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<blockquote data-quote="orsal" data-source="post: 2003069" data-attributes="member: 16016"><p>I agree with Darkness about the error in Pbartender's calculation. If you are doing anything with a DC which you will make with a 21 and fail with a 1, then an extra +1 to your bonus will add 5% to your chance of success. If the DC is out of that range, then an extra +1 will do nothing.</p><p></p><p>As Thanee noted, there are 20x20=400 possible outcomes, all with the same probability. That means the probability of any event is (# of outcomes)/400. Let's call the player with the higher bonus A, the player with the lower bonus B, and the difference in their bonuses x. Now A will lose iff (B's roll)-(A's roll) is greater than x. Count up the number of ways that can happen. There is 1 way to get a difference of 19, namely A gets 1 and B gets 20. There are 2 ways to get a difference of 18, namely (1,19) and (2,20). And so on -- there are (20-t) ways for B to exceed A's roll by exactly t. So there are</p><p></p><p>1 + 2 + 3 + . . . + (20-(x+1)) </p><p></p><p>ways for B to exceed A's roll by x+1 or more, and</p><p></p><p>1 + 2 + 3 + . . . + (20-x)</p><p></p><p>ways for B to exceed A's roll by x or more. The former is the number to use of A will win ties (official rules); the latter, if B will win ties (variant rule).</p><p></p><p>Now, 1 + 2 + . . . + k = k(k+1)/2, so:</p><p></p><p>If A wins ties, there are (19-x)(20-x)/2 outcomes which have x losing, with a total probability of (19-x)(20-x)/800.</p><p></p><p>If B wins ties, there are (20-x)(21-x)/2 outcomes which have x losing, with a total probability of (20-x)(21-x)/800.</p></blockquote><p></p>
[QUOTE="orsal, post: 2003069, member: 16016"] I agree with Darkness about the error in Pbartender's calculation. If you are doing anything with a DC which you will make with a 21 and fail with a 1, then an extra +1 to your bonus will add 5% to your chance of success. If the DC is out of that range, then an extra +1 will do nothing. As Thanee noted, there are 20x20=400 possible outcomes, all with the same probability. That means the probability of any event is (# of outcomes)/400. Let's call the player with the higher bonus A, the player with the lower bonus B, and the difference in their bonuses x. Now A will lose iff (B's roll)-(A's roll) is greater than x. Count up the number of ways that can happen. There is 1 way to get a difference of 19, namely A gets 1 and B gets 20. There are 2 ways to get a difference of 18, namely (1,19) and (2,20). And so on -- there are (20-t) ways for B to exceed A's roll by exactly t. So there are 1 + 2 + 3 + . . . + (20-(x+1)) ways for B to exceed A's roll by x+1 or more, and 1 + 2 + 3 + . . . + (20-x) ways for B to exceed A's roll by x or more. The former is the number to use of A will win ties (official rules); the latter, if B will win ties (variant rule). Now, 1 + 2 + . . . + k = k(k+1)/2, so: If A wins ties, there are (19-x)(20-x)/2 outcomes which have x losing, with a total probability of (19-x)(20-x)/800. If B wins ties, there are (20-x)(21-x)/2 outcomes which have x losing, with a total probability of (20-x)(21-x)/800. [/QUOTE]
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