Considering Extreme Cases

[mafou]

Two men are seated at a table of usual rectangular shape. One places a penny on the table, then the other does the same, and so on, alternately. It is understood that each penny lies flat on the table, and not on any penny previously placed. The player who puts the last coin on the table takes the money. Which player should win, provided that each plays the best possible game ?

[kaphirgoyim]

The one with the most pennies. :)

Or the guy that didn't go first.

[mafou]

Nice try, Kaphirgoyim ! ( Great Photo )Though, not that easy. Suppose that the table is so small it is covered by one penny. Then, the first player must win. Of course, if this were the situation, the first player would win only his own penny and the game would not be interesting, but considering this extreme case provides the clue needed to see the solution to the problem as originally stated. Imagine the size of the table being gradually increased. If the first player places the first penny precisely in the center, as soon as the table is large enough to hold a penny beside the first penny on any side, it will be large enough to hold another penny on the opposite side as well.

Then what?

[Mary Zimmermann]

P.S. I miss Kaphirgoyim's dog! (especially, the blow-up doll)

I can see that, irrespective of the size of the table, if the first player puts his first penny in the middle and after that always precisely matches what the second player does, but on the opposite side of the table, he will always win. ( Like bubblefish)
It is not essential to imagine the extreme case to solve this problem; an argument could be made from symmetry straightaway.