"Yes, but what about 27" - Oh, hello. Glad you asked. Don't stand up now. Under your chair there is a bomb. It will blow up if you stand up, or if the square you uncover in the next game you play is a bomb. The bomb under your chair will self-disarm if you successfully mark all the bombs on the board. Welcome to 4D minesweeper, where the highest number you can uncover is 80. Oh, also, if you ask about 81, the bomb under your chair will blow up.
That's childsplay: In 5D minesweeper with multiverses and time travel, you have to worry about uncovered mines in the past blowing you up. And don't even mention grandfather mines.
@@incription I used to be a big fan of 5D minesweeper with Multiverse Time travel, it's too bad things like the terminator gambit, Jurassic moves, and the exile strategy have made the game stale and unfun
I assume the part where it gets really bad is where you have two 3*3 groups that have a ring of mines on the outside, but you know based on the number of remaining mines that one has a mine in the middle, and the other doesn’t, basically making you make a blind 50/50 guess.
@@pokepress expert games are highly likely to end in a true 50:50 guess. more than 50% but probably not much more. I think the linear algebra solving technique can reach 45% solve success but I'm not sure if that includes successful 50:50 guesses. my own solver will halt and start a new game if it detects any 50:50 guess, but it only looks for those if there are no detectable safe squares. but my solver doesn't detect 7s exactly (because 7s are the same color as bombs if i only detect 3 bombs or less on a board I treat them all as 7s) and won't detect 8s at all because they aren't solve probative.
For the 8 in the shape of an 8, it made some sense to me. If you have one 8 already, you're using 8 of your mines in that 8, and these 8 mines are arranged in four edges of three consecutive mines. If you are to have another 8 on the board, starting with one of those edges gets you 3/8 of the way there, and requires you to use up three fewer of the total mine allotment. So it should be much more likely than random chance that, knowing you have an 8 on the board already, and knowing you have a second 8 somewhere, the second 8 would share an edge with the first
Yes, two 8s involve the positions of 16 mines, while a figure 8 only involves 13 mines, but with additional constraints on position, so it mrke sense that those two effects roughly cancel each other out
You (and other people saying they were not surprised) seem to have a well tuned probability intuition. Now, when I know the answer - yeah, I guess it makes sense. Bit when I saw the results for the first time, I was really suspicious there was a bug somewhere. Was totally in disagreement with my intuition, and even nwo it still rubs against it a little. For me it stands right next to the birthday paradox and Monty Hall paradox. (which fooled my intuition too the first time)
Yep. Despite the fact that there are a lot more ways to arrange two 8s on an intermediate board than one double 8, the 24 free mines on 238 cells of the two individual 8s gives ~5.3*10^32 remaining mine patterns per two 8s pattern, while the 27 free mines on 241 cells of the double 8 gives ~ 4.2*10^35 random mine patterns per double 8 pattern.
Back in highschool I used to always play a 83x38 custom board with 750 mines to pass the day instead of doing classwork. This was basically fullscreen minesweeper for my laptop and had a density of 23.78%, so just a little bit harder than expert. It took me a whole year to finally win a game lol. It took about 4 hours to get to the end game, but with a board and density of that size, the probability of getting way 4-6 pure 50/50 choices was basically guaranteed. It was a coinflip until you win at that point. Keep in mind you need to first get to that stage, which could have misclicks or even just a small but quick lapse in thinking could end your entire game.
@@MyWaifuNow it's technically a bit slower, but I'm quite partial to playing minesweeper with keyboard only (especially better than a laptop trackpad)
The 8-8 thing makes sense to me. To get two full 8's, you need 16 mines to be placed exactly right, but for one single 8-8, you only need 13 mines to be placed in the right pattern. Or, thinking about it more abstractly, once an 8 has already been placed, there are only 5 mines that need to be placed after that to make that 8 an 8-8.
It’s more than that. Reducing 8 to 5 definitely improves its chance, but it’s really down to the limited number of available mines that makes it so likely to occur. The fewer mines available, the more likely the two 8s will have to share them. If you have few enough mines (e.g. 14) regardless of the board size, the chance becomes 100% because there’s simply no other way that two 8s can physically form on the board.
@@doodeedah6409with 14, you could still have them off by one, so they only share 2 mines. Also, does being aligned horizontally instead of vertically still count as being in an 8 shape?
@@Jivvi Oh whoops, I miscounted, I meant 13 mines, i.e. 3 mines forced to be shared. Btw the fine print at the bottom of the video says he counts both vertical and horizontal shapes.
I've actually had an 8 before. I was looking for it by setting the difficulty settings to max, with there being like 100 mines. Took like 20 straight minutes. Was so hyped when I found it.
Actually, 27 IS possible if you're willing to up your Minesweeper game to four dimensions. In 4D Minesweeper it's actually possible to get numbers as high as 80. But what about 81?
Actually, 81 IS possible if you're willing to up your Minesweeper game to five dimensions. In 5D Minesweeper it's actually possible to get numbers as high as 242. But what about 243?
Minesweeper in VR is an absolute trip to go through, yes it's just 3d minesweeper but somehow it's immersive. I don't really say that but it's one of my favorite free VR games to use a way to pass time without having to move around to much. Kind of like putting a picture puzzle together, it's really interesting how it unfolds. I'd recommend checking it out if you're a fan of minesweeper itself.
The two 8s in an 8 configuration odds are actually fairly intuitive. Given an 8 on the board, there are two places (as long as it's not too close to an edge - and assuming you want them vertically orthogonally connected) where there are already 3 mines in a row. Requiring only 5 mines to be in specific places, to accomplish the goal of two 8s. Where a disconnected 8 on the same board as your 8 requires 8 specific mine placements. The reason it's only 1/3 as likely as finding a 2nd 8 floating in some unknown far off location is because there are significantly more than 2 places those 8 specific mines could be arranged than 2. But that only just over compensates for the 8th power vs. 5th power comparison.
@@justingolden21 no, not given, but the chance is not as small as expected because you only need 5 mines to be in a specific location instead of a whole new group of 8 mines
I liked that little tidbit about javelins resetting their records after some redesigning. I'm sure there are similar examples of other sports/games doing so, but it is quite refreshing to find these trivia in your vid!
Yeah, the funny thing though is that I have seen that in a video I watched just a few days ago, and now I wonder if that's why I got this video in my recommendation lol
For those interested. The javelin was redesigned since it was easier to make the javelin heavier than make every stadium larger. Therefore it became impossible to throw as far as before and thus all the records had to be reset
I got really good at this at one point, and I used to play a larger board. I think 24x30 with ~165 mines. I remember the cutoff was pretty sharp, where most games under that number of mines were easily solvable with good initial guesses. More than that and I had to start making too many good guesses to win. I never ran probabilities though.
In Super Mario 64 speedruns whenever something occurs that would make two sets of records (such as 120 vs 70 star, then 16 star, then 1 star, then 0 star) they just split the leaderboards. Super Mario World also has two common Any%s: regular gameplay vs using the Arbitrary Code Execution glitch to jump to the credits from level 1. I think minesweeper could do the same.. have a 8x8 leaderboard and a 9x9 leaderboard.
