# dnd 5e – When performing a group check, is it better to have an even number of creatures?

## 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:

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

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.

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

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

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