Problem 968

Solution:

Answer:
$$\frac{\binom{4030}{2016}}{2^{4030}} \text{ OR } \frac{\binom{4030}{2014}}{2^{4030}} \text{ OR } \frac{\binom{4032}{2016}-2\binom{4030}{2015}}{2^{4031}}$$

Solution 1. Note that total soups $=$ total salads $+1$ is equivalent to total soups $+$ total not-salads $=2016$. So there are precisely $\binom{2015+2015}{2016}$ possibilities, each occurring with probability $(1 / 2)^{2015+2015}$. Thus our answer is $\frac{\binom{4030}{2016}}{2^{4030}}$.

Solution 2. To count the number of possibilities, we can directly evaluate the sum $\sum_{i=0}^{2014}\binom{2015}{i}\binom{2015}{i+1}$. One way is to note $\binom{2015}{i+1}=\binom{2015}{2014-i}$, and finish with Vandermonde's identity: $\sum_{i=0}^{2014}\binom{2015}{i}\binom{2015}{2014-i}=\binom{2015+2015}{2014}=\binom{4030}{2014}$ (which also equals $\left.\binom{4030}{2016}\right)$.
(We could have also used $\binom{2015}{i}=\binom{2015}{2015-i}$ to get $\sum_{i=0}^{2014}\binom{2015}{2015-i}\binom{2015}{i+1}=\binom{2015+2015}{2016}$ directly, which is closer in the spirit of the previous solution.)

Solution 3 (sketch). It's also possible to get a handle on $\sum_{i=0}^{2014}\binom{2015}{i}\binom{2015}{i+1}$ by squaring Pascal's identity $\binom{2015}{i}+\binom{2015}{i+1}=\binom{2016}{i+1}$ and summing over $0 \leq i \leq 2014$. This gives an answer of $\frac{\binom{4032}{2016}-2\binom{4030}{2015}}{2^{4031}}$, which can be simplified by noting $\binom{4032}{2016}=\frac{4032}{2016}\binom{4031}{2015}$, and then applying Pascal's identity.