Empty Columns (a.k.a. holes)

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 😊 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.
How would you play this position?

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 don’t 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 protagonist’s terrible Dad joke.

Did I lose 5 points for a 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 🙂

The day we’ve all been waiting for

Today, every man dog and millipede on the planet celebrates National Solitaire Day. Solitaire is a card game that’s been around for over 200 years. But it was 1990 when Solitaire truly became a thing thanks to Microsoft. According to legend, Microsoft Solitaire was designed to teach computer users how to use … hey W,H,A,T,T,H,E,70,85,67,75 ?!?!?!?

In this day and age, it’s hard to imagine anyone credibly claiming to have invented a game with the intention of teaching people to use a mouse properly. But given that playing Klondike is beneath my dignity, all this is moot anyway. Of course I am going to celebrate today by playing a different game on my i-Phone instead. May the Four-Suit Spider Solitaire be w- okay that joke is lame, so I won’t bother completing it ☹

Okay, I confess to not knowing about National Solitaire Day until a few days ago, and I must thank Susan Sleggs for mentioning this on her own excellent blog. And she could probably give me some tips on writing short stories.

But for now I’m about to move one of the Heart Fours onto the 5 of clubs. Any 2,3,5,7,9,J or K will guarantee me at least one extra turnover. Let’s hope it’s a good one!

National (Spider) Solitaire Day

Seven of Hearts, hooray!!!!!!!! Life is good 😊

To be continued

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 😊

An example start position

Okay, so I ascii2word([70,85,67,75,69,68]) up. Apparently WordPress automatically converts xx-xx-xx-xx to hyperlinks (e.g. thinking it represents an 8-digit phone number). So instead of writing xx-xx-xx-xx I shall use the notation ascii2word([xx,xx,xx,xx]) instead.

EDIT: This is only relevant for mobile phone devices

By this stage the impatient reader probably wants to see some “action”. Here is a possible starting hand in Spider:

Let us try to find the best move in this position.

I recommend that a beginner player should start by asking the following questions: (i) how many cards are we guaranteed to turn over even if the worst possible cards turned up? Another useful question is (ii) What are the chances that the first new card turned over will be “good”? We will take good to mean “increasing the number of guaranteed turnovers”. Of course there is more to Spider Solitaire than counting guaranteed turnovers but if you’re a beginner then simplicity is the mother of self-improvement … or something like that.

Strictly speaking, it isn’t necessary to ask these questions to arrive at a good first move in the start of the game. If the first two columns are 3-4 of Hearts, then you could move the 3 onto the 4 regardless of the other eight columns: if it’s not the best move then the difference is small. But these questions will be good practice, and it will come in handy as the game progresses.

I hope you answered “Four cards” for the first question. Ignoring suits for now, we have J-0-9 for two turnovers, 7-6 for a third turnover and finally 4-3 for the fourth. Obviously we can’t count multiple turnovers for the three Sixes since we can’t stack them onto the same Seven without violating the laws of physics! Similarly, we only count one turnover for two Nines. Assuming we don’t ascii2word([70,85,67,75]) up the move order, we will turn over at least four cards before being forced to deal another row.

For the second question, there are 13 possibilities for the next exposed card (if we ignore suits). An Ace is clearly useless since we have no deuces, but a deuce I’d like to see since we have a three … okay that’s probably not the best way to start a rap song, but you get the gist.

Continuing in this fashion we get the following good cards: 25780Q. The chances of getting a good card is therefore 6/13. Note that 5 and 8 are especially good since we get two new cards instead of one. But the question defined good as “allowing at least one extra turnover” and it didn’t ask for “how good”. Assuming you have completed Year 3 or better in school, you should know by now that it is always wise to make sure you are answering the correct question! The observant reader may have noticed an error (okay, maybe one-and-a-half errors) in the above calculus. Before proceeding further I invite the reader to figure it out. To protect against accidentally reading spoilers I have inserted an image consisting of happy stars and blank spaces. Each happy star represents a point I obtained for a short story competition I entered some time last year, with a maximum score of 100. Unfortunately I didn’t win anything, not even a Honorable Mention. Perhaps the judges secretly docked 5 happy stars for the protagonist’s terrible Dad joke but we’ll never know. ascii2word([70,85,67,75])!!!!!

The first error is I have assumed each of the 13 cards from Ace to King occur with equal probability. This is not correct since we already know e.g. there are three Sixes and no Fives visible. Hence Fives are much more likely than Sixes. The probability of 6/13 is therefore only an approximation of the true probability of getting a good card. As a general rule, failing to take into account cards already exposed will almost always underestimate the true probability at the start of the game. With only 10 cards exposed, this error will probably not contribute much to All The Problems In The World As We Know It.

The other half-error is we must choose our move before seeing the next card and this may affect our chances for the worse. For instance, suppose we move the Ten in column 8 onto the Jack in column 5. Any Queen is no longer a good card unless the 10 and Jack are the same suit. Fortunately we are in luck here since they are both diamonds. This is why I only counted 1 and a half errors instead of 2. Clearly, moving the Ten onto the Jack is good because we “don’t lose any outs”.

Note that if we moved the Three onto the Four we don’t lose any good cards despite it being off-suit, since if we draw a Five it can still be played onto a Six. But obviously we want a Five to be “very good” (two new cards) instead of “just good” (only one card). This might sound overly technical, but this kind of deduction must become second nature if you aspire to kick ascii2word([65,82,83,69]) at Spider.

Okay, this example fails the Duh Test since one can arrive at the best move by observing it’s the only move that builds in-suit. But my point was to illustrate the concepts of counting guaranteed turnovers and calculating outs.

FUN FACT: Assuming perfect shuffling, a player should have on average 3.96 guaranteed turnovers at the start of every game.