Damn! It's not just MY birthday!

The number of outcomes in which among n people no two are born in two consecutive days is equivalent to the number of integral solutions to this equation $x_1+x_2+....+x_{n+1} = 365-n$ where

$x_1 \ge 0, x_2 \ge 1, x_3 \ge 1, .... x_n \ge 1, x_{n+1} \ge 0$

and $x_1+x_{n+1} \ge 1$

The reason is you can think of choosing one day for each person, and then $x_i$ determines the number of days between (i-1)'th birthday and i-th birthday. $x_1$ is the number of days before the first person's birthday, and $x_{n+1}$ is the number of days after the last person's birthday. Unlike other $x_i$ 's these two can be zero but not simultaneously. Call the number of integral solutions to the above equation A. This number must be multiplied by n! of course.

The probability that this does not happen is :

$1-\frac{A\times n!}{365^n}$

I took a more direct approach. If one modeled the calendar year as a circular ring with 365 spaces on it, designating any one slot as a birthday would thus 'X out' that square, plus the adjacent squares, if we're trying to avoid collisions.

In the same way that...

$A_n = \prod_{k=0}^{n-1} (365 - k)$

...generates the numerator for the classical birthday problem...

$A_n = \prod_{k=0}^{n-1} (365 - 3k)$

...generates the numerator for this problem. Letting $f(n) = 1 - \frac{A_n}{365^n}$ , we can simply calculate:

$f(13) < \frac{1}{2} < f(14)$