Speedrunning beginner is sort of pointless imo. As the creator pointed out, you can beat beginner in a single click with 0 on the clock. These zeroes are simply not counted on the leaderboard, because otherwise the top of the leaderboard would just be full of them. So instead, you have to click at least twice. That means people who want to grind a good beginner time just keep refreshing the board and randomly clicking twice. Each time randomly clicking twice doesn't win, they refresh. They try to click as fast as possible, maybe using a mouse that double clicks if you push the button right (usually an undesirable behavior). For instance, the minesweeper online #1 time is 0.02 seconds, apparently accomplished by dragging the mouse along the top of the board. Splitting the 8x8 and 9x9 times wouldn't change that. Beginner is just silly.
Interesting point. I actually agree about 2-click being mostly luck. What they did about it is they introduced Beginner 3BV category that limit minimal 3BV at 10. It helps somewhat, still they have a few disqualified attempts because "there was one lucky click". Which is now a bit weird. So what, lucky click? If at the shooting competition I shoot without looking and hit bull's eye, I still get the prize, right?
@@GamesComputersPlay Yeah, it also must be incredibly frustrating to try to grind. Every top time must be "lucky" to some extent, so it feels like if you get _too_ lucky, then it just doesn't count anymore. Imagine if whenever a golfer got two holes in one in a single round, that round got disqualified.
@@GamesComputersPlay What is the chance to beat a beginner board with a 3BV of at least 10 with only random clicks? If it's less than a million, then there's a real chance to piss off some pros. :D
When I was a kid I used to change the settings of the game to use the biggest board with the fewest mines so I could basically “farm” one click wins. Sure it was cheating, but I barely even considered minesweeper a game and the most fun I could have with it was this way lol
I think the reason the 8 of 8s is so likely is because it shares 3 mines between the two 8s. The configuration requires only 13 mines and the remaining mines can be anywhere else, while having two separate 8s would require a whole 16 mines split into two groups. I think it is misleading to compare the chance of getting an 8 of 8s to the chance of getting two separate 8s. Instead, the 8 of 8s should be viewed as its own kind of shape, and the chance of it appearing can be compared to the chance of two regular 8s that are not joined. It would be interesting to calculate what the chance of finding two 8s that share any number of bomb tiles is compared to finding two disconnected 8s. I would bet that it'd be more likely than not that if there are two 8-tiles on the grid, that they share at least one bomb.
ok, I ran a bigger simulations (220M), here's what I have: - just 2 eights: 122, 1/1.8M - 8s that touch each other at least in one spot: 88, 1/2.5M - 8 in a shape of 8: 71, 1/3.1M So: - Looks like my estimation of just having 2 eight maybe was a bit off in the video, it's looks to be more rare. However, it makes the "8 in the shape of 8" paradox even more prominent: difference between those are even smaller than 2 - You prediction that touching 8s are more common than non-touchin is quite true: almost 3/4 are touching each other. - estimation of 8 in the shape of eight still stands 3.1M and 3.2M - close enough.
@@GamesComputersPlay I bet the probability of two 8s that are immediately adjacent but don't use one set of mines in common is more like 100 or 1000 times less likely than any 2 eights. That would still use 16 mines
I’m so glad to see this vid come out! I played a game of minesweeper in 9th grade where the very first click was an 8. I felt like I won the most useless lottery ever lol
In University I worked on a machine learning algorithm to play Minesweeper (with... mixed results, I was a rookie), and 1-click games became a serious problem. Turns out my board wasn't dense enough and they were more common than they should have been, it fed the algorithm a ton of bad habits because that click ended up with a massive reward score. That one eluded me for a while...
The reason double 8 is not that much more unlikely is because it requires less mines in the formation than two 8s by themselves, allowing more leeway for the rest of the mines.
12:44 think of it this way: if you have 2 eights in a shape of an 8, that only forces 13 mines but having 2 eights on somewhere far away forces 16 mines. That makes the 2 eights generating in a shape of an 8 a little bit more likely.
Hmm. I always assumed that Minesweeper solved the "first click problem" by choosing an appropriate board after the first click. It would keep randomly producing boards until a compatible one revealed itself.
To be honest, I couldn't find any reliable info about latest minesweeper clones and how they handle it. But for old version - yes, they just move a mine. Also, I think they do create new board in "no guessing" mode in some implementations. (As just moving mine can create a guessing situation).
I have wondered about this "What if the first click is a mine"-issue and how they solve it, when you're suppose to be safe that one time. There 's actually a Game Boy versjon of "Minecraft", which «solves» the whole problem by letting you actually get a bomb on your first click... ua-cam.com/video/H-qMyyYB5kM/v-deo.html
The best way would be to shift the whole board down and to the right to the first safe spot, wrapping the bottom rows to the top and the rightmost columns to the left.
@@GamesComputersPlay almost every "no guess mode" version tells you which first click to make. they generate these boards by actually solving them and when a situation is reached that would be unsolvable it moves mines then to currently uncleared spaces that aren't associated with a cascade and continues solving. if it can't find a place in uncleared blocks, it puts it in any space that doesn't boarder any other mine. if a move was moved in this way it begins solving again from the beginning. seeds for these states with the starting square are pre generated and stored, often by brute force. my approach is slightly different. I run a solver that doesn't guess on my home computer and note seeds and starting clicks for any win. those are innately solvable too.
The reason the probability of the two eights being close to each other is so high is because it's easier for 5 mines with one space in the middle to spawn close to the 8 than it is for an entirely different ring of 8s to spawn. In essence, they are simply both 5s sharing 3 mines, not two completely separate 8s. As to them being on top of each other, not necessarily. The same logic applies sideways, where they could form the shape of an infinity symbol. I hope this explanation makes sense.
I mean, it makes sense. Having 2 8s anywhere on a board requires the correct permutation of 16 mines, where as when they’re bordering, you only need 13 correctly permuted mines
12:22 I immediately guessed that the conditional probability was going to be very likely, considering that it would only need an additional 5 mines for the second 8, instead of 8 for fully-separate 8s. You could also have 8s that share one or two mines, and I would expect either of those probabilities to be greater than fully-separate 8s too.
Thanks, man! There actually just some excel formulas, like this one (this is probability of 8 for expert): =(30-2)*(16-2)*(COMBIN(30*16-9,99-8))/COMBIN(30*16,99) Except I put board size, mines, and size and mines of the object in separate cells, to be able to re-use it for different things. It breaks for bigger numbers, but should work for everything in the video: 8, 9, 77, 8in8.
The correct way to calculate it would be a multi-step process. First step: Calculate the total number of different mine patterns, this is [number of squares] choose [number of mines], so 64 choose 10 or 151473214816 for the old beginner board. Next, calculate how many different ways your chosen pattern fits on the board. For the 8, that is 36 different ways - the 8's center square can lie on any of the 36 non-edge squares of the 8x8 board. Then, calculate how many different ways the non-pattern mines can be arranged for each of those ways - in the case of the 8, that is 55 choose 2 - two mines remaining, and we have used up 9 of the board squares for the 8, giving a result of 1485. Multiply the latter two numbers together, and divide by the first, to get in this case 36*1485/151473214816=3.53e-7 or about 1 in 2.83 million odds. This also shows why the 9/filled 8 is so much less likely in the beginner board: It only has one additional mine left, so there are in total only 1980 patterns that give you it, vs 53460 patterns that yield the normal 8. This also means that on the new 9x9 beginner board, the chance for an 8 is 49*2556/1878392407320=1 in ~15 million - yes, there are more than ten times as many total different mine patterns on the 9x9 board as there are on the 8x8 board!
