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Currently stuck in this game of Minesweeper. Can't seem to find any openings, so what would be the moves you would take and why?

Stuck in Minesweeper

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    $\begingroup$ I think, at this point, most people just take a guess and click a random block in the unexplored space $\endgroup$ – TrojanByAccident Jan 4 '17 at 23:30
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    $\begingroup$ Hmm, I'm also unable to find any reasonable move.. :S I would wait for another answer first. But if no one knows an answer, I would take a gamble with the red i, j, k, l. Only one of those four is a bomb, so you have a 75% chance to choose correctly. And welcome, btw. :) $\endgroup$ – Kevin Cruijssen Jan 4 '17 at 23:40
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    $\begingroup$ what happens if you click red q? $\endgroup$ – JMP Jan 5 '17 at 12:22
  • $\begingroup$ @JonMarkPerry you get a red 10-Q ? :-) $\endgroup$ – Carl Witthoft Jan 5 '17 at 16:30
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    $\begingroup$ I'd click the black X in the top right corner. $\endgroup$ – Mazura Jan 6 '17 at 1:02
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Red n is a mine. The pattern, subtracting off known flags, is 2-2-1 (above n-o-p) against a wall. The middle 2 must be next to 2 mines, but only one mine can be on o and p, so the second mine must be n.

Red q is not a mine. If it were, then the 3 above red p would be satisfied, which leaves only one square for the 2 above red o.

Also, as Kevin Crussijen points out, i, j, k, and l can have only one mine between them, so at worst, if nothing else comes up, guess there.

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  • $\begingroup$ In regards to the last part, out of those squares, one of them has a 50% chance of having a mine, while the others only have a 25% chance. $\endgroup$ – Justin Time Jan 5 '17 at 16:58
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    $\begingroup$ @KyleStrand That's on a per-square basis, not collectively. Since there's one set of four squares that share a single mine between them, each square in the set has a 1/4 chance of having that mine. However, one of those squares is also in a set of two squares that share a single mine between them, meaning it also a 1/2 chance of having a mine. In this case, the higher chance overrides the lower chance; ergo, each square in the set has a base chance of 25%, but one square in the set instead has a 50% chance due to also being in a different set. (Not a spoiler, 25% is 1/4 and 50% is 1/2.) $\endgroup$ – Justin Time Jan 5 '17 at 23:40
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    $\begingroup$ @JustinTime I'm going to go ahead and say that that's not how probability works. One square has a 1/2 half chance; the other three clearly have 1/6 each. $\endgroup$ – Kyle Strand Jan 5 '17 at 23:45
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    $\begingroup$ @KyleStrand neither of those are how it works. There are two scenarios- either those 5 squares account for 2 mines or one - the actual probability of which of those scenarios has occurred depends on how many bombs are remaining. In the extreme cases you could say it's probability 1 or 0 either way, in practice it is a complex calculation that depends on how many bombs remain unaccounted for, and probably has the effect of evening the probabilities out quite a lot. $\endgroup$ – Jeff Jan 6 '17 at 2:52
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    $\begingroup$ @KyleStrand He's looking at both sets, together, due to the overlap. Between the 4-square set and the 2-square one, they compose a total of 5 squares. Out of these, there's either one mine (on the square where the sets intersect), or two mines (one in each set, on non-intersecting squares). Which, in all honesty, is something I didn't even think about. $\endgroup$ – Justin Time Jan 6 '17 at 18:56
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Red q cannot be a mine, because if it were, the 2 above red o can only be surrounded by one mine on red n.

