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Review of Twilight: 2000: You’re on Your Own, Good Luck
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<blockquote data-quote="aramis erak" data-source="post: 8513179" data-attributes="member: 6779310"><p>to figure the expected number of successes on a die (keep it in fractions; it's easier than decimals): total the number of successes on the die for the numerator, and the sides as denominator - then just add the number of dice together.</p><p>So the standard YZE d6 is 1/6 success. It adds 0 to the minimum, and 1 to the maximum. The arithmetic mean should be the same as the expected, but real world dice are not truly fair most of the time.</p><p>Comparing the dice: [x, ..., y] brackets indicate array of sides, braces the range component. The fraction is the expected result contribution.</p><table style='width: 100%'><tr><td>Die</td><td>T2K</td><td>FL</td></tr><tr><td>d6</td><td>[0, 0, 0, 0, 0, 1] = 1/6 {0-1}</td><td>[0, 0, 0, 0, 0, 1] = 1/6 {0-1}</td></tr><tr><td>d8</td><td>[0, 0, 0, 0, 0, 1, 1, 1] = 3/8 {0-1}</td><td>[0, 0, 0, 0, 0, 1, 1, 2] = 4/8 {0-2}</td></tr><tr><td>d10</td><td>[0, 0, 0, 0, 0, 1, 1, 1, 1, 2] = 6/10 {0-2}</td><td>[0, 0, 0, 0, 0, 1, 1, 2, 2, 3] = 9/10 {0-3}</td></tr><tr><td>d12</td><td>[0, 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2] = 10/12 {0-2}</td><td>[0, 0, 0, 0, 0, 1, 1, 2, 2, 3, 3, 4]= 16/12 {0-4}</td></tr></table><p></p><p>Note that expected result is only valid for pools or repeated rolls; it roughly corresponds to the arithmetic mean expected on the initial roll. For the expected results for a pool, just add the dice's expected together. For the ranges, add the low ends together and add the high ends together.</p><p>With pushing, the T2K dice become</p><p>d6 [[0,0,0,0,0,0}, [0, 0, 0, 0, 0, 1],[0, 0, 0, 0, 0, 1],[0, 0, 0, 0, 0, 1],[0, 0, 0, 0, 0, 1],[1, 1, 1, 1, 1, 1]] = 10/36 each (vs 6/36 unpushed)</p><p>d6 with no negative on 1 is 11/36, since the first group becomes [0, 0, 0, 0, 0, 1]</p><p>d8 [[0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 1, 1, 1], [0, 0, 0, 0, 0, 1, 1, 1], [0, 0, 0, 0, 0, 1, 1, 1], [0, 0, 0, 0, 0, 1, 1, 1], [1, 1, 1, 1, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1, 1, 1]] 36/64=9/16</p><p>d10: [[0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 1, 1, 1, 1, 2], [0, 0, 0, 0, 0, 1, 1, 1, 1, 2], [0, 0, 0, 0, 0, 1, 1, 1, 1, 2], [0, 0, 0, 0, 0, 1, 1, 1, 1, 2], [1, 1, 1, 1, 1, 1, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1, 1, 1, 1, 1,], [1, 1, 1, 1, 1, 1, 1, 1, 1, 1],[2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2]]= 76/100 =19/25</p><p>d12 [[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2], [0, 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2],[0, 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2],[0, 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2], , [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], [2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2], [2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2], [2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2]] = (1×0)+(4×10)+(4×10)+3×20)=0+40+40+60=140/144.</p><p>Note that I'm making the inner dice reflect the final outcome; the first die is the outer set; each reroll (or the lack thereof for 1 and 6-12 is because no reroll, hence all the possibilities are the same result as the initial die would be.