The rules for group checks say:

To make a group ability check, everyone in the group makes the ability check. If at least half the group succeeds, the whole group succeeds.

So, for groups of 1 and 2, 1 needs to succeed. For groups of 3 and 4, 2 need to succeed. And so on.

Assuming all members of the group have an equal modifier (unlikely for PCs but de regur for monsters).

Running this anydice:

loop N over {1..8} {
  output ((N+1)/2)@Nd20


![enter image description here

Each of the odd-numbered groups has a symmetric distribution centred on 10.5 - this makes sense because we are looking at the middle dice in the group. That is, the median and the mean are always 10.5. As you add more creatures, the variance decreases which makes you less likely to succeed at hard things but more likely to succeed at easy things.

With the even-numbered groups, we are looking at the \$n\over2\$th die which will always be higher. So even-numbered groups are at an advantage over the next-smaller odd-numbered group. That means that 2 creatures can, say, Hide more easily than 1, 4 more easily than 3 and so on.

Wierd practical outcomes

For realistically sized parties, it's better to have an even number (the less the better) than any odd number but if you have to go odd, go as big as you can.

![enter image description here

However, when the target is 11, odd-numbered groups always perform the same but even-number groups are always better. For 12, odd-numbers get worse and even-numbers get also get worse but 8 is still better than one.

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?


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, at no cost.

This same phenomenon can be seen in a number of board games where a number scales with the number of players. The example that springs to mind is Eldritch Horror where one often need to collect thingies equal to half the number of players rounded up, to the point where our group would actively only play with an even number of players, including with a phantom player to bring us up to even.

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    \$\begingroup\$ Out of curiosity, did you consider changing the rule to be round down instead of adding a phantom player? \$\endgroup\$ – deepy Jun 9 at 12:32

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 success.

  • For \$k\$ odd, we need \$\displaystyle{\left\lfloor\frac{k}{2} \right\rfloor +1}\$ successes out of \$k\$ attempts.
  • For \$k+1\$, we need \$\displaystyle{\left\lfloor\frac{k}{2}\right\rfloor+1}\$ successes out of \$k+1\$ attempts.

It is obvious that requiring the same number of successes but with an extra attempt implies a higher probability of success (assuming the \$k+1th\$ attempt has a non-zero probability of success). A rigorous proof of this fact is left as an exercise to the reader (Hint: the events are independent, see here for a solution).

Finally, your intuition that larger parties have an easier time is correct. For an even number of adventurer's, we need half as many successes. For \$k\$ odd, we have:

$$\lim_{k\rightarrow\infty}\frac{1}{k}\displaystyle{\left(\left\lfloor\frac{k}{2} \right\rfloor +1\right)}=\frac{1}{2}$$

As \$k\$ grows, the discrepancy between odd parties and even parties becomes vanishingly small.

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    \$\begingroup\$ Are you sure that your limit is correct? \$\endgroup\$ – Eddymage Jun 8 at 21:33
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    \$\begingroup\$ @Eddymage Good catch, left out some parentheses. \$\endgroup\$ – Thomas Markov Jun 8 at 21:36

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 here.

The number of party members there are relates to the number of options you have for success (i.e. for a party of three with one failure you can have [Fail,Succeed,Succeed], [S,F,S] or [S,S,F] - three different options).

This is given by the binomial coefficient. For a party of \$N\$ with \$k\$ failures we have:

$$ {N\choose k}=\frac{N!}{k!(N-k)!}={}_N \mathrm{ C }_k $$

Different ways in which we can achieve that.

There are also the number of different combinations which make a success. For example with a party of 4 you can have 2,1 or 0 failures - for a party of 5 this doesn't change and likewise for every even number and the next odd number.