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    $\begingroup$ +1. Maybe add a picture to your answer to make this even clearer? $\endgroup$ – Rand al'Thor Jan 5 '17 at 1:03
  • $\begingroup$ Corresponding, the letter 3 before that has to be a mine. $\endgroup$ – Kevin Jan 5 '17 at 16:39
  • $\begingroup$ @Kevin Agreed. That's a 1-2-x pattern (note: not to be confused with red 'x'), so the square adjacent to x must be a mine. $\endgroup$ – Justin Time Jan 5 '17 at 21:28
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    $\begingroup$ @randal'thor : Clarity added without a picture: if Red Q were a mine, then the nearby 3 would indicate that both Red O and Red P are not mines. That would be required for the 3 to be only a 3 (and not a 4). However, if Red O and Red P are not mines, then there isn't enough space for the nearby 2 (above red O) to have enough mines to be a 2. Therefore, we run into a logic problem if Red Q was a mine; we can therefore conclude that Red Q is not a mine. Possibly see also: Tuan's answer. $\endgroup$ – TOOGAM Jan 7 '17 at 13:05
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Short answer:

Click the red q first.

Longer answer, no spoiler tags:

To be systematic about it, first we can update the numbers in squares that are adjacent to both blue squares and flags. By "update the numbers" I mean decrease the numbers by however many flags are adjacent to the square. So, for example, the square above the $\color{red}{A}$ decreases from $2$ to $1$ because it's already adjacent to $1$ flag. Doing this results in the following picture:

enter image description here

Now all the numbers we updated (they're all in black) tell us how many mines are in the adjacent blue squares. So we can deduce the following:

  1. Exactly one of $\color{red}{A}$ and $\color{red}{B}$ must be a mine.
  2. Exactly two of $\color{red}{A},\color{red}{B},\color{red}{C},\color{red}{d},\color{red}{e}$ must be mines.
    And $\#1$ tells us that one of these mines must be $\color{red}{A}$ or $\color{red}{B}$. So we can conclude $\#3$ below.
  3. Exactly one of $\color{red}{C},\color{red}{d},\color{red}{e}$ must be a mine.
  4. Exactly one of $\color{red}{e}$ and $\color{red}{f}$ must be a mine.
  5. Exactly one of $\color{red}{f}$ and $\color{red}{g}$ must be a mine.
  6. Exactly two of $\color{red}{f},\color{red}{g},\color{red}{h},\color{red}{i}$ must be mines.
  7. Exactly one of $\color{red}{h}$ and $\color{red}{i}$ must be a mine.
    (We can deduce this from the previous two or we can get this directly from the game board.)
  8. Exactly one of $\color{red}{i}, \color{red}{j}, \color{red}{k}, \color{red}{l}$ must be a mine.
  9. Exactly two of $\color{red}{m}, \color{red}{n},\color{red}{o}$ must be mines.
  10. Exactly two of $\color{red}{n}, \color{red}{o},\color{red}{p}$ must be mines.
  11. Exactly one of $\color{red}{o}, \color{red}{p}, \color{red}{q}$ must be a mine.

Since $11$ says we can't have a mine at both $\color{red}{o}$ and $\color{red}{p}$, then from $10$ we can deduce that $\color{red}{n}$ is definitely a mine. Therefore either $\color{red}{o}$ or $\color{red}{p}$ (but not both) is definitely a mine, which, combined with $\#11$ (again), tells us $\color{red}{q}$ is definitely not a mine. So let's update our board. I'll use a solid yellow square to indicate safety:

enter image description here

Now we can continue our list, but let's update some of the items first:

  1. Exactly one of $\color{red}{m}$ and $\color{red}{o}$ must be a mine.
  2. Exactly one of $\color{red}{o}$ and $\color{red}{p}$ must be a mine.
  3. Exactly one of $\color{red}{s}$ and $\color{red}{t}$ must be a mine.
  4. Exactly one of $\color{red}{t}$ and $\color{red}{x}$ must be a mine.
  5. Exactly two of $\color{red}{t},\color{red}{u},\color{red}{v},\color{red}{w},\color{red}{x}$ must be mines.
  6. Exactly one of $\color{red}{u},\color{red}{v},\color{red}{w}$ must be a mine. This is deduced from $\#15$ and $\#16$.
  7. Exactly one of $\color{red}{x},\color{red}{y},\color{red}{z}$ must be a mine.
  8. Exactly one of $\color{red}{y},\color{red}{z}, a$ must be a mine.
  9. Exactly one of $\color{red}{z}, a, b$ must be a mine.
  10. Exactly one of $a,b,c$ must be a mine.
  11. Exactly two of $b,c,d,e$ must be mines.
  12. At least one of $d$ and $e$ must be a mine. This follows from $\#21$ and $\#22$ together (since we can't have a mine at both $b$ and $c$ but we may have a mine at neither).
  13. Exactly one of $e$ and $f$ must be a mine.
  14. Exactly one of $f$ and $n$ must be a mine.
  15. Exactly two of $n,m,o$ must be mines.
  16. Exactly one of $o$ and $p$ must be a mine.
  17. Exactly one of $p,q,r$ must be a mine.
  18. Exactly two of $q,r,s,t,u$ must be mines.
  19. Exactly one of $t$ and $u$ must be a mine.
  20. Exactly one of $q,r,s$ must be a mine. This follows from $\#29$ and $\#30$ together.

