The Duplicate Spider Solitaire Club

“Minnie and her glasses did it again!” fumed Cy the Cygnet (*)

(*) Yes … I borrowed that idea from Frank Stewart’s excellent Bridge Columns

Minnie Mouse, the smallest member of the Duplicate Spider Solitaire club, wears second-hand bifocals that make her mix up same-colour suits, much to the chagrin of other players. Cy had been her chief victim.

“Now what?” I sighed. If I had a happy-face disc for every bad beat story someone told me then I swear I would never lose a game of Connect Four.

“The play had started well at my table. I had already turned over eight cards and I only needed one more good card to get an empty column.”

I nodded. Judging from the game state below, Cy hasn’t done anything majorly wrong yet.


“Alas, the next card in column 8 was the other Ten of Diamonds,” continued Cy. “Column 1 didn’t yield anything useful either, a Four of Spades underneath the Ace of Clubs.”

In this hand there is a stipulation saying no cards to be dealt from the stock. I presume this is to help students improve by focusing on one concept at a time.

“Game over, +100.” I said. “How did Minnie go?”

“Minnie started the same way, but then she moved the Ten of HEARTS in column nine onto column 2.”

“Thinking it was the Ten of Diamonds,” I said.

“Minnie turned over a Nine of Clubs in column 9 and that was all she wrote, if you pardon the terrible cliché. It wasn’t even close.”

“I’m okay with terrible clichés,” I replied. “I use them time and time again.”

“Minnie’s play was wrong on two counts,” insisted Cy. “Not only did she misread the suits, but her goal was to expose as many cards as possible, not build sequences in suit.”

Actually Minnie’s play was correct. There are three guaranteed turnovers in columns 1,8,9 even if the worst possible cards turned up – provided the cards were played in proper order. Cy’s impulsive play meant that he was no longer guaranteed to turnover a card in column 9. If he shifts the Js-0h in column 9 first then the turnover in column 8 will not run away.

One might even make an argument of shifting the Ace in column 1 first. This “kills” column 5, but column 7 contains a suited 2-A. Therefore, we will only regret this move if we turned over two Threes (whereas we only need one King in order to regret shifting the Js-0h). The important point is Minnie’s play was better than Cy’s.

“Has anybody managed to expose all the cards for a single hand yet?” coos the Smart 65,83,83.

“Don’t ask,” replies the Dumb Bunny.

“Shush!” I say. “There are still animals playing.”

Duplicate Spider Solitaire is a fun variant, particularly for lousy players who never get close to winning a game at the highest difficulty level. Certain stipulations are also provided such as “score 10 points per turnover” or “do not deal any cards from the stock.” Therefore, if you get into a complete mess you can always hope your measly score is enough to beat the others players who must play the same lousy hands. You gain match points whenever you perform better than anyone else.

Unfortunately I am not aware of any existing Duplicate Spider Solitaire clubs anywhere in the real world. Perhaps some of my Bridge friends would know of one (or are willing to start one!). If so, then please leave a comment below 😊

Continuing from the previous post

We established that there are 5 guaranteed turnovers and we could, if we wanted to, obtain the following position if we ignored the identity of the newly-turned cards.

Obviously in practice we would never ignore the new turnovers, but this shows a “worst-case scenario” and is a useful benchmark to keep in mind (on second thoughts if every face-down card is a politician then perhaps we should be aiming for the fewest guaranteed turnovers).

Bad jokes aside, we can immediately tell that if any of the new cards is an ace we get an extra turnover. Similarly, a 3, 5 or Q is a good card. Further thought shows that 7 is good. Are there any other good cards?

Suppose we go back to the start position. Recall there were three Sixes but only two Sevens, so we could choose only two out of the three columns containing Sixes. One of the columns involves exposing a Queen and moving it onto a King. Aha! That means a Jack is also a good card.

I believe there are no other good cards (check this!) In summary: going back to the start position

  • if we turn any one of A,3,5,7,J,Q then we are guaranteed more than 5 turnovers.
  • If we turn any one of 2,4,6,8,9,0,K then we only have exactly 5 turnovers

Well done if you came to the same conclusion. Even better if you found an error in my analysis 😊

Recall that we had to choose a move before looking at the next card. The most obvious option seems to be column 7 which builds two suited connectors (5/6 in diamonds and 6/7 in clubs), and this does not cost any outs, assuming I haven’t missed anything.

