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Talk:Voltorb Flip: Difference between revisions
→Easier way?: new section
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::I think it's dependent on what the chance is, based on what tiles are already uncovered. I haven't played in a while, but if I remember correctly that message came up most often when I was uncovering a tile which I had basically narrowed down to being a 50-50 chance of being a Voltorb or a good number. --[[User:AndyPKMN|Andy<sup>P</sup><sub>K</sub><sup>M</sup><sub>N</sub>]] [[User talk:AndyPKMN|(talk)]] 15:12, 25 April 2011 (UTC) | ::I think it's dependent on what the chance is, based on what tiles are already uncovered. I haven't played in a while, but if I remember correctly that message came up most often when I was uncovering a tile which I had basically narrowed down to being a 50-50 chance of being a Voltorb or a good number. --[[User:AndyPKMN|Andy<sup>P</sup><sub>K</sub><sup>M</sup><sub>N</sub>]] [[User talk:AndyPKMN|(talk)]] 15:12, 25 April 2011 (UTC) | ||
:::From my experience, the message pops up whenever there's only one numbered tile left in the row or column. [[User:Ztobor|Ztobor]] 14:06, 26 April 2011 (UTC) | :::From my experience, the message pops up whenever there's only one numbered tile left in the row or column. [[User:Ztobor|Ztobor]] 14:06, 26 April 2011 (UTC) | ||
== Easier way? == | |||
Is it not much simpler to say that if the number of voltorbs and the total of the tiles equal 5, the row is automatically dead, instead of incorporating all of this algebra? Or is this strategy not even mentioned and I didn't notice because I couldn't understand the algebraic part? [[User:Frenchhorn|Frenchhorn]] 01:32, 30 December 2011 (UTC) | |||