At this point we can't make any other immediate solid conclusions (unless I missed something). What I would do next is update the board in this same manner using the number you get from clicking on $\color{red}{q}$. See if anything falls out of that. If not, then make some guesses using the list above and see if any contradictions fall out of it, thus allowing you to rule out other possibilities.

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    $\begingroup$ Damn, this is a good answer. How much time did you spend on this? $\endgroup$ – Rand al'Thor Jan 5 '17 at 15:58
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    $\begingroup$ @randal'thor thanks. Hard to say since I was working on it in spurts while simultaneously doing a bunch of other stuff. If I was working on just this and nothing else then maybe 30-40 minutes? TBH the most tedious part was using MS Paint. $\endgroup$ – tilper Jan 5 '17 at 16:00
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    $\begingroup$ Best and most complete answer. $\endgroup$ – SQB Jan 6 '17 at 14:27
  • $\begingroup$ This process, which you call "update the numbers", is basically what I do in my head. Going step-by-step and showing the process with screenshots is excellent training. Excellent. (Although, it would have been easier if the added numbers (all 1's and 2's) overlaid overlaying the existing numbers only partially, so the actually-shown numbers would still be at least partially visible. Perhaps show the new numbers in little squares with a yellow background. Or maybe just use dots.) $\endgroup$ – TOOGAM Jan 7 '17 at 13:33
  • $\begingroup$ @TOOGAM thanks. I dunno if that would've worked so well in practice since each square would be very cramped and most of that frontier along the blue squares would be a cluster. $\endgroup$ – tilper Jan 7 '17 at 14:01
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This answer, although it does address the question itself, is more of an interesting observation coming from analysing the situation by expressing the system in terms of simple linear equations (eg/ using capitals for the reds, you might say A+B=1 to say that exactly one of A and B can be a mine, or F+G+H+I=2 because there must be exactly two mines amongst those four). I don't think it's reasonable to create these sorts of equations for every step.

When you ask something like Maxima to solve it, you get a lot of equations that don't help much... but you also get, for example (substituting away the arbitrary introduced variables), N=Q+1. This can only be satisfied in one way if all variables can be only 0 or 1 - namely, N=1 and Q=0. So there's a mine at red N and no mine at red Q.

Similarly, you get (using lower-caps for black) o=1+q+r, which can only be satisfied if o=1, and q=r=0. So there's a mine at black o and no mine at black q and black r. Re-checking has shown it is o=q+r, which isn't as helpful. Substituting the above result doesn't determine any further values with any confidence, unfortunately (in contradiction with a previous version of this answer).

Some interesting other observations that can be made from this analysis...

At least one of black d and black e must be a mine, and if both of them are, then so are black a and red S and X (if only one is, then black a, red S and X are not mines).

Once you eliminate duplicates and the two certain values, there are 26 equations in 44 unknowns.

