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.**

As an interesting question: Suppose the Jack of diamonds was replaced with the Jack of Spades. What would be your choice of opening move?

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