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Reading Mark Evanier's blog today got me to a posting at the New York Times (http://opinionator.blogs.nytimes.com/2012/10/01/its-my-birthday-too-yeah/) that dealt with the age-old conundrum of, "How many people have to be in a group for the odds of two (or more) of them sharing the same birthday be at least 50-50?" The answer is almost always a surprise.
The genesis of the article is the finding of an old clip from the Johnny Carson show that dealt with the issue in a mildly amusing manner. The all-time great talk show host bungled the issue a fair bit, trying to match exact birthdays by asking one person their birthday and then surveying the crowd of 500 for a match. That failed as did a second attempt. The third attempt, Carson's own birthday, resulted in two other co-celebrators on that day.
Still, the column and then watching the video at the Cornell Collective (the link is in the column), brought back my own introduction to the issue. And an astonishing result that even had my Grade 4 teacher, Mrs. Benham, absolutely speechless.
Mrs. Benham, one of my best teachers during my academic life, started math class by asking the familiar question. It wasn't me, but one smart aleck in the room answered immediately, "365." Which, of course, GUARANTEES there will be a match (barring a Feb. 29th outlier). In reality, as the column points out, the odds are ALMOST guaranteed with as few as 253 people. And that's because, birthdays are NOT scattered randomly over all 365 days. Birthdays occur in clusters approximately nine months after certain annual events ... Valentine's Day, the first cold weather day of the year, Christmas and New Year's, last harvest day of the season, etc.
However, Mrs. Benham got a bit of a shock on that day almost 47 years ago. I started laughing, even before the wise-ass went for the 365 answer. I KNEW she was going to say some ridiculously low number and she was going to be right, NO MATTER HOW LOW SHE WENT. She finally gave 23 (the exact size of the class with full attendance, as there was that day) as the 50-50 answer, drawing oohs from the kids who weren't wrapped up in my giggling.
Mrs. Benham gave me as close to a dirty look as she was capable of. And said, "Let's start the test of my theory with Gary here." I sat in the first seat in the first row immediately in front of the teacher's desk. "July 18th," I answered, collapsing in laughter. Mrs. Benham somehow failed to comment on, or ask, why I was laughing and asked the girl sitting behind me, my next door neighbour Susan. The reason behind my laughter became obvious immediately.
"July 18th," she answered.
Then I got a big shock myself. The third-to-be-asked was a girl named Nancy that I didn't know that well but had a bit of a crush on. She sort of gasped when Susan revealed her birthday. When Mrs. Benham, smiling like she had been totally vindicated, continued to Nancy, saying, "Well, let's see if we have more than one pair in the room."
Nancy's birthday? Obviously July 18th. Otherwise, there's no story.
I've tried this survey at various times over the years and have actually found a better than 50-50 record during that time. But that's outside of the real odds. I've even bet people and won, taking advantage of the fact that I knew the crowd included a set of non-identical twins (the East girls). But even when I haven't salted the results beforehand, it's still an interesting test to run at parties or bars. Get the group up to about 30 and bet the winning side and you might enjoy a night of drinks for free.
But don't try three out of three. The odds of that are more than a 100,000 to one.