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    $\begingroup$ black o and p can't both be a mine $\endgroup$ – Kruga Jan 5 '17 at 8:32
  • $\begingroup$ You say that, for black, o=1+q+r but lets assume there is no mine at o then there are mines at p, s, m, n and e and no mines at o, q, r and f. I've not seen any contradiction that would prevent such an arrangement - have I missed something? $\endgroup$ – MT0 Jan 5 '17 at 15:15
  • $\begingroup$ @Kruga - true. I'm re-doing the analysis, because I may have made a typo when typing the equations in. Expect an update soon. (looks like it should have said o=q+r) $\endgroup$ – Glen O Jan 6 '17 at 5:31
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Same answer as greenturtle3141 except with longer explanation.

Answer

q

Explanation

Among n/o/p, there are two bombs. The two bombs can't be on both o and p because then the 3 above p would have four bombs adjacent. The two bombs can't be on neither o and p because then the 2 above o could not have two bombs. Therefore one of o/p has a bomb and the other doesn't. This means that we have accounted for all three bombs below the 3 above p. Therefore q is safe.

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i would click on 'q'.

the 2 above the 'o' implies that two mine are between 'n', 'o', 'p'.

the 3 adjacent to it, means there is a mine on 'n', and another between 'o' and 'p'.

the 3 above 'p' has already two adjacent mines.

so, in conclusion, there should be no mine on 'q'.

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Note: Through this answer, red 'X' squares will be referred to as rX, and black 'X' squares as bX. For example, red 'A' is rA, and black 'A' is bA.

So far, there is one known mine:

rN: Starting from rP to the right, we have a 1-2-x pattern: There is one mine shared between rO and rP (it must be shared, because of the 2), which means that the other mine touching the 2 has to be rN.

There's also one known safe spot:

rQ: Thanks to rO and rP, we also have a 1-1-x pattern: There's one mine shared between those two, which means there can't be a mine touching rQ (because the 3 adjacent to rP reduces to a 1, thanks to having two exposed mines touching it).

If these don't allow further logic, then there are three optimal squares for guessing, which each have a 1/4 chance of being a mine.

rJ, rK, rL: Out of those three and rI, there can be only one mine, thanks to the 3 (reduced to 1). HOWEVER, because of the 2 (reduced to 1) adjacent to rI, there must be a mine shared between rH and rI, leaving rI with a 1/2 chance and the other three with 1/4.


Considering everything we know so far, we can calculate the probability of each unknown square being a mine, and group them together with related squares:
[Fractions are unreduced, with the number of mines in a set over the number of squares.]

  • rA: 1/2, shared between rA and rB: The 2 adjacent to rA reduces to 1.
    rB: 1/2, shared between rA and rB: The 2 adjacent to rA reduces to 1.

  • rC: 1/3, shared between rC, rD, rE (rE is 1/2): The 3 reduces to a 2, where one mine is in rA/rB, and thus the other one must be shared between these squares.
    rD: 1/3, shared between rC, rD, rE (rE is 1/2): The 3 reduces to a 2, where one mine is in rA/rB, and thus the other one must be shared between these squares.

  • rE: 1/2, shared between rE and rF (overrides rC/rD/rE): Both the 4 and the 2 reduce to 1, which gives a higher probability than rC/rD/rE. If this square has a mine, rG must also be mined.
    rF: 1/2, shared between rE and rF: Both the 4 and the 2 reduce to 1. Similarly, 1/2, shared between rF and rG: The 3 reduces to a 1.
    rG: 1/2, shared between rF and rG: The 3 reduces to a 1. If this square has a mine, rE must also be mined.

  • rH: 1/2, shared between rH and rI: The 2 adjacent to rI reduces to a 1.
    rI: 1/2, shared between rH and rI (overrides rI/rJ/rK/rL): The 2 adjacent to rI reduces to a 1.

  • rJ: 1/4, shared between rI, rJ, rK, and rL (rI is 1/2): The 3 reduces to a 1.
    rK: 1/4, shared between rI, rJ, rK, and rL (rI is 1/2): The 3 reduces to a 1.
    rL: 1/4, shared between rI, rJ, rK, and rL (rI is 1/2): The 3 reduces to a 1.