It's funny when thinking about 4D minesweeper, it occurred to me that if you play two 3D games at the same time, alternating turns on each board, you can conceptualize that they're the same board taken at different times where the mines have moved. You can even play some Schrodinger's cat with the boards by only placing mines once you've progressed on the other board because those internal spaces aren't available until you clear deeper into it.
12:45 Its because getting 2 8s next to each other uses less than 2x the number of mines; The total number of mines is not sixteen, but eight (from the first) plus five (eight from the second minus the overpap), so thirteen. This increases the probability a lot. This approximately cancels out with the chance that the two formations (one with eight and one with five mines, and each having one blank) appear right next to each other.
The eights paradox completely makes sense to me. If you have two eights on your board it can involve up to 16 mines. However, having two eights next to each other only depends on 13 mines since some are shared between the two eights. So it should actually be more likely to get the two eights next to each other than not sharing any mines between them, purely because having to independently and randomly placing more mines in specific spots exponentially reduces the odds with every added mine. It also makes perfect sense that out of all possible arrangements of two eights the most likely one is them being adjacent, for exactly the same reasons
12:25 it actually makes sense since youre only looking at five additional mines as oppossed to 8 entirely seperate ones. I honestly thought that it would be slightly more likely then not but 1 outta 3 aint bad
More careful simulation showed that it is closer to 1 out of 2. I agree, it makes sense to me now, I did get used to it. But my first reaction was pure disbelieve.
0-8, consecutively, all in a line. I was playing a minesweeper game that had a bunch of wacky items that did various things, and managed to slightly manipulate some mines to create 0-8.
@@oskain Of course it's possible. 1-2-3-4-5-6-7-8 Of course, 12345678 is not possible because the 7 and 8 touching means that the 8 can never be fully surrounded by mines. (the smallest consecutive line of numbers that are touching is 1-7)
for 1-7 X X X X X X X 1 2 3 4 5 6 7 X X X X X where X is a mine, so it needs 12 mines You can actually also save a mine if you zig zag the line at 4 and 2
@@naphackDT I more meant 12345678, with it being possible to have a mine on a number spot. Obviously impossible if every number has to be visible. I posted an imgur link with what I did, but I guess youtube auto-deleted it? that's kinda annoying. I'll try posting it again. if you don't see another post with the link then just know that it was deleted for some reason.
Well I just won my first sub-30 seconds game of Minesweeper on a 9x9 after watching this video! I used to think it was more nerve-wracking than fun, because it'd take me like 2 minutes to get close to winning, and I'd almost always lose. Thanks for the probability talk!
7:54 although, I have seen some unofficial implementations (they were meant for demonstrating the abilities of various programming languages, if memory serves) that were missing this feature, so if you remember clicking a mine on the first click, this might be the reason why
I've had many one-click wins; I just set the width and height to Expert and set the mines to ten! 🤣 (I also used to set the various sounds to something silly because reasons)
At 8:28 your adjust for 1st click will have more than one value, if the first click is at the corner of the board there is only 1 possible 8 that you ruin, however if you click in the middle that same click could destroy an 8 in 8 different positions
Hm, I don't follow. Here's my logic: For every eight you have, you have 8/64 = 1/8 chance of destroying it with your first click. Doesn't matter where 8 is - it's still going to be 1/8. So multiply the probability by 7/8 to get the corrected result. Am I missing something?
I think it makes sense that the chance of getting two 8s next to each other is relatively likely compared to getting two 8s anywhere. It’s because there are only a limited number of mines on the board - the fewer there are the more likely the two 8s will have to share mines. In fact, if you have few enough mines (e.g. 14) on any board size, the chance of the two 8s to be next to each other is 100%, i.e. they can’t physically form anywhere else.
I got an 8 once on google minesweeper in class. Took a screenshot but then my computer got wiped and I couldn't save it. Was quite upsetting now that nobody believes me.
Someone needs to make 5d minesweeper with multiverse time travel, just 5d minesweeper you could get a 242 tile, but with the added multiverse time travel, well that number would probably skyrocket.
For the double 7, am I mistaken or is it pretty simple (without all this text) to solve with math on the begginer difficulty? In a 8×8 grid, there are obviously 64 cells. To get the desired formation of mines, there are exactly 10 mines used. By counting how many different positin the center can be in, it is 6 ways along the axis with the fewer mines and 5 ways with the more mines. Therefore there are (5×6)×2 positions for the formation to exist. Because there are exactly 10 cells we must fill, and each cell filled makes one less cell possible to fill, which gives: First mine: 10 availible cells in the grid (64 empty cells) Second mine given the first mine is correct: 9 availible cells int he grid which has now 63 empty cells Third 8 positions with 62 positions availible And so on. It may be written like 10!/(64!/(64-10)!) = 10!×54!/64! This gives the probability of the desired formation with a number between 0 and 1. To get it expressed by 1 in some number, simply inverse it. Because of the 60 positions the formation can take, multiply by 60. Final equation: 60 × 64!/(10!×54!) = 2.525 × 10^9 ≈ 2.5 × 10^9 Hope it was not too difficult, I recommend googling "nPr" and "nCr", which tells us about the number of combinations desired.
It's because the 2 8s together only require 13 mines (in a certain arrangement) to create, whereas 2 separate 8s need 16 mines in a certain configuration. Basically when you already have one 8, you just need to get lucky with the placement of a "5" to get the figure-8/infinity shape, rather than having 2 incredibly unlikely 8s occur independent of each other
"first click is always safe" Well given that windows seems to have different versions for minesweeper throughout the years I dont know if this is also a version case but I do have actually hit mines first click essentially found cuz I was bored and just click once and start new board just for the hell of it.
the two 8 being more likely then you thought makes sense given that the second 8 only needs 5 additional mines to be arranged exactly instead of an additional 8 if it were to exist somewhere else on the board
I think the double 8 in an 8 shape is not as unlikely as you think because the chance for 2 separate 8s is 2 independent events. But a combined double 8 while still unlikely already has 3 of the mines generated. If the events were mutually exclusive, like if you generated 2 boards and then superimposed them on each other to get the combined 8s it would be highly unlikely as you suspect, but an 8 already existing on the board makes a combined 8 more likely because they are not exclusive of each other. Another way to think of it, is the chance to get 2 8s is the chance of *16* mines generating a certain way. the chance of combined 8s is the chance of *13* mines generating in a certain way.
There have been 2 occasions where I got an 8 in expert mode. The first time, I unluckily hit a mine before I got to the part of the board with the 8. The second time, I solved the whole board, but the game unexpectedly quit a fraction of a second later. I also had one instance of 2 sevens with no shared mines in expert mode (i.e. a total of 14 mines around the sevens). They were by no means near each other on that board. I have never had an 8 in intermediate, but have had 7 in intermediate three times.
