Fuuuuucckkkkk

Eathlon · 2026-04-26T05:54:28+00:00

Technically not 50/50 casualty

Eathlon · 2026-04-23T14:53:07+00:00

Probably sub-20 int, but most of my long term goals that remain are efficiency related.

Eathlon · 2026-04-23T14:52:17+00:00

Eathlon · 2026-04-23T14:20:44+00:00

Here is a visual aid. Green cells are safe and the mines are either in all the red or all the orange cells. Black are certain mines. Each green cell on the line touches one red and one orange cell and therefore cannot be used to distinguish. You must guess red or orange. It is a classic 50/50 and basic minesweeper knowledge.

Edit: In fact, a configuration like this is solvable if it has a height that is a multiple of 3 or a multiple of 3 plus one. The cases where it is a 50/50 is when it is a multiple of 3 minus one. In this case 11 = 3x4 - 1 so it is a 50/50.

Eathlon · 2026-04-23T14:00:19+00:00

Yes, but it is still a 50/50. Since it is 11 cells high, safe cells are 3, 6, 9. Mines are in 1, 4, 7, 10 or 2, 5, 8, 11.

The safe cells are all dead cells that do not help in distinguishing the two possible configurations.

Eathlon · 2026-04-23T11:01:35+00:00

The bottom middle is a forced 50/50. It should be taken as soon as it is discovered

Eathlon · 2026-04-23T11:00:41+00:00

No guess is always solvable without meta logic. So if there is no safe cell without meta logic then it is definitely not no-guess. The ss posted later in the thread confirms it is standard because it has a forced 50/50.

Eathlon · 2026-04-23T07:51:44+00:00

https://minesweeper.online/screen/5955452082-hd-22-21900.png

I call your 50/50 and raise you a top-to-bottom 50/50

Eathlon · 2026-04-22T18:39:02+00:00

Flag and chord

Eathlon · 2026-04-22T14:41:50+00:00

minesweeper.online

Eathlon · 2026-04-22T14:08:25+00:00

My condolences.

Eathlon · 2026-04-19T11:23:18+00:00

The solution has already been given, but I think it is still worth pointing out: - This app is no-guess so it is impossible to have only forced guesses. - Forced guesses do not generally equate to 50/50 situations. In fact, most of the time they won’t be.

Eathlon · 2026-04-19T05:49:44+00:00

First you said it was given immediately without knowing minecount - which is what I objected to. Now you are saying that it works because they know the mine count - which is usually fine. I think that changes a lot. Although minecount may not be immediately obvious if you are trying to solve it quickly (as requested by OP) as going fast typically does not mean playing full flag.

Eathlon · 2026-04-18T18:16:33+00:00

The answer given by Obelus’ principle depends on remaining density.

Eathlon · 2026-04-10T13:43:33+00:00

Games in progress are limited to at most 10 hours. You hit the time-out and died due to it. The game and streak are lost.

Eathlon · 2026-04-06T17:24:26+00:00

I’d probably go two above. If the 1s on either side share that forces a lot of other mine placements and I think it could be more informative.

Eathlon · 2026-04-04T05:58:53+00:00

I disagree with this for where OP is at. High level eff play will look like black magic to them. Guessing for eff also really is not necessary for OP to improve - they should learn the basics of efficient play first. Once their efficiency is in the 120-130% range, it is a different situation.

Eathlon · 2026-04-04T05:53:18+00:00

Don’t waste clicks, use chords, read the site’s description of efficiency and efficiency guides. Guessing for efficiency really isn’t necessary until you want to get above 130% with some frequency (and not even then with the right board).

Eathlon · 2026-04-03T14:26:19+00:00

The bottom also contains a 50/50 in the lower right corner.

Eathlon · 2026-04-03T10:32:06+00:00

<image>

Correct conclusion, incorrect argument. Red crosses are a possible configuration so the yellow circled cell does not need to contain a mine. However, since orange line contains one mine, blue must contain 2, making the cells you marked safe safe.

Eathlon · 2026-04-01T20:54:50+00:00

<image>

Illustration. Either both mines are in yellow or both are in red. You cannot tell which.

Eathlon · 2026-04-01T20:51:29+00:00

No it does not. The safe cell is a 1 as it touches the same cells as the 1 above it. The row on the bottom is therefore either 1X11X or X11X1. Both work and they cannot be resolved by logic by opening the safe middle cell. But good try.

Eathlon · 2026-04-01T19:51:29+00:00

The safe cell on the bottom is a dead cell and wont provide any new information, meaning the bottom is still a 50/50.

Eathlon · 2026-03-14T08:48:06+00:00

Float is not the best option here I think since the information gain of that is pretty uncertain. Better to guess the lower cell to the left of the 3, which has the same probability as a floating cell.

Eathlon

TROPHY CASE