Math story: Ten marriage candidates and 'greatest of all time' I remember a story about a famous mathematician who was offered ten marriage candidates and had to pick one of them, with the condition he had to meet them in turn and propose during that meeting, with no changing his mind to an earlier one. If he went through all ten without proposing then it was too bad for him.
This caused him to develop a mathematical formula to maximise the odds of knowing when he met the best one based on how many times he met a 'best candidate so far', and he proposed to the seventh (I think).
Could someone remind me who he was?
 A: You are thinking of Kepler. Scroll down to the section called "second marriage".
There is a very interesting exposition of this story in Chapter 10 "Computing A Bride" in one of my favorite books, Arthur Koestler's The Sleepwalkers. 
A: This is the secretary problem. If there are $n$ suitors, the optimal strategy is to reject the first $r-1$ of them and then accept the first one that is better than all of those $r-1$.  In general $r \approx n/e$. In the $n = 10$ case the best choice is $r = 4$. That is, reject the first three suitors and accept the first suitor that is better than all three of those. 
This paper of Ferguson outlines some of the history of the problem, and doesn't seem to mention the version of the story that you gave. It seems possible to me that this version of the story could have just been used to dress up the underlying mathematics and is not historically accurate.
A: Perhaps you are interested in the work on stopping rule problems that was done by Herbert Robbins?  http://en.wikipedia.org/wiki/Robbins%27_problem
A: As Michael noted, this is the Standard Secretary Problem. I have several papers on the subject, and I recall reading that Lindley used the 1/e policy to select his wife. Having said that, the questioner may have heard about Kepler. 