The 2 8-shape form being more common can be intuitively deduced: some of the mines are already there so there is a higher chance that the 2nd 8 will use some mines of the first one. There is also a fixed amount of mines so the first 8 can always be a "spawn" point for the 2nd one contrary to randomly scattered mines.
10:40 About the asterisk, poisson distribution says that if you expect 1.3 hits, getting 3 hits happens about 10% of the time, and you'll get even more than 3 about 5% of the time. That's only a ~1:7 outlier, and there are so many simulations here that it'd be weird if you *didn't* get something that outlying.
4:50 "What else are we going to do, switch to 9x9 and reset all the records?" let me respond to your rhetorical question with my own. Is adding a separate "Beginner (9x9)" table to the scoreboards all that difficult?
Sorry to get technical (channeling my inner Sheldon Cooper here), but 99% density will give you 5 non-mine cells on Expert and 3 non-mine cells on Intermediate - games, which are astronomically unwinnable. Making the number of mines X*Y-1 would, indeed, guarantee a 1-click win.
What about 7-7-7, or boards with numbers only higher than 4? A 4x4 square of mines? 4's in a square? 5's in a square? A chain of 8's? What if the numbers average out to approximately pi, or e or golden ratio? How likely is it to get the same board twice in a row?
I would guess the 8s spawn in an 8 pattern because that setup utilizes less mines than having two separate mine circles, so it's more likely to occur with the amount of mines placed on an expert board
Yeah, I guess I figured it's not pure trolling at some point, but kept the game up, cause some of that I think was indeed a lighthearted troll-like fun.
the reason for the eight in shape of eight is barely less likely than two separate eight is comes from the used number of mines. two eight need 16 mines to be placed in a right positions, yes, you can move them around on the board, so you have a bit more chance, but the eight shaped eight need only 5 more mines than a single eight or 13 mines to be placed in the right spot, so you save 3 mines to be in random spot for a more constricted placement for the other 13.
Great stuff! Really enjoyed the video! The probability of two "8's" in the shape of an 8 did surprise me. Video format seems really close to @TwoMinutePapers
The highest digit in any dimensional minesweeper is 3ⁿ-1 where n is the dimension. So in 2D this is 3x3-1=8, and in 3D this is 3x3x3-1=26. The highest number in 12D minesweeper is 531'440. The highest number in 1D minesweeper is 3-2=2. The highest number in 0D minesweeper is 1-1=0.
When I think of "9" I think of a bomb surrounded completely by other bombs. So you would have no way of knowing what's in the middle until the very end of the game.
To anyone who actually doesn't realize why there can't be a nine, each spot is surrounded by 8 tiles. If it were a nine, that would mean that you clicked on a mine that was surrounded by 8 mines
Fact check true: I do know what a gerund is - knew that for about 25 years by now. However knowing what gerind is, is one thing. Using it in your speaking, let alone youtubing - is just asking too much.
the 2 8's in an 8 configuration has the advantage of already sharing 3 of the mines amongst one another, so it only takes 13 mines to make it, whereas two seperate 8's takes 16 mines. I suspect that helps with it being relatively common within the cases that have 2 8's.
Lol, i once got a circle of mines and i was so happy that i got an 8 which up to that point i thought was somehow prevented from happening. So i pressed it and there was a mine there, it really pissed me off, but now i know that it was even rarer and i got a screenshot of it so yayy i guess.
Would be interesting to have a multiplayer minesweeper kinda like battleship. Each side places their mines and then takes turns clicking each other's board.
I've been playing Minesweeper for 30 years now. Out of the thousands and thousands of games I've played today is the first time I got double 7s. Feels like the odds of getting it are lower than 1-20000 lol
"Yes, but what about 27" - Oh, hello. Glad you asked. Don't stand up now. Under your chair there is a bomb. It will blow up if you stand up, or if the square you uncover in the next game you play is a bomb. The bomb under your chair will self-disarm if you successfully mark all the bombs on the board. Welcome to 4D minesweeper, where the highest number you can uncover is 80. Oh, also, if you ask about 81, the bomb under your chair will blow up.
That's childsplay: In 5D minesweeper with multiverses and time travel, you have to worry about uncovered mines in the past blowing you up. And don't even mention grandfather mines.
@@incription I used to be a big fan of 5D minesweeper with Multiverse Time travel, it's too bad things like the terminator gambit, Jurassic moves, and the exile strategy have made the game stale and unfun
What about 81? What about 243? What about...?
@@JalebJay SILENCE MORTAL. YOU WISH TO PLAY....
D MINESWEEPER?
what about 69E^69 billionsD minesweeper
In Minesweeper, a "9" is a 3x3 array of mines. It's not a real number you can get. However, it is exceptionally rare.
I kinda suspected that, but never got a proof. Now, looking back, the nine questions were too persistent to be pure trolling.
@@GamesComputersPlay yeah its pretty much universal to call that a 9 in the minesweeper community, its also universally hated by everyone
I assume the part where it gets really bad is where you have two 3*3 groups that have a ring of mines on the outside, but you know based on the number of remaining mines that one has a mine in the middle, and the other doesn’t, basically making you make a blind 50/50 guess.
@@pokepress expert games are highly likely to end in a true 50:50 guess. more than 50% but probably not much more. I think the linear algebra solving technique can reach 45% solve success but I'm not sure if that includes successful 50:50 guesses. my own solver will halt and start a new game if it detects any 50:50 guess, but it only looks for those if there are no detectable safe squares.
but my solver doesn't detect 7s exactly (because 7s are the same color as bombs if i only detect 3 bombs or less on a board I treat them all as 7s) and won't detect 8s at all because they aren't solve probative.
My first thought was that you actually could have numbers greater than 9 in a minesweeper board with hyperbolic geometry
For the 8 in the shape of an 8, it made some sense to me. If you have one 8 already, you're using 8 of your mines in that 8, and these 8 mines are arranged in four edges of three consecutive mines. If you are to have another 8 on the board, starting with one of those edges gets you 3/8 of the way there, and requires you to use up three fewer of the total mine allotment. So it should be much more likely than random chance that, knowing you have an 8 on the board already, and knowing you have a second 8 somewhere, the second 8 would share an edge with the first
Yes, two 8s involve the positions of 16 mines, while a figure 8 only involves 13 mines, but with additional constraints on position, so it mrke sense that those two effects roughly cancel each other out
You (and other people saying they were not surprised) seem to have a well tuned probability intuition.
Now, when I know the answer - yeah, I guess it makes sense.
Bit when I saw the results for the first time, I was really suspicious there was a bug somewhere. Was totally in disagreement with my intuition, and even nwo it still rubs against it a little.
For me it stands right next to the birthday paradox and Monty Hall paradox. (which fooled my intuition too the first time)
yup. The fact that 8s can re-use the same mines is why its much more likely!
@@GamesComputersPlay Honestly you were already there talking about the array of 9, I think you just hadn't connected it to this occasion.
Yep. Despite the fact that there are a lot more ways to arrange two 8s on an intermediate board than one double 8, the 24 free mines on 238 cells of the two individual 8s gives ~5.3*10^32 remaining mine patterns per two 8s pattern, while the 27 free mines on 241 cells of the double 8 gives ~ 4.2*10^35 random mine patterns per double 8 pattern.
