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<blockquote data-quote="Vinyafod" data-source="post: 507091" data-attributes="member: 6206"><p>For such things it is best to write down the chances/ combinations of possible die rolls and simply start counting then:</p><p></p><p>possible outcomes of 2d10 (not summed up!! but percentile roll)</p><p></p><p>[code]</p><p></p><p> 0 1 2 3 4 5 6 7 8 9 </p><p>0 00 [b][color=red]01 02 03 04 05 06 07 08 09[/color]</p><p>1[color=red] 10 11 12 12 14 15 16 17 18 19[/color]</p><p>2[color=red] 20 21 22 23 24 25[/b][/color] 26 27 28 29 </p><p>3 30 31 32 33 34 35 36 37 38 38</p><p>4 40 41 42 43 44 45 46 47 48 49 </p><p>5 50 51 52 53 54 55 56 57 58 59</p><p>6 60 61 62 63 64 65 66 67 68 69</p><p>7 70 71 72 73 74 75 76 77 78 79</p><p>8 80 81 82 83 84 85 86 87 88 89</p><p>9 90 91 92 93 94 95 96 97 98 99[/code]</p><p></p><p>We have a hundred different combinations here. 25 of those gives scores of 25 or lower. 25 out of 100 =25/100 = 1/4 = 25%</p><p></p><p>d4:</p><p></p><p>1</p><p>2</p><p>3</p><p>4</p><p></p><p>We have 4 "combinations". One of those gives a 1. 1 out of 4 = 1/4 = 25% </p><p></p><p>That is absolutely correct and nothing to argue about. Chances are <strong>exactly</strong> the same. Believe it or not. if that won't convice you, nothing will.</p><p></p><p>To take the approach you tried in your last post:</p><p></p><p>Assuming numbers 1 to 25 are not wanted...</p><p>The first die always displays the 10-digit, the second the 1-digit.</p><p>First roll = 0 (chance to happen: 10%), second roll must be 0 in order to succeed (chance to happen: 10%) -> 0.1*0,1=0.01 -> 1% chance to succed on a first roll of a 0</p><p></p><p>First die shows a 1 (10% chance), nothing can save you now -> plain 0% chance to suceed -> 0%</p><p></p><p>First die shows a 2 (10% chance), 6 to 9 will succeed (40% chance) -> 0.1*0.4=0.04 -> 4% chance to succeed on a first roll of a 2</p><p></p><p>First die show 3 to 9 (70% chance), all those will automatically succeed -> plain 70% chance</p><p></p><p>1% + 4% + 70% = 75% 3 out of 4 succeed 75% chance to succeed.</p><p></p><p>Thats the maths whith which you calculate it!</p><p>Hope that will convince you. <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f642.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":)" title="Smile :)" data-smilie="1"data-shortname=":)" /></p></blockquote><p></p>
[QUOTE="Vinyafod, post: 507091, member: 6206"] For such things it is best to write down the chances/ combinations of possible die rolls and simply start counting then: possible outcomes of 2d10 (not summed up!! but percentile roll) [code] 0 1 2 3 4 5 6 7 8 9 0 00 [b][color=red]01 02 03 04 05 06 07 08 09[/color] 1[color=red] 10 11 12 12 14 15 16 17 18 19[/color] 2[color=red] 20 21 22 23 24 25[/b][/color] 26 27 28 29 3 30 31 32 33 34 35 36 37 38 38 4 40 41 42 43 44 45 46 47 48 49 5 50 51 52 53 54 55 56 57 58 59 6 60 61 62 63 64 65 66 67 68 69 7 70 71 72 73 74 75 76 77 78 79 8 80 81 82 83 84 85 86 87 88 89 9 90 91 92 93 94 95 96 97 98 99[/code] We have a hundred different combinations here. 25 of those gives scores of 25 or lower. 25 out of 100 =25/100 = 1/4 = 25% d4: 1 2 3 4 We have 4 "combinations". One of those gives a 1. 1 out of 4 = 1/4 = 25% That is absolutely correct and nothing to argue about. Chances are [b]exactly[/b] the same. Believe it or not. if that won't convice you, nothing will. To take the approach you tried in your last post: Assuming numbers 1 to 25 are not wanted... The first die always displays the 10-digit, the second the 1-digit. First roll = 0 (chance to happen: 10%), second roll must be 0 in order to succeed (chance to happen: 10%) -> 0.1*0,1=0.01 -> 1% chance to succed on a first roll of a 0 First die shows a 1 (10% chance), nothing can save you now -> plain 0% chance to suceed -> 0% First die shows a 2 (10% chance), 6 to 9 will succeed (40% chance) -> 0.1*0.4=0.04 -> 4% chance to succeed on a first roll of a 2 First die show 3 to 9 (70% chance), all those will automatically succeed -> plain 70% chance 1% + 4% + 70% = 75% 3 out of 4 succeed 75% chance to succeed. Thats the maths whith which you calculate it! Hope that will convince you. :) [/QUOTE]
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