</p><p></p><p>Not showing the work as detailed on all the FL dice</p><p>d6: 1/6 unpushed, 10/36 pushed</p><p>d8: 4/8=1/2 unpushed, [1×[0,0,0,0,0,0,0,0], 4×[0,0,0,0,0,1,1,2], 2×[1,1,1,1,1,1,1,1], 1×[2,2,2,2,2,2,2,2]]=(0+(4×4)+(2×8)+(1×16))/8²=48/64, = 3/4</p><p>d10: 9/10 unpushed,. [[1×[0,0,0,0,0,0,0,0,0,0], 4×[0,0,0,0,0,1,1,2,2,3], 2×[1,1,1,1,1,1,1,1,1, 1], 2×[2,2,2,2,2,2,2,2,2,2], 1×[3,3,3,3,3,3,3,3,3,3]]=(0+(4×9)+(2×10)+(2×20)+(1×30))/10²=(0+36+20+40+30)/100=126/100 =63/50 = 1 13/50</p><p>d12: 16/12 = 4/3 unpushed. [1×[0,0,0,0,0,0,0,0,0,0,0,0], 4×[0,0,0,0,0,1,1,2,2,3,3,4], 2×[12×1], 2×[12×2], 2×[12×3], 1×[12×4]]=(0+(4×16)+(2×12)+(2×24)+(2×36)+(1×48))/12²=(0+64+24+48+72+48)/144=256/144=16/9 = 1 7/9</p><p></p><p><strong>Important Note: The expected result normally doesn't tell you the odds of success. It will tell you only if you're above or below 50%...</strong></p><p></p><p>finding it provides the needed information to work out actual odds. As a GM, I've only needed to eyeball the odds. Is it likely they're going to succeed? At C+C, unpushed? 3/4 expectation is below the 1 needed, so likely to fail. C+B, 3/8 +6/10= 15/40+24/40= 39/40 - pretty close to 1 success, so just under 50%. Pushed C+C is 9/16+19/25... time for prime factors... LCD=400, so... (25×9)+(16×19)/400=(225+304)/400=529/400= 1 129/400. Pretty good but still plenty of chance for failure. Note the range is 0-4...</p><p></p><p>It's also a case of do the hard part once, and be able to eyeball it. In YZE games with variable numbers of successes needed, instead of variable pools, the expected result lets you eyeball very quickly. 7d? Expected= 1 1/6... but a difficulty of 2s? that's 1/6 expected. Long odds. (and not the same odds as 1d needing 1s.) It's a rough eyeball route.</p></blockquote><p></p>
[QUOTE="aramis erak, post: 8513179, member: 6779310"] to figure the expected number of successes on a die (keep it in fractions; it's easier than decimals): total the number of successes on the die for the numerator, and the sides as denominator - then just add the number of dice together. So the standard YZE d6 is 1/6 success. It adds 0 to the minimum, and 1 to the maximum. The arithmetic mean should be the same as the expected, but real world dice are not truly fair most of the time. Comparing the dice: [x, ..., y] brackets indicate array of sides, braces the range component. The fraction is the expected result contribution. [TABLE] [TR] [TD]Die[/TD] [TD]T2K[/TD] [TD]FL[/TD] [/TR] [TR] [TD]d6[/TD] [TD][0, 0, 0, 0, 0, 1] = 1/6 {0-1}[/TD] [TD][0, 0, 0, 0, 0, 1] = 1/6 {0-1}[/TD] [/TR] [TR] [TD]d8[/TD] [TD][0, 0, 0, 0, 0, 1, 1, 1] = 3/8 {0-1}[/TD] [TD][0, 0, 0, 0, 0, 1, 1, 2] = 4/8 {0-2}[/TD] [/TR] [TR] [TD]d10[/TD] [TD][0, 0, 0, 0, 0, 1, 1, 1, 1, 2] = 6/10 {0-2}[/TD] [TD][0, 0, 0, 0, 0, 1, 1, 2, 2, 3] = 9/10 {0-3}[/TD] [/TR] [TR] [TD]d12[/TD] [TD][0, 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2] = 10/12 {0-2}[/TD] [TD][0, 0, 0, 0, 0, 1, 1, 2, 2, 3, 3, 4]= 16/12 {0-4}[/TD] [/TR] [/TABLE] Note that expected result is only valid for pools or repeated rolls; it roughly corresponds to the arithmetic mean expected on the initial roll. For the expected results for a pool, just add the dice's expected together. For the ranges, add the low ends together and add the high ends together. With pushing, the