Back in highschool I used to always play a 83x38 custom board with 750 mines to pass the day instead of doing classwork. This was basically fullscreen minesweeper for my laptop and had a density of 23.78%, so just a little bit harder than expert.
It took me a whole year to finally win a game lol. It took about 4 hours to get to the end game, but with a board and density of that size, the probability of getting way 4-6 pure 50/50 choices was basically guaranteed. It was a coinflip until you win at that point. Keep in mind you need to first get to that stage, which could have misclicks or even just a small but quick lapse in thinking could end your entire game.
Same but with 10 mines.
my best is 99 mines and I'm a freshman haha (that also means I've pretty much beaten the Google version since there aren't custom boards)
i have ended many, many games with a miss click
my middle school dream
@@MyWaifuNow it's technically a bit slower, but I'm quite partial to playing minesweeper with keyboard only (especially better than a laptop trackpad)
The 8-8 thing makes sense to me. To get two full 8's, you need 16 mines to be placed exactly right, but for one single 8-8, you only need 13 mines to be placed in the right pattern.
Or, thinking about it more abstractly, once an 8 has already been placed, there are only 5 mines that need to be placed after that to make that 8 an 8-8.
Welcome to the club, me and Mallomon also posted on this, but I'm pretty sure none of us actually knew of the others.
@@Alterion57 i actually wanted to see if anybody got that first... so i'm not in the club although i wanted to post that too XD
It’s more than that. Reducing 8 to 5 definitely improves its chance, but it’s really down to the limited number of available mines that makes it so likely to occur.
The fewer mines available, the more likely the two 8s will have to share them.
If you have few enough mines (e.g. 14) regardless of the board size, the chance becomes 100% because there’s simply no other way that two 8s can physically form on the board.
@@doodeedah6409with 14, you could still have them off by one, so they only share 2 mines. Also, does being aligned horizontally instead of vertically still count as being in an 8 shape?
@@Jivvi Oh whoops, I miscounted, I meant 13 mines, i.e. 3 mines forced to be shared.
Btw the fine print at the bottom of the video says he counts both vertical and horizontal shapes.
I've actually had an 8 before. I was looking for it by setting the difficulty settings to max, with there being like 100 mines. Took like 20 straight minutes. Was so hyped when I found it.
Actually, 27 IS possible if you're willing to up your Minesweeper game to four dimensions. In 4D Minesweeper it's actually possible to get numbers as high as 80.
But what about 81?
5D minesweeper with multiverse time travel
Actually, 81 IS possible if you're willing to up your Minesweeper game to five dimensions. In 5D Minesweeper it's actually possible to get numbers as high as 242.
But what about 243?
@@jhawk2402 well, just enter the 6th dimension. But what about 729?
@Bobbeguille but is your version of 4d minesweeper using TIME as its 4th dimension? :-B
@@chandir7752 but what about 1024? or 16384... or huge numbers?
I rate this video a 9 out of 8.
I rate 27/26
81/80 for me
243/242 for me
Still 9/8 is higher than 27/26 etc.. 😆
729/728 for me
Minesweeper in VR is an absolute trip to go through, yes it's just 3d minesweeper but somehow it's immersive. I don't really say that but it's one of my favorite free VR games to use a way to pass time without having to move around to much. Kind of like putting a picture puzzle together, it's really interesting how it unfolds. I'd recommend checking it out if you're a fan of minesweeper itself.
Lamers, 2D and 3D is sooooo 2020, today 'retro' is cool and real players do Minesweeper in 1D.
I would actually love to code this
The two 8s in an 8 configuration odds are actually fairly intuitive. Given an 8 on the board, there are two places (as long as it's not too close to an edge - and assuming you want them vertically orthogonally connected) where there are already 3 mines in a row. Requiring only 5 mines to be in specific places, to accomplish the goal of two 8s. Where a disconnected 8 on the same board as your 8 requires 8 specific mine placements. The reason it's only 1/3 as likely as finding a 2nd 8 floating in some unknown far off location is because there are significantly more than 2 places those 8 specific mines could be arranged than 2. But that only just over compensates for the 8th power vs. 5th power comparison.
OOOOH because it's GIVEN that there's already an 8 and a limited number of mines and an additional 8 should be next to 8 mines
Me, Kpsla, and Mallomon have all also posted on this without knowing of each other funny enough.
@@justingolden21 no, not given, but the chance is not as small as expected because you only need 5 mines to be in a specific location instead of a whole new group of 8 mines
I liked that little tidbit about javelins resetting their records after some redesigning. I'm sure there are similar examples of other sports/games doing so, but it is quite refreshing to find these trivia in your vid!
Yeah, the funny thing though is that I have seen that in a video I watched just a few days ago, and now I wonder if that's why I got this video in my recommendation lol
For those interested. The javelin was redesigned since it was easier to make the javelin heavier than make every stadium larger. Therefore it became impossible to throw as far as before and thus all the records had to be reset
@@yuralis123 also, it was done because someone threw too far
I got really good at this at one point, and I used to play a larger board. I think 24x30 with ~165 mines. I remember the cutoff was pretty sharp, where most games under that number of mines were easily solvable with good initial guesses. More than that and I had to start making too many good guesses to win. I never ran probabilities though.
In Super Mario 64 speedruns whenever something occurs that would make two sets of records (such as 120 vs 70 star, then 16 star, then 1 star, then 0 star) they just split the leaderboards. Super Mario World also has two common Any%s: regular gameplay vs using the Arbitrary Code Execution glitch to jump to the credits from level 1. I think minesweeper could do the same.. have a 8x8 leaderboard and a 9x9 leaderboard.
Speedrunning beginner is sort of pointless imo. As the creator pointed out, you can beat beginner in a single click with 0 on the clock. These zeroes are simply not counted on the leaderboard, because otherwise the top of the leaderboard would just be full of them. So instead, you have to click at least twice. That means people who want to grind a good beginner time just keep refreshing the board and randomly clicking twice. Each time randomly clicking twice doesn't win, they refresh. They try to click as fast as possible, maybe using a mouse that double clicks if you push the button right (usually an undesirable behavior). For instance, the minesweeper online #1 time is 0.02 seconds, apparently accomplished by dragging the mouse along the top of the board.
Splitting the 8x8 and 9x9 times wouldn't change that. Beginner is just silly.
Interesting point. I actually agree about 2-click being mostly luck.
What they did about it is they introduced Beginner 3BV category that limit minimal 3BV at 10. It helps somewhat, still they have a few disqualified attempts because "there was one lucky click".
Which is now a bit weird. So what, lucky click? If at the shooting competition I shoot without looking and hit bull's eye, I still get the prize, right?
@@GamesComputersPlay Yeah, it also must be incredibly frustrating to try to grind. Every top time must be "lucky" to some extent, so it feels like if you get _too_ lucky, then it just doesn't count anymore. Imagine if whenever a golfer got two holes in one in a single round, that round got disqualified.
@@GamesComputersPlay What is the chance to beat a beginner board with a 3BV of at least 10 with only random clicks?