Okay, I agree this analysis was probably a bit over the top for experienced players, but I think it’s useful for a newbie to get used to this kind of thinking. Of course there is more to Spider than counting guaranteed turnovers (especially when you are not close to being forced to deal another 10 cards). In some cases it is wise to “sacrifice” turnovers if we can gain some other advantage, such as getting an empty column or completing a suit. But that lesson comes later😊

Note that Cy the Cynic or Unlucky Louie (borrowing characters from Frank Stewart’s excellent Bridge website) would probably choose to turnover column 7 without doing any calculation – two easy suited connectors and a turnover can hardly be a serious error at this early stage. But I think the calculation exercise is useful because

  • we have some idea of whether the game is going well or badly.
  • You never know if some trick play can be used to gain an edge over the obvious play.

Here is a possible outcome after replacing politicians with actual cards. One can readily compute we get 2 more turnovers on top of our guaranteed 5.

EXERCISE: Start a new game by dealing 10 extra cards from the stock. Calculate the guaranteed turnovers and the probability that the first newly turned card will yield at least one extra turnover. To keep things simple, (i) assume you get to look at the new card before making your first move, (ii) any rank occurs with probably 1/13 (iii) when you see the new card you are allowed to call the suit (but not the rank).

That’s it for now. Next lesson: Empty columns (aka holes)

A trickier example

Okay, I’m feeling cocky. Having won three games in a row, I’m gonna start a new game but automatically dealing 10 cards from the stock regardless of the initial deal. Again we will look at the same questions as last time.

  • How many cards are we guaranteed to turn over even if all face-down cards happen to contain faces of random politicians? (and the rules say we can’t do anything with them).
  • What are the chances of drawing a good card (assuming each rank occurs with probability 1/13 and politicians occur with probability zero).
I’m feeling lucky, let’s DO THIS!!!

This is much harder than the previous example. As I mentioned earlier, one of the non-trivial aspects of Spider Solitaire is exposed cards are not necessarily in descending sequence (now you can see why).

To answer the first question, I recommend the beginner should break the problem down by considering each column in isolation. Assuming the columns are numbered 1-10 from left to right is it possible to turn over a card in column 1, even if we were willing to trash the rest of the tableau in every way possible? Then repeat the same question for column 2, column 3 etc. The answers for each column are summarised below:

  1. Clearly impossible: to shift the Deuce we need another Three, but the only other Three is covered by a king. Note that if the 2-3 were in-suit then yes, it could be done.
  2. Clearly impossible: the Jack can be shifted in two moves, but again there is no way to shift the Deuce
  3. This column contains a king. Don’t. Even. Think. About. It.
  4. Yes, even my Dad can do this.
  5. Yes
  6. Yes
  7. Yes
  8. Yes
  9. No. There is no spare Ten to shift the Nine.
  10. Yes, but only because we are lucky enough to have a suited connector.

So our initial analysis shows that we can achieve a turnover in six out of 10 columns. But we haven’t established that it is possible to guarantee all six turnovers simultaneously. For instance, there could be supply/demand issues (remember in the earlier “non-cocky” example we had three sixes and only one seven) or there could be complex timing issues. But at least we know that the answer to Question 1 is at most 6.

Unfortunately there is no easy way to answer this question and we have to visualise (or experiment with 85,78,68,79). The beginner may find the latter helpful at first, but I strongly recommend he she or ze try to discard this dangerous habit as soon as possible.

You should find that 5 turnovers are possible. Further experimentation might reveal the reason why we can’t get all 6 turnovers: three of the “good” columns contain Sixes and we only have two Sevens. Therefore we can only obtain turnovers in two out of the three columns, (but at least we get to choose which ones). It is pure coincidence in both examples, we had too many Sixes and not enough Sevens.

The diagram below shows the game state after obtaining 5 turnovers. I have replaced the newly exposed cards with random politicians so the reader can easily verify that (i) No further turnovers are possible without knowledge of these cards (ii) This position can in fact be reached from the original position. (In the actual game, the new cards gave me 2 extra turnovers).

Five turnovers is not bad, so my cockiness was justified!

Whew! That was a complex analysis: hopefully I have inserted more than enough corny jokes to prevent the reader from being totally confused and/or bored and I haven’t even attempted to answer the second question, so that will be for the next post. If you understood all of this perfectly and are really bored, perhaps you could try writing your own rap song for this position 😊

That’s it for now, happy spidering 😊