However, if the strategic goal is to maximize the chances for winning two or more squares, it has been determined that your squares should all be in the same column corresponding to the visitor's team. I have studied this particular strategy and have found that you should pick as many squares as possible in that column (all 10 if you can). If you can pick even more squares, start filling them up in another column too. Not surprisingly, if you want a higher chance of winning, pick more squares. What's not obvious is that the strategy of picking squares in the same column holds for these additional squares.

Now, let's consider picking 1-10 squares in a column. In tabulating the results, let us normalize to picking 10 squares. That is, if you pick one square, multiply the results by 10. For two squares, multiply by five. Multiply by fractions for most of the other numbers so that we can compare the strategic differences on the same footing. Here's a figure:

The figure shows that taking the optimal strategy to win 2 or more squares by selecting squares in a column, you still have to choose how many squares to select. Over the long term, there is no difference in what you do, for every 10 squares selected, you have a 10% chance of winning in any one Q1-Q3,final score. However, for any one game, you can trade off a have a lower probability for winning at least one square (this blue line goes up on the graph as you pick fewer and fewer squares) for a larger probability of winning two or more squares (red line that is highest for picking all 10 squares in a column). I also show the probability (axis on the right of the graph) of winning all four squares (very rare) that shows a very large advantage for picking your squares 10 in a row. In terms of numbers and comparing relative to picking 10 single square games, picking 10 squares in a single column will see your likelihood of winning anything at all drop by 26.4%, but your likelihood for winning two or more squares will increase by 59.3% with your chance of winning all four squares increasing by 402% - over 4x more likely!

I was a little surprised to notice that the lines in the figure are smooth and, in particular, there is no jump at selecting four squares. After all, in any one game, there are four periods to determine winning squares. Of the 4953 games under study, 12 have four winning squares in a single square, 155 in a vertical column of two squares, 154 in a vertical column of 3 squares, and only 40 in a vertical column of four squares. Most of the time that you win all four squares comes not from four in a column and four distinct rows, but having enough selected squares in a column such when a game has winners in two or three in a column, you happen to have the rows for those squares covered.

One interesting configuration is what to do if you were going to pick 13 squares. I consider four cases.

- 10 in one column and 3 in another column
- two columns of 9 and 4
- two columns of 7 and 6
- an 'L' shaped configuration with 10 in one column and the other 3 in a row

Finally, if you choose a strategy that wants to minimize the chance of not winning anything, then the strategy is to pick squares in unique rows and columns. What's not obvious is whether you'd be better off picking four such squares in a game OR a single square in four separate games. This has been studied and it is found that you are better picking the four squares in a single game. The difference is small: 13.4% to 13.1% (2.3% difference) although, as expected, the choice of picking a single square in four separate games has a 8.3% larger chance of winning more than two squares. In fact, by entering more four games with a single square, you even have a very very small chance of about 1 chance in 20,000 of winning 5 or more squares between your entries.

Jan 2017: I just re-read this post and there is some confusing language. I should never use "chance of winning in any one Q1-Q3,final score" when I mean "expectation value". In fact, in the figure, I computed prob(winning >= 2 squares), prob(winning > 0 ), and prob(winning=4) squares and divided by 10/(number of squares).

ReplyDeleteBetter is to treat multiple games not independently (i.e. not just dividing by number of games). If you know P=Prob(winning > 0) for a certain configuration, then 1-P=Prob of winning no squares. For M multiple games, your probability of winning > 0 squares is not MP; rather it is 1-(1-P)^M. This is how I correctly correctly calculate the 13.1% in the last paragraph.