.6923×, so that same 2.00× offer would be keeping almost everything. Second, if your depth isn’t on the ladder above, rebuild the fair figure from the general form of the count you just ran by hand, C(25 − m, k) ÷ C(25, k) for m mines and k picks, then invert it. Third, run the method forwards as a sanity check: an operator advertising 99% RTP should be paying roughly 8.0420 × 0.99 = 7.9616× at twelve picks against three mines, so a materially smaller number at that depth is a wider margin than the one advertised.
How does the number of mines change your odds?
Adding mines cuts your chance of completing a long run far faster than it cuts your chance of a single safe pick.
Set ten mines and a single pick is still safe 60% of the time, which is why the grid keeps feeling playable. Five in a row at that setting happens once in about eighteen rounds. The distance between those two feelings is the whole reason mine count behaves as a variance dial rather than a difficulty dial.
Underneath that falling ladder there is a symmetry worth knowing about. Three mines with five safe picks and five mines with three safe picks land on the same number, 49.5652%, and the match is exact rather than close.
C(25 − m, k) ÷ C(25, k) explains it. Written out in full, the chance of a k-pick run against m mines is (25 − m)! × (25 − k)! ÷ [25! × (25 − m − k)!]. Swap m and k in that expression and nothing moves. So 26,334 ÷ 53,130 and 1,140 ÷ 2,300 are two different pieces of arithmetic arriving at the same 0.4956521739, and every mine count and depth pair has a mirror image somewhere else on the grid. It is a property of the distribution, not a game designer’s decision.
Is there a pattern to where the mines land?
No, and the math gives a stronger reason than “the layouts are random.” The rules guide takes the practical side of this apart at length. What the arithmetic adds is that click order was never in the calculation to begin with.
Look back at where 49.5652% came from. It was C(22,5) ÷ C(25,5) — a count of five-tile combinations, and a combination has no order inside it. Clicking the same five tiles clockwise, diagonally or at random produces the identical count and therefore the identical probability. A sequence isn’t a weak edge here; it is not the kind of thing the number responds to at all.
You control exactly two inputs: how many mines you set and how deep you go before cashing out. Those two decide every probability and every fair multiplier in this article. Nothing else in the round does.
Mines is a PWR Original, so the layout is committed to before your first click and you reproduce that commitment afterwards. That is a different assurance from the one carried by independently certified third-party games, where a lab tests the generator and you read the verdict rather than check your own round. The commitment is a SHA-256 hash of the server seed; the step-by-step method is on the provably fair page, and it works the same way on Mines as on the other Originals.
What do these odds mean for how you play?
They mean cashing out is a trade you can price, and that no single depth is the right one.
Every extra tile raises your payout by exactly the factor by which it cuts your chance of surviving, so the expected value doesn’t move. Go from five picks to six against three mines and the fair multiplier climbs from 2.0175× to 2.3736× while your chance of getting there falls from 49.57% to 42.13%. Neither depth is worth more over a long run. They differ only in how bumpy the ride between the two ends up being.
Frequently asked questions
What are the odds of winning Mines?
There’s no single number, only a number per setting. Against three mines, five safe picks in a row land 49.5652% of the time. Add mines or push the depth and that figure falls; take either one back and it climbs.
How is the Mines multiplier calculated?
It is the inverse of your chance of reaching that pick depth, reduced by the operator’s margin. Five safe picks against three mines land 49.5652% of the time, so the break-even multiplier is 2.0175×. Divide what your game actually pays by that 2.0175x and subtract the result from one, and what is left is the house edge.
Do your odds get worse with each pick in Mines?
Yes. Uncovering a safe tile takes that tile out of the pool but leaves every mine in it, so the ratio of safe tiles to remaining tiles falls with each pick you make. Against three mines the per-pick chance slides from 88.00% on the first click to 78.57% by the twelfth.
How do you work out the house edge on a Mines round?
Divide the multiplier you are offered by the fair multiplier for that mine count and depth, then subtract the result from one. The fair multiplier is one divided by the cumulative probability of reaching that depth — the ladder above for three mines, and C(25 − m, k) ÷ C(25, k), inverted, for any other setting.
How many mines should you set?
Whichever count matches the variance you’re willing to sit through, because the long-run value holds steady across settings: more mines buy a larger multiplier at a rarer hit rate. The symmetry above doubles as a sanity check — three mines at five picks and five mines at three picks both land 49.5652%, so two settings that look nothing alike can be the identical bet.
Sources (Citations)
- Hypergeometric Distribution, Wolfram MathWorld (linked above) — the distribution governing sampling without replacement, which is the model for a Mines grid
- Combination, Wolfram MathWorld (linked above) — the C(n,k) notation used to derive the five-pick probability
- NIST FIPS 180-4, Secure Hash Standard (linked above) — the SHA-256 specification behind the provably fair commitment
- Read More: RTP, House Edge, and Variance: Casino Math for Humans
Take the Lead, Gamble Responsibly
A Mines round lasts a few seconds, so a short session can push a surprising amount of turnover through. Fix three numbers before you start: your mine count, your exit depth, and what you’re prepared to spend. Then treat every figure above as a description of how often a plan finishes, not a promise about tonight. Deposit limits, wagering limits, self-exclusion and Track your Activity are all on the responsible gambling page. If the game stops being entertaining, stop playing and use the support options listed there.
By the Power.win Editorial Team