This is an interesting hand that arose when I was playing Spider Solitaire on my mobile phone. I aim to win without undoing any moves, but the number of moves required is irrelevant.
WARNING: I apologise if this question is too long for most folk. But I consider this kind of detailed analysis necessary to achieve a respectable win rate at Four-Suit Spider Solitaire
Notation: Moves are identified by specifying source/destination columns. Columns are labelled a-j, starting from the left. For instance, moving the 5432A of Diamonds in column 4 onto the other Six of Diamonds is denoted “da”.
Some general considerations:
- Standard Spider Solitaire rules apply. Googling is left as an exercise for the reader
- I am guaranteed at least two turnovers. For instance, I can reveal at least two face-down cards in column 6 at the expense of both empty columns even if the worst possible cards turned up
- I want to consider the chances of obtaining more than two turnovers. For instance, any Queen would provide a parking place for the Jack in Column 6 or 10, yielding an extra turnover.
- Every card in Spades and Hearts is visible except the Two of Spades and Ace of Hearts
- Beware the gap in column 1 (no Eight between the 9-7-6). And yes, I’ve been caught before …
Following these considerations, my next goal is to turnover a card in column 2 at the expense of only one empty column. I may be willing to give up a few “suited connectors” but I definitely wanna improve my guaranteed minimum 2 turnovers if possible. It’s true that sticking the King of Hearts onto an empty column has its disadvantages but sooner or later I’m gonna have to think about completing a suit.
I will define an “action” as a sequence of one or more moves that terminates as soon as any face-down card in the tableau is exposed. For purposes of this question, I assume we never deal from the stock. To plan ahead, I want to choose two actions. More specifically:
- The first action turns over a new card in the tableau. The aim is to maximise our chances of improving on our guaranteed minimum two turnovers.
- The second action does not turn over any new cards. IMPORTANT: we cannot take advantage of any information gained in the previous action. The aim is to tidy up (increase suited connectors) assuming the worst case scenario when the newly-revealed card is as useless as a random list of dating apps
To compute overall happiness, I do the following:
- There are 13 possible ranks from Ace to King. For simplicity I assume that when the next card in column 2 is revealed, each of the thirteen ranks occur with probability 1/13.
- I also assume for each of the 13 ranks you get the worst possible suit.
- Given Position 1, for each of the 13 ranks, I get 10 happy stars if I improve on my guaranteed minimum two turnovers (it doesn’t matter if I improve to three or improve by a lot). The contribution to overall happiness is therefore some multiple of 10 between 0 and 130.
- Given Position 2 for each “suited connector” I get one happy star. A run of length N counts as (N-1) suited connectors.
EXAMPLE: Consider the following: First action = “jh,jc”, Second action = “fg”. (note that this does not fulfil my goal of revealing a card in column 2). We lost both empty columns but we can recover one empty column with “hc”. This in turn allows us to reveal another card in column 6 regardless of what was revealed in column 10. Therefore, we still have a guaranteed 2 turnovers. If, after Position 1, we reveal a Five in column 10 then we get an extra turnover since the Five can move to column 1, allowing another turnover in column 6. This gives us 10 happy stars. If we reveal a King in column 10 then we do not improve on 2 turnovers. Determining what happens to our minimum guaranteed turnovers for any rank except Five or King is left as an exercise for the reader. In Position 2 we gained a suited connector with “fg”. The original diagram had 32 suited connectors so we now have 33 happy stars, plus whatever multiple of 10 we computed for Position 1.
- Question (easy): find any “first action” that successfully reveals a card in column 2 and spends only one empty column. Do not worry about the second action or the number of happy stars.
- Question (hard): find two actions that results in a target score of 105 happy stars.
- Question (probably impossible): can you exceed 105 happy stars and prove I missed something during the game?