T2K dice become d6 [[0,0,0,0,0,0}, [0, 0, 0, 0, 0, 1],[0, 0, 0, 0, 0, 1],[0, 0, 0, 0, 0, 1],[0, 0, 0, 0, 0, 1],[1, 1, 1, 1, 1, 1]] = 10/36 each (vs 6/36 unpushed) d6 with no negative on 1 is 11/36, since the first group becomes [0, 0, 0, 0, 0, 1] d8 [[0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 1, 1, 1], [0, 0, 0, 0, 0, 1, 1, 1], [0, 0, 0, 0, 0, 1, 1, 1], [0, 0, 0, 0, 0, 1, 1, 1], [1, 1, 1, 1, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1, 1, 1]] 36/64=9/16 d10: [[0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 1, 1, 1, 1, 2], [0, 0, 0, 0, 0, 1, 1, 1, 1, 2], [0, 0, 0, 0, 0, 1, 1, 1, 1, 2], [0, 0, 0, 0, 0, 1, 1, 1, 1, 2], [1, 1, 1, 1, 1, 1, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1, 1, 1, 1, 1,], [1, 1, 1, 1, 1, 1, 1, 1, 1, 1],[2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2]]= 76/100 =19/25 d12 [[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2], [0, 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2],[0, 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2],[0, 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2], , [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], [2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2], [2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2], [2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2]] = (1×0)+(4×10)+(4×10)+3×20)=0+40+40+60=140/144. Note that I'm making the inner dice reflect the final outcome; the first die is the outer set; each reroll (or the lack thereof for 1 and 6-12 is because no reroll, hence all the possibilities are the same result as the initial die would be. Not showing the work as detailed on all the FL dice d6: 1/6 unpushed, 10/36 pushed d8: 4/8=1/2 unpushed, [1×[0,0,0,0,0,0,0,0], 4×[0,0,0,0,0,1,1,2], 2×[1,1,1,1,1,1,1,1], 1×[2,2,2,2,2,2,2,2]]=(0+(4×4)+(2×8)+(1×16))/8²=48/64, = 3/4 d10: 9/10 unpushed,. [[1×[0,0,0,0,0,0,0,0,0,0], 4×[0,0,0,0,0,1,1,2,2,3], 2×[1,1,1,1,1,1,1,1,1, 1], 2×[2,2,2,2,2,2,2,2,2,2], 1×[3,3,3,3,3,3,3,3,3,3]]=(0+(4×9)+(2×10)+(2×20)+(1×30))/10²=(0+36+20+40+30)/100=126/100 =63/50 = 1 13/50 d12: 16/12 = 4/3 unpushed. [1×[0,0,0,0,0,0,0,0,0,0,0,0], 4×[0,0,0,0,0,1,1,2,2,3,3,4], 2×[12×1], 2×[12×2], 2×[12×3], 1×[12×4]]=(0+(4×16)+(2×12)+(2×24)+(2×36)+(1×48))/12²=(0+64+24+48+72+48)/144=256/144=16/9 = 1 7/9 [B]Important Note: The expected result normally doesn't tell you the odds of success. It will tell you only if you're above or below 50%...[/B] finding it provides the needed information to work out actual odds. As a GM, I've only needed to eyeball the odds. Is it likely they're going to succeed? At C+C, unpushed? 3/4 expectation is below the 1 needed, so likely to fail. C+B, 3/8 +6/10= 15/40+24/40= 39/40 - pretty close to 1 success, so just under 50%. Pushed C+C is 9/16+19/25... time for prime factors... LCD=400, so... (25×9)+(16×19)/400=(225+304)/400=529/400= 1 129/400. Pretty good but still plenty of chance for failure. Note the range is 0-4... It's also a case of do the hard part once, and be able to eyeball it. In YZE games with variable numbers of successes needed, instead of variable pools, the expected result lets you eyeball very quickly. 7d? Expected= 1 1/6... but a difficulty of 2s? that's 1/6 expected. Long odds. (and not the same odds as 1d needing 1s.) It's a rough eyeball route. [/QUOTE]
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