tag:blogger.com,1999:blog-613387965721174471.post6605376600614912409..comments2023-05-31T06:48:28.561-04:00Comments on the Wood between Worlds: Opposed Checks in D&D are the Same as Coin FlipsReecehttp://www.blogger.com/profile/10691425406901685427noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-613387965721174471.post-33674076208776689652015-08-01T12:48:19.874-04:002015-08-01T12:48:19.874-04:00Probabilistically, they're the same. That doe...Probabilistically, they're the same. That doesn't mean that your players will respond the same way, though :PReecehttps://www.blogger.com/profile/10691425406901685427noreply@blogger.comtag:blogger.com,1999:blog-613387965721174471.post-47069763812638226172015-08-01T10:46:53.940-04:002015-08-01T10:46:53.940-04:00Cool, thanks for that. You got the question right ...Cool, thanks for that. You got the question right yeah. So if you are rolling to determine the difficulty of something, you might as well just use the median value of the die. Thats pretty interesting.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-613387965721174471.post-15161422628720712122015-08-01T00:13:11.187-04:002015-08-01T00:13:11.187-04:00Apparently the math does encode!
Let me take away...Apparently the math does encode!<br /><br />Let me take away all of the fun of doing the sum yourself:<br /><br />$$\sum_{x=1}^n \Pr(X_2x)\cdot\Pr(X_1=x) = \sum_{x=1}^n \left(\sum_{k=1}^{x-1} \frac{1}{n}\right)\cdot \frac{1}{n} = \frac{1}{n^2} \sum_{x=1}^n \sum_{k=1}^{x-1} 1 = \frac{1}{n^2} \sum_{x=1}^n (x-1) = \frac{1}{n^2} \frac{n(n-1)}{2} = \frac{1}{2} - \frac{1}{2n}.$$<br /><br /><br />For $nReecehttps://www.blogger.com/profile/10691425406901685427noreply@blogger.comtag:blogger.com,1999:blog-613387965721174471.post-33929196495420241662015-07-31T23:58:56.134-04:002015-07-31T23:58:56.134-04:00Let me make sure I understand the question.
Your ...Let me make sure I understand the question.<br /><br />Your attacking skill is 19, so you need to roll 19 or under to hit? Or your block skill is 12, so you need to roll under 12 to block? Is that the gist?<br /><br />If you are rolling a 20-sided die, then the probability of rolling less than $x$ is $p(x-1)$ (I hope the math encodes correctly there). That's based on the PC, and doesn'Reecehttps://www.blogger.com/profile/10691425406901685427noreply@blogger.comtag:blogger.com,1999:blog-613387965721174471.post-43343050550474954242015-07-31T22:25:45.862-04:002015-07-31T22:25:45.862-04:00This is very interesting, I also never thought abo...This is very interesting, I also never thought about the sum first n positive integers series like that. But I was wondering, if instead of the two rolls trying to beat each other, they are instead trying to succeed at beating a target number (akin to Basic roleplaying where to make a successful attack you roll under your attacking skill and to block a successful attack you can attempt to roll Anonymousnoreply@blogger.com