Game theory question

[Atti0626]

Beginner and Second are taking turns writing numbers into a 19x19 table, either 0 or 1. When the table is completed, they sum up the rows and the columns. If the biggest rowsum is bigger then biggest columnsum, Beginner wins. If the biggest columnsum is bigger then the biggest rowsum, Second wins. If the two biggest sums are equal, it is a tie. Is there a winning strategy?

What I did is try to play the game with a smaller table. In the cases of a 2x2 or 3x3, I am pretty sure it is always going to be a tie, if both play rationally. I tried up to 7x7, playing each table aprroximately 5 times, and my results were all ties, but I am not so sure about that, because I played against myself, and I may have made some strategic mistakes.
I can't describe my exact thoughtprocess while playing the games, but for example, when I were Second, I tried to place ones in columns where there already were ones and preferably no zeros, but in a row that is either empty or has multiple zeros in it. I also "ruined" a row with a zero if it had too many ones compared to the size of the table.
I couldn't figure out a general strategy while playing these games, and I am not sure if I could play a game with a 19x19 table and make it a tie, because there are more possible mistakes to be made.
I appreciate any kind of help, thank you.
(Sorry for the possible mistakes in my English.)

 

[Cubist]

This won't answer your question "is there a winning strategy" but might it give you some ideas.

I have thought of a very simple strategy that limits the score of the opponent...
Assume the opponent is playing for the biggest column sum and you are playing for biggest row sum.

IF opponent has just moved in column "x", with counter "c" AND there is still space in column x
THEN make your move in column "x" but with counter colour "1-c".
ELSE play randomly

This should limit their final score to approx half the grid size (depending on whether it is odd or even grid size and who moves last). This is because columns will fill up with 1-0 and 0-1 pairs of counters.

--

This strategy could be improved by adding a second priority rule that chooses a favourable row from those available in the column of play. If playing colour "y" then look for the row that has the highest { sum of "y" counters MINUS sum of "1-y" counters }. This should help to create rows of predominately 1, and also rows of 0, helping to maximise your score.

 

[JeffM]

Consider the 2 by 2 game.

It should be clear that we can swap the positions of the rows and columns without changing the maximum value of the row totals and without changing the maximum of the column totals. So let's assume B places his first move in cell (1, 1).

If B puts 0 into (1, 1), S can try for a win by putting 1 into (1, 2). If B then puts 1 into (2, 2), S guarantees a win by putting 0 into 2, 1. (The row totals will each be 1 whereas the total of column 2 will be 2.) If B puts 0 into (2. 1), S guarantees a win by putting 1 into (2, 2). However, B can guarantee a tie by putting 0 into (2, 2). Thus, if B selects 0 for a first move, best play on both sides will end up in a tie.

If B puts 1 into (1, 1), S can try for a win by putting 1 into (2, 1). If B then puts 0 into either of the remaining cells S will win by putting 0 into the remaining cell. If B puts 1 into cell (1, 2), there is a tie no matter what S puts in the remaining cell.

Thus, you can prove that best play by both B and S results in a tie.

That may not be quite enough. I suspect you may have to show that every row column equals every column total to extend to 3 by 3 and so on.

In other words, I am suggesting a complete analysis of the 2 by 2 game and then an inductive process.