NOW LIVE! Today's the day you meet your new best friend. You don’t have to leave Wolfy behind... In 'Pets & Sidekicks' your companions level up with you!

Reply to thread

Message: <blockquote data-quote="billd91" data-source="post: 5156559" data-attributes="member: 3400">That's one way to interpret it.But when you look at the instructions about Party B surprising party A on 5 in 6 less 1 in 6 and translate that to other dice, it's not that clear. It's easy to see that, compared to the normal 2 in 6 of being surprised, that party A's 1 in 6 better at avoiding the surprise. But then, the dice size is the same.If group C was surprised 1 in 8, party B's 5 in 6 reduces to what? If 2 in 6 is normal, what do we subtract from the 5 in 6 to arrive at the new surprise value?You interpret it in the other direction. 5 in 6 is 3 in 6 better than the base assumption so you're saying they surprise Party C by adding 3 in 8 to get 4 in 8. At least that's what I'm interpreting from your post. You just concern yourself with the numerators.But the operations, as described by the DMG, can't be followed without doing a significantly different calculation. I can't reduce 5 in 6 by the difference between normal and 1 in 8 without putting them in the same terms and changing the die. I can take the common denominator approach and see that Party B's 20 in 24 is reduced by 5 in 24 (normal's 8 in 24 - Party C's 3 in 24) to 15/24 or 5 in 8. Notice how that's a different outcome from your interpretation at 4 in 8.The difference in our interpretations gets even weirder with the monk whose surprise die is d%. Party B's massive 5 in 6 surprise advantage becomes virtually worthless against monks. A 3rd level monk is surprised 30% of the time. 30 in 100. Are you really suggesting adding just 3 to that and turning a surprise percentage that's normally 50% higher into merely 3% higher simply because the enemies are 3rd level monks? This is why 1e surprise was such a mess.</blockquote>

Verification