All Spider players know that empty columns (aka holes) are one of the most valuable commodities in the game. Any card can legally move onto a hole (not just Kings). Having a hole means so many extra options for manoeuvring cards. Of course, more options also imply more chance of making a sub-optimal play đ In fact, I believe the hallmark of a winning player is the ability to take maximum advantage of holes.

There are three main use cases for holes:

- Turning over a new card
- Moving a sequence that is not in-suit
- Tidying cards so they are in-suit.

Examples of the three use cases are:

- we can turn-over a new card in columns 4 or 5. Further thought shows that column 6 is also possible since the 6-5-4 in diamonds can fill the empty column and the Q can go on one of two kings. Similarly column 3 is another option.
- We can also turnover column 1. A little thought that if we have at least one empty column any length-2 sequence can be shifted to a non-empty column regardless of suits. This is the simplest example of a supermove.
- We can also swap the deuces in columns 1 and 8. This increases the number of in-suit builds (3-2 of clubs).

You should immediately notice that options 2 and 3 mean we improve our position for free since we keep the empty column. Therefore option 1 is the worst. It might be tempting to turn over Column 3 so we can get a straight flush in Diamonds but that is a serious error. Not only is the Six-high straight flush the second-weakest of all possible flushes in poker, but having a King in an empty column means we would be a long way from securing another empty column (we need a minimum of three good cards).

Option 3 is also completely safe because it is reversible, so an experienced player will make this move immediately. I am assuming we are only playing to win without regard to score (e.g. -1 penalty per move), otherwise more thinking would be required. Option 2 is the only way to turnover a card without using the hole.

Although option 1 is the worst, most of the time it is available as a fallback option. In other words a hole (usually) implies you always have at least 1 turnover.

**EXERCISE:** Assume you get 1 brownie point for every suited-connector (e.g. 6-5-4 in diamonds is worth 2 brownie points). From the diagram position above, what is the maximum brownie points you can get without losing the empty column or exposing any new cards?

Again I will use the happy-star method for avoiding the reader unintentionally reading spoilers. For those who donât recall from a previous post: each happy star represents 1 point in a short story comp, and I have no idea if the judges docked 5 points for the protagonistâs terrible Dad joke.

Answer: we currently have 8 brownie points and get 2 more (3-2 of clubs and 9-8 of spades).

Well done if you answered correctly (or found an error in my counting). If you aspire to kick 65,82,83,69 at Spider Solitaire, finding opportunities to tidy up suits “for free” must become second nature.

That’s all for now, toodle pip and piddle too đ