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<blockquote data-quote="willows" data-source="post: 4862183" data-attributes="member: 79929"><p>Here's a relatively easy way to do this without probability gymnastics.</p><p></p><p>Make several stat arrays using the same point-buy total. It doesn't matter how many.</p><p></p><p>Make a list of the ability scores in whatever order you want. It doesn't matter what order they're in.</p><p></p><p>Finally get two sets of d6 and d4, in two colors you can tell apart easily. Designate one color 'high' and one 'low'.</p><p></p><p>To create a character, first roll a d-whatever (depending on how many arrays you have) to select a stat array.</p><p></p><p>Then roll 2d6. Assign the highest value in the array to the stat indicated by the high die, and the second-highest according to the low die. If you roll doubles, use the indicated stat for the high value and the player gets to choose the second value.</p><p></p><p>Cross those ability scores off your list.</p><p></p><p>Use the same procedure, with the matching d4s, for the third and fourth-highest stats.</p><p></p><p>Just flip a coin for the last two.</p></blockquote><p></p>
[QUOTE="willows, post: 4862183, member: 79929"] Here's a relatively easy way to do this without probability gymnastics. Make several stat arrays using the same point-buy total. It doesn't matter how many. Make a list of the ability scores in whatever order you want. It doesn't matter what order they're in. Finally get two sets of d6 and d4, in two colors you can tell apart easily. Designate one color 'high' and one 'low'. To create a character, first roll a d-whatever (depending on how many arrays you have) to select a stat array. Then roll 2d6. Assign the highest value in the array to the stat indicated by the high die, and the second-highest according to the low die. If you roll doubles, use the indicated stat for the high value and the player gets to choose the second value. Cross those ability scores off your list. Use the same procedure, with the matching d4s, for the third and fourth-highest stats. Just flip a coin for the last two. [/QUOTE]
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