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General Tabletop Discussion
*Dungeons & Dragons
Blending individual checks into group checks
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<blockquote data-quote="Li Shenron" data-source="post: 9831570" data-attributes="member: 1465"><p>This is a small idea that came to my mind, although I haven't needed to use it yet.</p><p></p><p>It's based on my general fandom of group checks to resolve several exploration tasks, and on the potential worry of players "rounding up" after one of them attempts something and fails.</p><p></p><p>Consider the iconic case of a stuck door: the strongest PC tries to open it, the DM has decided that the result should be random, and calls for an ability check. The strongest PC fail, so another player has the idea of their PC try but fails as well... at this point the DM has already opened the Pandora box of letting others try, and sure enough every other PC is now going to, largely increasing the odds.</p><p></p><p>The problem here is that what matters to me as a DM is the probability (not a precise value but at least a ballpark). If I grant the first PC a check, I set the DC so that there is a certain probability of success. Letting the second PC try is roughly similar to granting advantage, as in allowing the Working Together option (which is DM's call, not player's entitlement). This is already generous but let's say that if I would have allowed Working Together, then I am also probably ok with a second PC try. But this kinds of set the precedent, and how am I now going to justify to a third or fourth PC that I don't allow them? If I do allow them however, my probability ballpark is gone, success almost guaranteed, and why did I even ask a check to the first PC? BUT sometimes I DO want some outcomes to be random and not predetermined by me.</p><p></p><p>So here came my small idea...</p><p></p><p>What if, after the second PC either trying on their own or granting advantage to the first with Working Together, the resolution <strong>blends into a group check? </strong>It means, all the checks now count together and half of them need to succeed. The third PC trying won't be able to succeed alone because the first 2 already failed, so both the 3rd and a 4th need to succeed. If only one of them does, an additional 5th and 6th would be needed. This kinds of smooths the probability out again.</p></blockquote><p></p>
[QUOTE="Li Shenron, post: 9831570, member: 1465"] This is a small idea that came to my mind, although I haven't needed to use it yet. It's based on my general fandom of group checks to resolve several exploration tasks, and on the potential worry of players "rounding up" after one of them attempts something and fails. Consider the iconic case of a stuck door: the strongest PC tries to open it, the DM has decided that the result should be random, and calls for an ability check. The strongest PC fail, so another player has the idea of their PC try but fails as well... at this point the DM has already opened the Pandora box of letting others try, and sure enough every other PC is now going to, largely increasing the odds. The problem here is that what matters to me as a DM is the probability (not a precise value but at least a ballpark). If I grant the first PC a check, I set the DC so that there is a certain probability of success. Letting the second PC try is roughly similar to granting advantage, as in allowing the Working Together option (which is DM's call, not player's entitlement). This is already generous but let's say that if I would have allowed Working Together, then I am also probably ok with a second PC try. But this kinds of set the precedent, and how am I now going to justify to a third or fourth PC that I don't allow them? If I do allow them however, my probability ballpark is gone, success almost guaranteed, and why did I even ask a check to the first PC? BUT sometimes I DO want some outcomes to be random and not predetermined by me. So here came my small idea... What if, after the second PC either trying on their own or granting advantage to the first with Working Together, the resolution [B]blends into a group check? [/B]It means, all the checks now count together and half of them need to succeed. The third PC trying won't be able to succeed alone because the first 2 already failed, so both the 3rd and a 4th need to succeed. If only one of them does, an additional 5th and 6th would be needed. This kinds of smooths the probability out again. [/QUOTE]
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