  • rM: 1/2, shared between rM and rO: Considering rN, the 3 reduces to a 1. If this square is mined, rP must also be mined.
    rO: 1/2, shared between rM and rO: Considering rN, the 3 reduces to a 1. Similarly, 1/2, shared between rO and rP: Considering rN, the 2 reduces to a 1.
    rP: 1/2, shared between rO and rP: Considering rN, the 2 reduces to a 1. If this square is mined, rM must also be mined.

  • rR: UNKNOWN. Depends on rQ.

  • rS: 1/2, shared between rS and rT: The 3 and the 2 both reduce to 1.
    rT: 1/2, shared between rS and rT: The 3 and the 2 both reduce to 1. Similarly, 1/2, shared between rT and rX: The 3 touching 2 exposed mines reduces to 1.
    rX: 1/2, shared between rT and rX (overrides rX/rY/rZ): The 3 touching 2 exposed mines reduces to 1.

  • rU: 1/3, shared between rU, rV, and rW: Considering rT/rX, the 3 jutting out reduces to 1.
    rV: 1/3, shared between rU, rV, and rW: Considering rT/rX, the 3 jutting out reduces to 1.
    rW: 1/3, shared between rU, rV, and rW: Considering rT/rX, the 3 jutting out reduces to 1.

  • rY: 1/3, shared between rX, rY, and rZ (rX is 1/2): The 3 reduces to a 1. Similarly, 1/3, shared between rY, rZ, and bA: The 2 reduces to a 1.
    rZ: 1/3, shared between rX, rY, and rZ (rX is 1/2): The 3 reduces to a 1. Similarly, 1/3, shared between rY, rZ, and bA: The 2 reduces to a 1. Similarly, 1/3, shared between rZ, bA, and bB (bB is 2/4), due to the 1.
    bA: 1/3, shared between rY, rZ, and bA: The 2 reduces to a 1. Similarly, 1/3, shared between rZ, bA, and bB (bB is 2/4), due to the 1.
    [Note: If bA is safe, then rX is safe, rT is mined, and rS is safe.]

  • bB: 2/4, shared between bB, bC, bD, and bE (overrides rZ/bA/bB; bD and bE are 4/6): The 3 reduces to a 2.
    bC: 2/4, shared between bB, bC, bD, and bE (bD and bE are 4/6): The 3 reduces to a 2.
    [Note: If bA is safe, then 2/4 set bB/bC/bD/bE breaks into 1/2 sets bB/bC and bD/bE.]

  • bD: 4/6, shared between bD, bE, bF, bG, bH, and bI (overrides bB/bC/bD/bE): Can't reduce the 4.
    bE: 4/6, shared between bD, bE, bF, bG, bH, and bI (overrides bB/bC/bD/bE): Can't reduce the 4.
    bF: 4/6, shared between bD, bE, bF, bG, bH, and bI: Can't reduce the 4.
    bG: 4/6, shared between bD, bE, bF, bG, bH, and bI: Can't reduce the 4.
    bH: 4/6, shared between bD, bE, bF, bG, bH, and bI: Can't reduce the 4.
    bI: 4/6, shared between bD, bE, bF, bG, bH, and bI: Can't reduce the 4. Similarly, 4/6, shared between bI, bJ, bK, bL, bM, and bN: Can't reduce the 4.
    [Note: bE/bF is also 1/2. bF/bN is also 1/2 (2 reduces to 1). bF/bH/bI/bJ/bN is also 3/5.]

  • bJ: 4/6, shared between bI, bJ, bK, bL, bM, and bN: Can't reduce the 4.
    bK: 4/6, shared between bI, bJ, bK, bL, bM, and bN: Can't reduce the 4.
    bL: 4/6, shared between bI, bJ, bK, bL, bM, and bN: Can't reduce the 4.
    bM: 4/6, shared between bI, bJ, bK, bL, bM, and bN: Can't reduce the 4. Similarly, 2/3, shared between bM, bN, and bO: The 3 reduces to a 2.
    bN: 4/6, shared between bI, bJ, bK, bL, bM, and bN: Can't reduce the 4. Similarly, 2/3, shared between bM, bN, and bO: The 3 reduces to a 2.
    bO: 2/3, shared between bM, bN, and bO: The 3 reduces to a 2.
    [Note: bF/bN is also 1/2 (2 reduces to 1). bF/bH/bI/bJ/bN is also 3/5.

