## 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.