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26

You are trying to kill a fly with a cannon. This answer established that when 3 people make a group check, you need two successes to call it a successful group check. We don't need a deep statistical analysis or AnyDice programs here. A simple heuristic proof sketch should be sufficient to establish that for \$k\$ odd, \$k+1\$ has a higher probability of ...


12

Yes you're right but I'll try to explain why The TLDR is that with an even number and its next odd number you have the same number of failures accepted. Party of 4 can have at most 2 failures but so can a party of 5. This doesn't outpace the increase in options you get with more party members. There are two different parts of a binomial distrubution at play ...


11

This is correct The cause of this oddity is the rounding. Specifically here we are rounding up the number of successes necessary. So, let's look at it from number of attempts to get enough successes and starting from say 5. We then need 3 successes with 5 attempts. But since this is already rounded up, getting a 6th participant gets us an additional attempt, ...


3

Now, 2 people are always going to be better than one because the rolling mechanism is identical to advantage (i.e. identical to one helping the other). But for 4 to be worse than 1; the target has to be at least 17. Is this analysis correct? Yes. As other answers have noted, adding an extra member to an odd-sized group can never reduce the group's chance of ...


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