Shoutbox

I just noticed a strange piece of logic in [o] which may sacrifice a lot of his chances:


In that screenshot [o] was the last to move, and I had opened up that corner square bordered in red.  Surely that square to the south-west of where I went would have a much greater chance of being a mine than where [o] went, because it would satisfy 4 of the 9 effective statements made by the 4 already-revealed squares (each one being "there is a mine next to me".  Instead [o] always tries to open a square which has as few opened squares touching it as possible, as in the screenshot.  Now unless there is a huge gaping gap somewhere else on the field, it is less likely that each of these 4 squares have mines adjacent to them that aren't shared by other squares, because it means there are more mines.

I'm too tired to figure out the exact math that needs to go in, but to successfully account for the combined probability of each of these squares, the entire situation would have to be supposed.  For each entire theoretical situation in which all the shown squares are satisfied, the number of squares without mines must be compared to the number of squares with mines, and then compared to the average number squares to every one mine, which is 256/51 = 5.0196078431372549019607843137255 .  I think that the probability of each combination of mines and non-mines would then be given a deviation from that 5.01... and the standard deviation worked out, then each deviation translated into a standardised score, which is the comparitive value of each combination, and then for each combination the standardised score being applied to each square involved in it, and for each square its probability in relation to an adjacent square such as a "2" being multiplied by the standardised score.  If all this hasn't already completely confused me, the average standardised score * probability across all the combinations would be found for each square which would result in a true probability.

Hope that's at least vaguely understandable... incidentally neither the square I thought was more likely nor the one [o] thought was more likely was a mine though as a general rule it is.