The probability of a single event for a party of \$N\$ with \$k\$ failures is: $$ P(N,k) = P_{s}^{N-k} \times P_{f}^{k} $$

Where \$P_{s}\$ is the probability of success and \$P_{f}\$ is the probability of failure. If we're going from an even number to an odd number, however, we would have the same possibilities for \$k\$ but an extra \$N\$ to take into account.

$$ P(4,2) = P_{s}^{4-2} \times P_{f}^{2}= P_{s}^{2} \times P_{f}^{2} $$ $$ P(5,2) = P_{s}^{5-2} \times P_{f}^{2}= P_{s}^{3} \times P_{f}^{2}=P(4,2)* P_{s} $$

Because our probability of success is below \$1\$ (\$\frac{11}{20}\$ in our case) this means the probability of getting that single event is 55% less.

Now - that was for a single event, when we bring the number of different ways we can achieve the \$k\$ failures in a party of \$N\$ (the bit we covered at the top). The probability looks like: $$ P_{T}(N,k) ={}_N \mathrm{ C }_k \times P(N,k) $$

Carrying on with our example of \$N=4\$ vs \$N=5\$ for \$N=4\$ we have 6 ways of getting 2 failures and 2 successes, for \$N=5\$ we get 10. So whilst the number of options has increased it isn't at a rate that beats the additional 55% multiplier added on.

$$ P_{T}(4,2) ={}_4 \mathrm{ C }_2 \times P(4,2)= 6 \times P(4,2) $$

$$ P_{T}(5,2) = {}_5 \mathrm{ C }_2 \times P(4,2) \times P_{s} = 10 \times P(4,2) \times 0.55 = 5.5\times P(4,2) $$

The same logic then applies for \$k=1\$, \$k=0\$ - you have to sum up each of these options to get the total probability.

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    \$\begingroup\$ The 155mm howitzer round has critically hit the fly. I salute you. (Thanks for taking the time and effort to be thorough, +1) \$\endgroup\$ – KorvinStarmast Jun 7 at 13:41
  • \$\begingroup\$ @KorvinStarmast Haha thanks - I was half way through my response when I saw yours and began to reconsider my life choices :D I was already committed though. \$\endgroup\$ – Lio Elbammalf Jun 7 at 13:47
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    \$\begingroup\$ You may mention the Binomial distribution, which is exactly what you described above. Just a curiosity: why do you write \$=_N C^k\$ for the binomial coefficient and not \$C_N^k\$? \$\endgroup\$ – Eddymage Jun 7 at 14:25
  • \$\begingroup\$ @Eddymage I believe there are many different versions of the notation. My guess would be its how I was taught it but its been so long since then. All it can say is that is just how I remember it. N Choose k. \$\endgroup\$ – Lio Elbammalf Jun 7 at 14:34
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    \$\begingroup\$ @LioElbammalf I edited that particular notation into something more standard. \$\endgroup\$ – Thomas Markov Jun 7 at 14:36

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?


As other answers have noted, adding an extra member to an odd-sized group can never reduce the group's chance of passing a group check: you still need the same number of successes to pass, but the extra member contributes a possible extra success. In fact, this holds regardless of how likely each member of the group is to pass the check individually. At worst, even if the extra member is guaranteed to fail, the group's chance of passing the check merely remains the same whether the extra member is in the group or not.

So how hard is it to pass a group check as opposed to an individual check, assuming that all characters in the group need to roll the same target number to pass?

For odd-sized groups such a group check is equivalent to rolling multiple dice and taking the middle roll. This tends to concentrate the results towards the middle of the d20 range, making easy checks (target number < 11) easier to pass and hard checks (target number > 11) harder to pass. The break-even point for all odd group sizes is 11, as this AnyDice script demonstrates:

AnyDice screenshot
Figure 1: Probability of passing a group check (vertical axis) for an odd-sized group as a function of the target number needed to pass an individual check (horizontal axis).

For even-sized groups a group check is equivalent to rolling multiple dice and taking the higher of the two middle rolls. This has (for groups of 4 or more members) the same effect of concentrating the results towards the middle of the range as for odd-sized groups, but taking the higher of the two middle rolls also biases the results upwards, making the check easier, as this AnyDice script demonstrates:

AnyDice screenshot
Figure 2: Probability of passing a group check (vertical axis) for an even-sized group as a function of the target number needed to pass an individual check (horizontal axis).

