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by qiemem
3420 days ago
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I don't follow. This would only work if the distribution of numbers was predetermined and influenced the distribution of mines. I was under the impression that the mines were uniformly distributed though. So in a "T" scenario, it really is 50-50. The fact that 4s are more rare doesn't matter at that point. Sort of like in a series of coin flips, of you've flipped 5 heads in a row, tails isn't more likely to come since 6 heads are rare: it's still just 50-60. |
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Try to apply the tactic I mentioned to solve the minefield in the article.