If it's less than a million, then there's a real chance to piss off some pros. :D
When I was a kid I used to change the settings of the game to use the biggest board with the fewest mines so I could basically “farm” one click wins. Sure it was cheating, but I barely even considered minesweeper a game and the most fun I could have with it was this way lol
good
I think the reason the 8 of 8s is so likely is because it shares 3 mines between the two 8s. The configuration requires only 13 mines and the remaining mines can be anywhere else, while having two separate 8s would require a whole 16 mines split into two groups.
I think it is misleading to compare the chance of getting an 8 of 8s to the chance of getting two separate 8s. Instead, the 8 of 8s should be viewed as its own kind of shape, and the chance of it appearing can be compared to the chance of two regular 8s that are not joined.
It would be interesting to calculate what the chance of finding two 8s that share any number of bomb tiles is compared to finding two disconnected 8s. I would bet that it'd be more likely than not that if there are two 8-tiles on the grid, that they share at least one bomb.
hey, let me run that. I'll report back.
ok, I ran a bigger simulations (220M), here's what I have:
- just 2 eights: 122, 1/1.8M
- 8s that touch each other at least in one spot: 88, 1/2.5M
- 8 in a shape of 8: 71, 1/3.1M
So:
- Looks like my estimation of just having 2 eight maybe was a bit off in the video, it's looks to be more rare. However, it makes the "8 in the shape of 8" paradox even more prominent: difference between those are even smaller than 2
- You prediction that touching 8s are more common than non-touchin is quite true: almost 3/4 are touching each other.
- estimation of 8 in the shape of eight still stands 3.1M and 3.2M - close enough.
Everyone is posting on this lol
@@GamesComputersPlay I bet the probability of two 8s that are immediately adjacent but don't use one set of mines in common is more like 100 or 1000 times less likely than any 2 eights. That would still use 16 mines
I’m so glad to see this vid come out! I played a game of minesweeper in 9th grade where the very first click was an 8. I felt like I won the most useless lottery ever lol
In University I worked on a machine learning algorithm to play Minesweeper (with... mixed results, I was a rookie), and 1-click games became a serious problem. Turns out my board wasn't dense enough and they were more common than they should have been, it fed the algorithm a ton of bad habits because that click ended up with a massive reward score. That one eluded me for a while...
The reason double 8 is not that much more unlikely is because it requires less mines in the formation than two 8s by themselves, allowing more leeway for the rest of the mines.
12:44 think of it this way: if you have 2 eights in a shape of an 8, that only forces 13 mines but having 2 eights on somewhere far away forces 16 mines. That makes the 2 eights generating in a shape of an 8 a little bit more likely.
These probabilities make me want to have all ~7 billion people on earth to play minesweeper once just to see what boards have these rare things lol
Hmm. I always assumed that Minesweeper solved the "first click problem" by choosing an appropriate board after the first click. It would keep randomly producing boards until a compatible one revealed itself.
To be honest, I couldn't find any reliable info about latest minesweeper clones and how they handle it. But for old version - yes, they just move a mine.
Also, I think they do create new board in "no guessing" mode in some implementations. (As just moving mine can create a guessing situation).
I have wondered about this "What if the first click is a mine"-issue and how they solve it, when you're suppose to be safe that one time. There 's actually a Game Boy versjon of "Minecraft", which «solves» the whole problem by letting you actually get a bomb on your first click... ua-cam.com/video/H-qMyyYB5kM/v-deo.html
The best way would be to shift the whole board down and to the right to the first safe spot, wrapping the bottom rows to the top and the rightmost columns to the left.
@@GamesComputersPlay almost every "no guess mode" version tells you which first click to make. they generate these boards by actually solving them and when a situation is reached that would be unsolvable it moves mines then to currently uncleared spaces that aren't associated with a cascade and continues solving. if it can't find a place in uncleared blocks, it puts it in any space that doesn't boarder any other mine. if a move was moved in this way it begins solving again from the beginning.
seeds for these states with the starting square are pre generated and stored, often by brute force.
my approach is slightly different.
I run a solver that doesn't guess on my home computer and note seeds and starting clicks for any win. those are innately solvable too.
The reason the probability of the two eights being close to each other is so high is because it's easier for 5 mines with one space in the middle to spawn close to the 8 than it is for an entirely different ring of 8s to spawn. In essence, they are simply both 5s sharing 3 mines, not two completely separate 8s. As to them being on top of each other, not necessarily. The same logic applies sideways, where they could form the shape of an infinity symbol. I hope this explanation makes sense.
I will admit, Mallomon posted on this first, so credit where credit is due, however, I had no idea they had posted until just now.
I love this kind of stuff. Finding out the answers to these "what if" questions purely for the sake of being curious and the programming pleasure
I mean, it makes sense. Having 2 8s anywhere on a board requires the correct permutation of 16 mines, where as when they’re bordering, you only need 13 correctly permuted mines
12:22 I immediately guessed that the conditional probability was going to be very likely, considering that it would only need an additional 5 mines for the second 8, instead of 8 for fully-separate 8s. You could also have 8s that share one or two mines, and I would expect either of those probabilities to be greater than fully-separate 8s too.
Everyone is posting on this lol
Excellent video, I would love even brief slides showing your probability calculations but that’s just because I’m a math nerd. Love the content
Thanks, man!
There actually just some excel formulas, like this one (this is probability of 8 for expert): =(30-2)*(16-2)*(COMBIN(30*16-9,99-8))/COMBIN(30*16,99)
Except I put board size, mines, and size and mines of the object in separate cells, to be able to re-use it for different things.
It breaks for bigger numbers, but should work for everything in the video: 8, 9, 77, 8in8.
The correct way to calculate it would be a multi-step process. First step: Calculate the total number of different mine patterns, this is [number of squares] choose [number of mines], so 64 choose 10 or 151473214816 for the old beginner board. Next, calculate how many different ways your chosen pattern fits on the board. For the 8, that is 36 different ways - the 8's center square can lie on any of the 36 non-edge squares of the 8x8 board. Then, calculate how many different ways the non-pattern mines can be arranged for each of those ways - in the case of the 8, that is 55 choose 2 - two mines remaining, and we have used up 9 of the board squares for the 8, giving a result of 1485. Multiply the latter two numbers together, and divide by the first, to get in this case 36*1485/151473214816=3.53e-7 or about 1 in 2.83 million odds. This also shows why the 9/filled 8 is so much less likely in the beginner board: It only has one additional mine left, so there are in total only 1980 patterns that give you it, vs 53460 patterns that yield the normal 8.
This also means that on the new 9x9 beginner board, the chance for an 8 is 49*2556/1878392407320=1 in ~15 million - yes, there are more than ten times as many total different mine patterns on the 9x9 board as there are on the 8x8 board!
It's funny when thinking about 4D minesweeper, it occurred to me that if you play two 3D games at the same time, alternating turns on each board, you can conceptualize that they're the same board taken at different times where the mines have moved. You can even play some Schrodinger's cat with the boards by only placing mines once you've progressed on the other board because those internal spaces aren't available until you clear deeper into it.
