Spider Solitaire - maximum number of cards in a single column

Question

What is the maximum number of cards you can have in a single column of four-suit Spider Solitaire?

The diagram below has 11 cards in column 2, more than any other column. Obviously, the correct answer is much bigger than 11. Assume standard rules: start with 54 cards in the tableau with only 10 face-up, plus 50 cards in the stock. The four “surplus cards” are in the left-most four columns. Further information can be found here:

Bonus question: does anything change if we have only 1 or 2 suits?

EDIT: as requested by @bass, here are the rules for 4-suit

• A single card can only be moved to another pile if the card being moved is one less than the card it will be placed on. For example any 9 can be placed on any 10.
• Groups of cards can only be moved if they are all in same suit and are in perfect descending order. For example you could move a 10, 9 and 8 of diamonds as a group onto any open Jack.
• If a card that is face down in a column is open it must be turned over.
• Any group or single card you might be able to move can be placed on an empty column.
• You can deal 10 cards from the cards remaining, one to each column if you cannot make any moves. However there must be at least one card in each column when you do this.
• If you have a complete group of cards in one suit in perfect descending order it can be removed from play. For example King of Spades all the way down to the Ace of Spades. (Most software implementations enforce a complete suit be removed). Remove all the cards to win the game.

NOTE: this question was inspired by a friend's comment on my personal Spider Solitaire blog.

Answer 1

Since we are looking for a theoretical maximum, we can get up to

83 cards

in a single column.

This follows from our great advantage of being able to

always choose the best possible outcome for any unknown card that gets revealed;

we are, after all, looking for the theoretical maximum, not a practical one.

Since we are armed with perfect luck, there's nothing stopping us from always having the next card available, so when the first face-up card in the leftmost column is a king, we can just stack the entire suit on top of it. We can repeat this trick five more times by dealing a new king from the deck to the left column, so we end up with $6\times13=78$ face-up cards on top of the $5$ face-down ones.

Since there are no other ways to get larger cards on top of smaller ones (in Spider, you can only move strictly descending sequences), this must be optimal, and the number of distinct suits is inconsequential. Unless I'm missing something obvious, of course.