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<blockquote data-quote="Dega Moo" data-source="post: 1102648" data-attributes="member: 19930"><p>You can reduce likelihood of a tie using one or more 'twists' on the basic game:</p><p></p><p>1) multiply the the die - will get you more possibles</p><p>2) use colored die and declare one color the first number and the other color the second number. So a red and white die with red the first number and white the second red comes up 4 and white comes up 2 then the result is 42. If red is a 1 and white is a 2 then the result is 12.</p><p>3) apply a factor for each stop and multiply results by that factor at each stop</p><p>4) pick a total number they need and the closest to that number wins. Let them call a die negative or positive, before the roll of each. Multiply die results, add to total.</p><p></p><p>Really though, I like the beer solution best. <img src="data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7" class="smilie smilie--sprite smilie--sprite1" alt=":)" title="Smile :)" loading="lazy" data-shortname=":)" /></p></blockquote><p></p>
[QUOTE="Dega Moo, post: 1102648, member: 19930"] You can reduce likelihood of a tie using one or more 'twists' on the basic game: 1) multiply the the die - will get you more possibles 2) use colored die and declare one color the first number and the other color the second number. So a red and white die with red the first number and white the second red comes up 4 and white comes up 2 then the result is 42. If red is a 1 and white is a 2 then the result is 12. 3) apply a factor for each stop and multiply results by that factor at each stop 4) pick a total number they need and the closest to that number wins. Let them call a die negative or positive, before the roll of each. Multiply die results, add to total. Really though, I like the beer solution best. :) [/QUOTE]
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