Short answer: no.
An article in the Guardian suggests that because Mo Farah, Chris Hoy and Steve Redgrave share a birthday with Jason Kenny, Roger Bannister, Joe Calzaghe and Mike Atherton, March 23rd must be a Good Day to have Olympian kids.
It's unusual, right enough, for three gold medallists from these Games - Farah, Kenny and Hoy- to share March 23rd as a birthday. You can work out the odds against it, roughly, using a binomial distribution: The probability of any one athlete having March 23rd as a birthday is - give or take - 1/365. There were 44 British gold medallists at the games. So the probability of at least three gold medallists sharing March 23rd as a birthday works out to be
1 - 44C0 (1/365)0(364/365)44 - 44C1 (1/365)1(364/365)43 - 44C2 (1/365)2(364/365)42
(That is, the probability of there not being 0, 1 or 2 athletes with that birthday).
That works out to be 1 - 0.8863 - 0.1071 - 0.0063, which is about 1 in 4000. Tiny odds!
Trouble is, that's a deeply, deeply flawed experiment. We've started with the knowledge that three of the athletes share that particular birthday. The probability of three athletes sharing any birthday is much higher.
Roughly speaking, it's 365 times as high, about one in 11, nowhere near enough to be statistically significant.
Actually working it out properly at 8am is a little beyond me, but writing a little simulation isn't. Taking 44 random birthdays, repeatedly, the results look like this:
- No two people share a birthday about 7% of the time
- No more than two people share any birthday about 84% of the time
- Three people share a birthday about 8% of the time
- Four people share a birthday about 0.3% of the time
- Five people share a birthday about 0.005% of the time
If, as the article does, you extend it to previous medallists and athletes from other sports, you're bound to be able to cherry-pick more people with the same birthday.
Really, you'd expect the Guardian to run these things past a mathematician before they go live.