We got some good luck here. We obtained an empty column and also some useful cards such as the Ten of Spades and Five of Clubs. This means we are significantly better than the minimum guaranteed turnovers calculated from last week. Now is a good time to evaluate options.
We have two guaranteed turnovers at this stage. The empty column is clearly worth one turnover and we can also expose a card in column 2 or 9 (using the fact column 1 is missing a Five). It is of course desirable to keep the empty column as long as possible, so we identify two basic choices: (i) build 0s-9c-8c, 5c-4c or (ii) insert 5c between 6c-4h-3d-2d-Ad.
Note that the first option will kill column 3: it will no longer be possible to build Qc-Jd-0s and we would “regret” seeing a Nine. A third option is to shift the Jd-0s onto the Qc first before turning over column 2 – but then we would regret seeing a Jack. The other problem is that we also lose the option of shifting the 9h-8c in column 6. If we expose a King then that would really 83,85,67,75.
Unfortunately it is not possible to procrastinate by shifting the Five of Clubs directly onto the empty column since we cannot recover the empty column if a bad card turns up (but it would work if it was the Five of Hearts).
There is another sneaky option available: Suppose that we shift the 7h-6c in column 1 onto the 9c-8c in column 2 (thus breaking an in-suit build!) before inserting the 5c into column 1. The logic is that we are guaranteed to recover another in-suit build with 9h-8h (at the expense of the empty column). If you spotted this possibility then you are already above beginner level. But for most folk, it is sufficient to know that the number of options to consider increase dramatically when you have an empty column.
What would be your play here?
Edifying Thoughts of a Spider Solitaire Addict 