  • bP: 1/2, shared between bO and bP (bO is 2/3 & 4/6): Both the 2 and the 3 reduce to 1s.

  • bQ: 1/3, shared between bP, bQ, and bR (bP is 1/2): The upper 2 reduces to a 1. Similarly, 1/3, shared between bQ, bR, and bS: Considering bT/bU, the lower 2 reduces to a 1.
    bR: 1/3, shared between bP, bQ, and bR (bP is 1/2): The upper 2 reduces to a 1. Similarly, 1/3, shared between bQ, bR, and bS: Considering bT/bU, the lower 2 reduces to a 1.
    bS: 1/3, shared between bQ, bR, and bS: Considering bT/bU, the lower 2 reduces to a 1.

  • bT: 1/2, shared between bT and bU, due to the 1.
    bU: 1/2, shared between bT and bU, due to the 1.

Not sure if any other answers analysed the probabilities for each square. Looked at the question, did some checks, took a look at a couple answers, noticed that they only looked at one or two squares, and decided to make a more complete one.

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  • $\begingroup$ Nice answer, but re: your last sentence, you must not've looked at mine since I definitely looked at more than only one or two squares. :P $\qquad$ $\endgroup$ – tilper Jan 5 '17 at 21:22
  • $\begingroup$ @tilper Nope. xD Only looked at the two I commented on, noticed that one of them didn't get the probabilities quite right, and then decided to figure out the probability for each square. Nice answer, though, it more-or-less rendered mine redundant before the fact. xD $\endgroup$ – Justin Time Jan 5 '17 at 21:28
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Red q location

Same answer as greenturtle3141, except with a picture.

Red q cannot be a mine, because if it were, 3 above p would be "full". Hence, red o or red p (next to the left of red q) cannot be a mine. However, 2 above red o would have only one mine, red n.

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Your mistake was not clicking 5 random squares in 4 quadrants and somewhere in the center. If you can hit all 5 and pass, you're much more likely to make a strong final run of it. If in doubt, manually write up the squares and perform math calculations for each square and determine the least probable fail square. If you're a minesweeper pro, you do that in your head on the go with each and every click. You don't use the question mark clicker or you'd make a better run of it. Red I,J,K, & L are obviously the next clicks. 1 in 4 seems to be your best odds at the moment. The 3/2/3 under Red NOP might have a solution but you may need to grid that out. Cheers.

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    $\begingroup$ I've heard that starting in the corners is better to maximize the chance of not losing to random chance. $\endgroup$ – greenturtle3141 Jan 5 '17 at 4:39
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    $\begingroup$ @greenturtle3141: Some people play to maximize winning percentage, while others play to achieve the fastest win without regard for how many times they get blown up along the way. If one makes five clicks in two seconds and has a 90% chance of getting blown up, but will likely have learned information that would have taken a minute to work out by hand, then even if it takes another two seconds to restart the game after each kaboom one, repeatedly restarting and trying five quick clicks will likely yield one a good (unexploded) position faster than it could have been achieved by careful play. $\endgroup$ – supercat Jan 5 '17 at 19:05
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    $\begingroup$ I always click the 4 corners first. I hate getting all the way to the end where my final move is a 50/50 guess. What a waste of the previous n > 0 minutes. $\endgroup$ – John Jan 5 '17 at 19:25
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    $\begingroup$ 1 in 4 is not your best odds at the moment. Assuming a standard minesweeper game with 99 bombs, the squares about which you have no information have an estimated 14.6% chance to be a bomb, which is significantly better than 1 in 4 (25%). The best choice, assuming the squares already identified in other answers did not provide more/better information, would be a square about which we know nothing, but which would provide additional information about squares for which we already have some information. $\endgroup$ – Makyen Jan 7 '17 at 17:21

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