Let $n$ be a number of people. At least two of them may be born on the same day of the year with probability: $1-\prod_{i=0}^{n-1} \frac{365-i}{365}$
But what is the probability that at least two of them are born on two consecutive days of the year (considering December 31st and January 1st also consecutive)? It seems a good approximation is: $1-\prod_{i=0}^{n-1} \frac{365-2 \times i}{365}$
However, simulating pseudo-random integers with Python, the 99%-confidence intervals may be slightly different. So do you have any closed formula?
Results of the simulation with Python. Here are 99%-confidence intervals below:
Number of people: 1 Lower bound: 0.0 Upper bound: 0.0 Number of people: 2 Lower bound: 0.00528 Upper bound: 0.00567 Number of people: 3 Lower bound: 0.01591 Upper bound: 0.01657 Number of people: 4 Lower bound: 0.03185 Upper bound: 0.03277 Number of people: 5 Lower bound: 0.0528 Upper bound: 0.05397 Number of people: 6 Lower bound: 0.07819 Upper bound: 0.07959 Number of people: 7 Lower bound: 0.10844 Upper bound: 0.11006 Number of people: 8 Lower bound: 0.14183 Upper bound: 0.14364 Number of people: 9 Lower bound: 0.17887 Upper bound: 0.18086 Number of people: 10 Lower bound: 0.21816 Upper bound: 0.2203 Number of people: 11 Lower bound: 0.25956 Upper bound: 0.26183 Number of people: 12 Lower bound: 0.30306 Upper bound: 0.30544 Number of people: 13 Lower bound: 0.34678 Upper bound: 0.34925 Number of people: 14 Lower bound: 0.39144 Upper bound: 0.39397 Number of people: 15 Lower bound: 0.43633 Upper bound: 0.4389 Number of people: 16 Lower bound: 0.48072 Upper bound: 0.48331 Number of people: 17 Lower bound: 0.52476 Upper bound: 0.52734
I give here some results with a tweaked approximation formula, using Wolfram Alpha: $\left( 1 - \frac{n-1}{2 \times 365 + n-1} \right) \times \left( 1-\prod_{i=0}^{n-1} \frac{365-2 \times i}{365} \right)$
However, this is just a tweak, ans is clearly wrong for $n=33$ since:
Number of people: 33 My guess: 0.91407 Number of people: 33 Lower bound: 0.94328 Upper bound: 0.94447
Thanks to Jacopo Notarstefano, leonbloy, and Moron, here is the (correct) formula: $ 1-\sum_{k=1}^{n}\frac{1}{365^{n-k}k}\left(\prod_{i=1}^{k-1}\frac{365-\left(k+i\right)}{365\times i}\right)\sum_{j=0}^{k-1}\left(-1\right)^{j}C_{k}^{j}\left(k-j\right)^{n} $
And here are the results of the computations using this formula with Python:
Number of people: 1 Probability: 0.0 Number of people: 2 Probability: 0.005479452 Number of people: 3 Probability: 0.016348283 Number of people: 4 Probability: 0.032428609 Number of people: 5 Probability: 0.053459591 Number of people: 6 Probability: 0.079104502 Number of people: 7 Probability: 0.108959718 Number of people: 8 Probability: 0.14256532 Number of people: 9 Probability: 0.179416899 Number of people: 10 Probability: 0.218978144 Number of people: 11 Probability: 0.260693782 Number of people: 12 Probability: 0.304002428 Number of people: 13 Probability: 0.34834893 Number of people: 14 Probability: 0.393195856 Number of people: 15 Probability: 0.438033789 Number of people: 16 Probability: 0.482390182 Number of people: 17 Probability: 0.525836596 Number of people: 18 Probability: 0.567994209 Number of people: 19 Probability: 0.608537602 Number of people: 20 Probability: 0.647196551 Number of people: 21 Probability: 0.683756966 Number of people: 22 Probability: 0.718059191 Number of people: 23 Probability: 0.749995532 Number of people: 24 Probability: 0.779509664 Number of people: 25 Probability: 0.806569056 Number of people: 26 Probability: 0.831211564 Number of people: 27 Probability: 0.853561895 Number of people: 28 Probability: 0.873571839 Number of people: 29 Probability: 0.892014392 Number of people: 30 Probability: 0.906106867 Number of people: 31 Probability: 0.919063161 Number of people: 32 Probability: 0.928791992 Number of people: 33 Probability: 0.944659069