In particular, as others have noted, a group check for a two-member group (with equal target numbers) is equivalent to rolling a single check with advantage, and thus is always easier than an individual check.

For even-sized groups with four or more members, there's a threshold target number below which the group check is easier to pass than an individual check. As the size of the group grows, this threshold goes down, approaching 11 for large enough groups. We can use this AnyDice script to determine where that threshold lies:

Group size Group check is easier than individual check if target is at most
2 20
4 16
6 14
8 13
10, 12, 14 12
even ≥ 16 11
odd ≥ 3 10 (equal to individual check if target = 11)

Addendum: How to fix it?

Obviously, that depends on what one means by "fix" and "it". There's no way to eliminate the centralizing tendency of the D&D 5e group check mechanic without replacing it with something completely different. For better or worse, group checks done using this mechanic will always tend to be easier than individual checks for low target numbers and harder for high target numbers.

However, we can fix the bias favoring even-sized groups. A simple modification that accomplishes this is to change the rule to read as follows (changes in bold):

To make a group ability check, everyone in the group makes the ability check. If more than half the group succeeds, the whole group succeeds. If less than half the group succeeds, the group fails. Otherwise, if exactly half the group succeeds, flip a coin to determine whether the group succeeds or fails.

This simple rules tweak eliminates the bias in favor of group success in the case that exactly half the members succeed. Interestingly, it turns out to make the success rate for an even-sized group exactly equal to that of an odd-sized group with one less member (assuming, again, that everyone in the group needs to meet or exceed the same target number to pass).

  • \$\begingroup\$ I don’t mean anything by “fix” and “it”. I have no interest in fixing it; I just wanted to know what it was. \$\endgroup\$ – Dale M Jun 9 at 10:22
  • \$\begingroup\$ @DaleM: Fair enough. Rephrased to avoid using the word "you" there. \$\endgroup\$ – Ilmari Karonen Jun 9 at 10:23
  • \$\begingroup\$ Another (untested) way to fix it, which keeps the bias towards success: in an odd-sized group, have everyone roll twice. Then it works exactly like an even-sized group with 2n characters. You need an average of 1 of 2 successes from everyone, but some can contribute 2 and some contribute 0. (Makes things less swingy, closer to the statistical norm, which is maybe good for something your party spends a long time doing, maybe less good for a 1-off group athletics check for a short but hard climb, for example.) \$\endgroup\$ – Peter Cordes Jun 9 at 12:06
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    \$\begingroup\$ @PeterCordes: That ends up favoring odd-sized groups for easy checks with low target numbers (since having more rolls is better in that case) while favoring even-sized groups even more than the official system for difficult checks with high target numbers (since having a small even number of rolls is most advantageous in that case). \$\endgroup\$ – Ilmari Karonen Jun 9 at 15:45

Yeah, this analysis appears to be correct for a group of similar creatures making a group check.

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    \$\begingroup\$ Why is it correct? Came from the LQ queue, and as the post stands it doesn't offer any meaningful interpretation that confirms the assumptions. \$\endgroup\$ – Akixkisu Jun 7 at 11:27
  • \$\begingroup\$ I found it hard to answer this question... the post shows all its work. All of what it shows is correct. Not sure what anyone could say that isn't just repeating the post itself. \$\endgroup\$ – Erik Jun 7 at 11:36
  • \$\begingroup\$ I think there are different approaches to this problem. One could be establishing that Dale's choice and use of analysis is correct and that Dale indeed didn't make any mistakes (e. g.: by showing the formula or a different approach), if you are the expert that can answer this, then you probably can come up with something. \$\endgroup\$ – Akixkisu Jun 7 at 11:58
  • \$\begingroup\$ Yes/no question, yes/no answer. I don't see the problem here. \$\endgroup\$ – SeriousBri Jun 10 at 12:32

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