"I'm thinking of a number from 1 to 100. Whoever guesses closest to the number gets to ride in the front seat."

While I was still a young lad of single digits, I realized that the correct strategy for the above offer was to choose either 50 or 51. That would guarantee me a 50% chance of winning, assuming that my sister would choose the other. If I were to choose some other number, for example 30, then my sister could choose 31, leaving me with just a 30% chance of winning, compared to her 70%. Not a good plan!

A much more interesting question is, "What is the correct strategy is if there are three players instead of two"?

We need some rules to make this a well stated problem, so here they are:

     1) Each player must choose, in turn, a whole number in the range of 1 to 100 inclusive;

     2) Each player hears all of the choices preceding his or her own, and must choose a number that has not been previously chosen;

     3) If the unknown number is the same distance from the two closest guesses, then a coin flip will determine which of those two is the winner;

     4) Each player is perfectly rational, and his or her only goal is to maximize the chance of winning;

     5) If a player has more than one optimal play, he or she will make an equally likely random choice among them.

Assuming all of that, what are the best choices for the first, second, and third players, and what are their chances of winning?

Extra credit:
     a) What is the best strategy if we have four players?
     b) How about any number of players?
     c) What is the best strategy if we use numbers from 1 to 101, instead of 1 to 100?
     d) How about other top numbers?