Same Birthday

There is actually a rather complicated-looking  summation formula that calculates the probability and looks like this: 1 - ∑365!100! / {n! (100-2n)! (365-100 +n )! 2^n 365^100}, n=0 to 50

I summed up the first 25 terms and this is what I got:

0.0000003072 4892785157 735709                       
0.0000057176 0222881694 70586
0.0000508909 4268458230 219
0.0002886351 9731554141 541
0.0011725134 2701322632 6
0.0036356601 5220545585 22
0.0089549931 7932524642 57
0.0180040514 1305516981
0.0301304065 5435743848 6
0.0425931051 2915248603 5
0.0514369825 9415105677 2
0.0535378343 2065788517 8
0.0483676643 9980012911 8
0.0381427347 9231609518 2
0.0263757108 7405165721 6
0.0160515040 4620857996 3
0.0086219711 4581710867 67
0.0040969650 1255139206 62
0.0017251648 9044419630 59
0.0006445389 9539204962 057
0.0002138286 3864673084 781
0.0000630164 3196881678 332
0.0000164976 4999120274 6726
0.0000038355 0437597406 85323
0.0000007913 2162082163 957176
________________________________
1 - 0.354135321464259511361245756459
=64.59% - This  is the probability of at least 3 people sharing the same birthday !!!.

For those who may be skeptical of the "Guest's" calculation, or rather the accuracy of the calculation, can refer to Wolfram/Alpha's calculation here:

https://www.wolframalpha.com/input/?i=birthday+problem+100+people