12:45
Its because getting 2 8s next to each other uses less than 2x the number of mines; The total number of mines is not sixteen, but eight (from the first) plus five (eight from the second minus the overpap), so thirteen. This increases the probability a lot. This approximately cancels out with the chance that the two formations (one with eight and one with five mines, and each having one blank) appear right next to each other.
i think that when you get an 8 they shouldve made it so its animated and looks like its on fire
The eights paradox completely makes sense to me. If you have two eights on your board it can involve up to 16 mines. However, having two eights next to each other only depends on 13 mines since some are shared between the two eights. So it should actually be more likely to get the two eights next to each other than not sharing any mines between them, purely because having to independently and randomly placing more mines in specific spots exponentially reduces the odds with every added mine. It also makes perfect sense that out of all possible arrangements of two eights the most likely one is them being adjacent, for exactly the same reasons
The double 8 thing makes sense, because it only has to have 13 mines in a set position, rather than 2 sets of 8 in set positions.
Lol, everyone is posting on this
that 3d minesweeper is a real thing it is especially amazing if you can play it in vr and i personally find it very fun
12:25 it actually makes sense since youre only looking at five additional mines as oppossed to 8 entirely seperate ones. I honestly thought that it would be slightly more likely then not but 1 outta 3 aint bad
More careful simulation showed that it is closer to 1 out of 2.
I agree, it makes sense to me now, I did get used to it. But my first reaction was pure disbelieve.
Lol, everyone is posting on this
This is an amazing video! Looking forward to see you do more of these.
I've never had a double-7, but I did once get a triple-6. I know that's not nearly as rare, but it was still a fun find.
I was thinking that's got to be more rare and then I realised 666 not 767.
Ha ha ha I got a 7 5 5 5
9 would probably be also possible in a non euclidean minesweeper.
But good luck comprehending that mess.
0-8, consecutively, all in a line. I was playing a minesweeper game that had a bunch of wacky items that did various things, and managed to slightly manipulate some mines to create 0-8.
not possible
@@oskain Of course it's possible.
1-2-3-4-5-6-7-8
Of course, 12345678 is not possible because the 7 and 8 touching means that the 8 can never be fully surrounded by mines.
(the smallest consecutive line of numbers that are touching is 1-7)
for 1-7
X X X X X X X
1 2 3 4 5 6 7 X
X X X X
where X is a mine, so it needs 12 mines
You can actually also save a mine if you zig zag the line at 4 and 2
@@naphackDT I was talking bout 12345678
@@naphackDT I more meant 12345678, with it being possible to have a mine on a number spot. Obviously impossible if every number has to be visible. I posted an imgur link with what I did, but I guess youtube auto-deleted it? that's kinda annoying. I'll try posting it again. if you don't see another post with the link then just know that it was deleted for some reason.
Well I just won my first sub-30 seconds game of Minesweeper on a 9x9 after watching this video! I used to think it was more nerve-wracking than fun, because it'd take me like 2 minutes to get close to winning, and I'd almost always lose. Thanks for the probability talk!
Now i wonder, what's your record at this moment?
i got a 9 once.. yes in 2d. also 4d minesweeper is fun
disclaimer: the time i got 9 was when i did changed an open source minesweeper to include it
7:54 although, I have seen some unofficial implementations (they were meant for demonstrating the abilities of various programming languages, if memory serves) that were missing this feature, so if you remember clicking a mine on the first click, this might be the reason why
What are the odds to have both an 8 and a 3x3 clumb of mines, meaning u would end up with a 50/50 click to win the board
This just showed up in my recommendations and it is so nerdy that I’m more impressed than I ever thought I could be
In a good way
I've had many one-click wins; I just set the width and height to Expert and set the mines to ten! 🤣
(I also used to set the various sounds to something silly because reasons)
1. This is childish
2. I did that too several times
@@GamesComputersPlay chill
@@bruh.5620 cmon, I'm just kidding. Everybody did that, once discovered that you can set any number of bombs.
@@GamesComputersPlay oh ok
You got 8 in Minesweeper?
Well done. Now try to get 26 in 3D Minesweeper
What is the hardest possible board? By hardest I mean "requires the most forced guesses"?
At 8:28 your adjust for 1st click will have more than one value, if the first click is at the corner of the board there is only 1 possible 8 that you ruin, however if you click in the middle that same click could destroy an 8 in 8 different positions
Hm, I don't follow. Here's my logic:
For every eight you have, you have 8/64 = 1/8 chance of destroying it with your first click. Doesn't matter where 8 is - it's still going to be 1/8.
So multiply the probability by 7/8 to get the corrected result.
Am I missing something?
🇰🇳 guess this flag@@GamesComputersPlay
How to have a 100% chance of getting an 8
1. Set the board size 3x3
2. Click on the middle and there you go
Сразу слышно земляка) спасибо за офигенный видос, мало такого годного контента по сапёру
Тоже услышал русский акцент
One time I randomly clicked and found an 8
I think it makes sense that the chance of getting two 8s next to each other is relatively likely compared to getting two 8s anywhere.
It’s because there are only a limited number of mines on the board - the fewer there are the more likely the two 8s will have to share mines.
In fact, if you have few enough mines (e.g. 14) on any board size, the chance of the two 8s to be next to each other is 100%, i.e. they can’t physically form anywhere else.
Hey, I made it in this video! 4:40 :D
wow
I got an 8 once on google minesweeper in class. Took a screenshot but then my computer got wiped and I couldn't save it. Was quite upsetting now that nobody believes me.
I think nine is a 3x3 field full of bombs
I got a 6 a couple weeks back and was shocked, but an 8? Now THAT is rare.
Someone needs to make 5d minesweeper with multiverse time travel, just 5d minesweeper you could get a 242 tile, but with the added multiverse time travel, well that number would probably skyrocket.
Me: watches 16 minute video about minesweeper probability
Video: calls me a nerd
Me: fine, I’ll subscribe.
It's ... Ninesweeper.
People discussing if a beginner board is 8x8 or 9x9, then there's google over there with a easy board that is 6x11
All fun and games till you get a nine
awesome video :D
you are also awesome!
There is no 9 because seven ate nine
That's the kind of joke I should have thought of.
Alright, here I go:
But why did it ate nine? Because it needed three square meals a day!
Bro i Had to delete this joke
For the double 7, am I mistaken or is it pretty simple (without all this text) to solve with math on the begginer difficulty?
In a 8×8 grid, there are obviously 64 cells. To get the desired formation of mines, there are exactly 10 mines used. By counting how many different positin the center can be in, it is 6 ways along the axis with the fewer mines and 5 ways with the more mines. Therefore there are (5×6)×2 positions for the formation to exist.
Because there are exactly 10 cells we must fill, and each cell filled makes one less cell possible to fill, which gives:
First mine:
10 availible cells in the grid (64 empty cells)
Second mine given the first mine is correct:
9 availible cells int he grid which has now 63 empty cells
Third 8 positions with 62 positions availible
And so on.
It may be written like 10!/(64!/(64-10)!) = 10!×54!/64!
This gives the probability of the desired formation with a number between 0 and 1. To get it expressed by 1 in some number, simply inverse it.
Because of the 60 positions the formation can take, multiply by 60.
Final equation:
60 × 64!/(10!×54!) = 2.525 × 10^9
≈ 2.5 × 10^9
Hope it was not too difficult, I recommend googling "nPr" and "nCr", which tells us about the number of combinations desired.
That's exactly right.
It's because the 2 8s together only require 13 mines (in a certain arrangement) to create, whereas 2 separate 8s need 16 mines in a certain configuration. Basically when you already have one 8, you just need to get lucky with the placement of a "5" to get the figure-8/infinity shape, rather than having 2 incredibly unlikely 8s occur independent of each other
People: "talk about 9!"
GCP: "nein"
"first click is always safe"
Well given that windows seems to have different versions for minesweeper throughout the years I dont know if this is also a version case but I do have actually hit mines first click
essentially found cuz I was bored and just click once and start new board just for the hell of it.
This minesweeper is different from the original one. 0/10 lies were spouted out
all jokes aside, wow, this is the sequel I needed
12:55 it happens because to get 2 disjoint 8s you need 16 bombs, but for a double 8 in the shape of an 8, you only need 13 bombs
This video inspired me to try playing minesweeper, and I'm glad it did. It was very fun.
the two 8 being more likely then you thought makes sense given that the second 8 only needs 5 additional mines to be arranged exactly instead of an additional 8 if it were to exist somewhere else on the board
I think the double 8 in an 8 shape is not as unlikely as you think because the chance for 2 separate 8s is 2 independent events. But a combined double 8 while still unlikely already has 3 of the mines generated.
If the events were mutually exclusive, like if you generated 2 boards and then superimposed them on each other to get the combined 8s it would be highly unlikely as you suspect, but an 8 already existing on the board makes a combined 8 more likely because they are not exclusive of each other.
Another way to think of it, is the chance to get 2 8s is the chance of *16* mines generating a certain way. the chance of combined 8s is the chance of *13* mines generating in a certain way.
There have been 2 occasions where I got an 8 in expert mode. The first time, I unluckily hit a mine before I got to the part of the board with the 8. The second time, I solved the whole board, but the game unexpectedly quit a fraction of a second later. I also had one instance of 2 sevens with no shared mines in expert mode (i.e. a total of 14 mines around the sevens). They were by no means near each other on that board. I have never had an 8 in intermediate, but have had 7 in intermediate three times.
"Prepare for the answer" is my new favorite way to disclose info-hazards
I live for cool minesweeper content, thanks man! Also have a nice day to all fellow sweepers out there
The 2 8-shape form being more common can be intuitively deduced: some of the mines are already there so there is a higher chance that the 2nd 8 will use some mines of the first one. There is also a fixed amount of mines so the first 8 can always be a "spawn" point for the 2nd one contrary to randomly scattered mines.
10:40 About the asterisk, poisson distribution says that if you expect 1.3 hits, getting 3 hits happens about 10% of the time, and you'll get even more than 3 about 5% of the time. That's only a ~1:7 outlier, and there are so many simulations here that it'd be weird if you *didn't* get something that outlying.
Yep, you are quite right. My probability intuition is quite impefect, so an outlier like this stil triggers all sorts of alarm bells.
3D minesweeper is what a vulcan child would play
4:50 "What else are we going to do, switch to 9x9 and reset all the records?" let me respond to your rhetorical question with my own. Is adding a separate "Beginner (9x9)" table to the scoreboards all that difficult?
If you make the density 99% you always get a one click win. Screwing around with this is how I learned the double 8 rule as a kid
Sorry to get technical (channeling my inner Sheldon Cooper here), but 99% density will give you 5 non-mine cells on Expert and 3 non-mine cells on Intermediate - games, which are astronomically unwinnable.
Making the number of mines X*Y-1 would, indeed, guarantee a 1-click win.
What about 7-7-7, or boards with numbers only higher than 4? A 4x4 square of mines? 4's in a square? 5's in a square? A chain of 8's? What if the numbers average out to approximately pi, or e or golden ratio? How likely is it to get the same board twice in a row?
I would guess the 8s spawn in an 8 pattern because that setup utilizes less mines than having two separate mine circles, so it's more likely to occur with the amount of mines placed on an expert board
I think what people meant with "What about 9?" is just a 3x3 area with all mines. No numbers, all mines. I THINK that is what people meant
Yeah, I guess I figured it's not pure trolling at some point, but kept the game up, cause some of that I think was indeed a lighthearted troll-like fun.
I liked and subscribed because I REALLY enjoyed this Video.
the reason for the eight in shape of eight is barely less likely than two separate eight is comes from the used number of mines. two eight need 16 mines to be placed in a right positions, yes, you can move them around on the board, so you have a bit more chance, but the eight shaped eight need only 5 more mines than a single eight or 13 mines to be placed in the right spot, so you save 3 mines to be in random spot for a more constricted placement for the other 13.
Great stuff! Really enjoyed the video! The probability of two "8's" in the shape of an 8 did surprise me. Video format seems really close to @TwoMinutePapers
The highest digit in any dimensional minesweeper is 3ⁿ-1 where n is the dimension. So in 2D this is 3x3-1=8, and in 3D this is 3x3x3-1=26.
The highest number in 12D minesweeper is 531'440.
The highest number in 1D minesweeper is 3-2=2.
The highest number in 0D minesweeper is 1-1=0.
When I think of "9" I think of a bomb surrounded completely by other bombs. So you would have no way of knowing what's in the middle until the very end of the game.
Yeah, turns out it is a slang name for that.
suggestion: board without 4's, 5's, 6's etc.
3D one it so cool! Gotta try my hands on it.
16:05 EASILY one of the best endings of a UA-cam video I have ever seen!
suggestion: what about a 2x2 square of 5's?
To anyone who actually doesn't realize why there can't be a nine, each spot is surrounded by 8 tiles. If it were a nine, that would mean that you clicked on a mine that was surrounded by 8 mines
I love how hard gerunds are for non native speakers, and how few native speakers know the word "gerunds"
Fact check true: I do know what a gerund is - knew that for about 25 years by now. However knowing what gerind is, is one thing. Using it in your speaking, let alone youtubing - is just asking too much.
the 2 8's in an 8 configuration has the advantage of already sharing 3 of the mines amongst one another, so it only takes 13 mines to make it, whereas two seperate 8's takes 16 mines. I suspect that helps with it being relatively common within the cases that have 2 8's.
The algorithm has decided his fate is playing minesweeper
Lol, i once got a circle of mines and i was so happy that i got an 8 which up to that point i thought was somehow prevented from happening. So i pressed it and there was a mine there, it really pissed me off, but now i know that it was even rarer and i got a screenshot of it so yayy i guess.
Would be interesting to have a multiplayer minesweeper kinda like battleship.
Each side places their mines and then takes turns clicking each other's board.
The winning strategy would be to place as many 50/50 as possible. Unless it were against the rules and only "no guessing" mines were allowed.
Riririeiieieieiririririttoir
12:50 - I figure it is because one wall is already made on that, so it is less you are having to account for.
I've been playing Minesweeper for 30 years now. Out of the thousands and thousands of games I've played today is the first time I got double 7s. Feels like the odds of getting it are lower than 1-20000 lol