Bart Wright claims a win rate of 25% on 4-suit sans ROT13(haqb). This is nothing to sneeze at if you excuse the terrible cliché. He also mentions he did get through the book Spider Solitaire Winning Strategies, so Bart is doing something right.

Bart cites the following:

If you have a choice of several cards to reveal in the initial deal, suppose you have two columns that have 6 at the end, then turning over a 5 is an excellent choice. I think of it as the “market value” of 5s is low so it’s a good time to buy — there are two columns that demand it and only one to supply. That would be my choice over an in-suit move. Even after you move that 5, you still have space for another 5 if it comes up.

Here is an example to illustrate Bart’s point (Microsoft refuses to give an example with two Sixes and a Five, so this will have to do):

There are two options: shift the 9d onto one of the Black Tens or to build in-suit. Personally I prefer to build in-suit because the chances of getting a good card do not decrease.

To be more specific: suppose a card is good if it increases our minimum guaranteed turnovers (2 in this case). In this example, the good cards are A34679JQ (regardless of which option we choose). I usually prefer to build in-suit. You never know – perhaps the next two cards will be the Nine of Spades and Nine of Clubs. But I agree in this particular example, the difference is small. With three tens exposed, I might be swayed towards shifting the Nine.

The reason this difference is small is because most of the time we will shift both the 9 and 8 of diamonds in some order. But in an alternative universe we might find only one option is available, but not both (this usually happens after at least one row from the stock is dealt). Then it becomes a judgment call. For instance we might choose option A because it is closer to obtaining an empty column, or we might choose option B because that avoids exposing an Ace.

2 thoughts on “In depth Spider 4-suit strategy discussion”

The reddit topic seems to be getting zero interest, so I’m glad to have found a live human being interested in the game. Glad you’re back! Do you have a name? Should I call you “spidergm”?

I agree that you are very likely to make both moves at some point, and that they are very close in value. So we are talking about a small effect and a pretty advanced point in terms of mastering Spider. I’m certainly not sure I’m right about this. But here’s my thinking… After you move the 8 onto the 9, 8 is no longer a “good card”, while if you move the 9 first, 9 is still a “good card”. Whichever method we choose, we now have a column with just 3 hidden cards, so that’s a high priority to “dig” there. Suppose the next 3 cards up in that column are two queens and an ace, which we will eagerly move, as we are hot on the trail of getting a space — we’ve exposed all the cards and if we can move this last card we have a space. Now if we’ve used the Bart method and the next card up is an 8 of spades, I can automatically move that 8 of spades onto the 9 of diamonds (it’s reversible), I haven’t made an in-suit move in diamonds, but I have a space. If on the other hand it’s a 9, then I still have a place to move it. On your method, then if that 8 comes up, you are stuck. Of course different cards might come up, but in terms of where the market price comes in, it’s more likely you’ll need to move a single 8 than that you’ll have to move two 9s. Maybe there are competing cases where your method is better that I might have overlooked.

Here’s also an easier, fun little problem. What is the maximum number of cards you could possibly have in a single column at some point during a game (counting both turned and unturned cards)?

I have other little strategy points to discuss, and is email the best way to get them to you?

Interesting analysis, but I think there is a small error. If you shift the 9 first, the 8 is still a “bad card” because it doesn’t increase your minimum guaranteed turnovers. Yes, you can shift the 8 but only at the cost of not shifting the other 8.

If you have other strategy points to discuss, emailing me is fine.

The reddit topic seems to be getting zero interest, so I’m glad to have found a live human being interested in the game. Glad you’re back! Do you have a name? Should I call you “spidergm”?

I agree that you are very likely to make both moves at some point, and that they are very close in value. So we are talking about a small effect and a pretty advanced point in terms of mastering Spider. I’m certainly not sure I’m right about this. But here’s my thinking… After you move the 8 onto the 9, 8 is no longer a “good card”, while if you move the 9 first, 9 is still a “good card”. Whichever method we choose, we now have a column with just 3 hidden cards, so that’s a high priority to “dig” there. Suppose the next 3 cards up in that column are two queens and an ace, which we will eagerly move, as we are hot on the trail of getting a space — we’ve exposed all the cards and if we can move this last card we have a space. Now if we’ve used the Bart method and the next card up is an 8 of spades, I can automatically move that 8 of spades onto the 9 of diamonds (it’s reversible), I haven’t made an in-suit move in diamonds, but I have a space. If on the other hand it’s a 9, then I still have a place to move it. On your method, then if that 8 comes up, you are stuck. Of course different cards might come up, but in terms of where the market price comes in, it’s more likely you’ll need to move a single 8 than that you’ll have to move two 9s. Maybe there are competing cases where your method is better that I might have overlooked.

Here’s also an easier, fun little problem. What is the maximum number of cards you could possibly have in a single column at some point during a game (counting both turned and unturned cards)?

I have other little strategy points to discuss, and is email the best way to get them to you?

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Interesting analysis, but I think there is a small error. If you shift the 9 first, the 8 is still a “bad card” because it doesn’t increase your minimum guaranteed turnovers. Yes, you can shift the 8 but only at the cost of not shifting the other 8.

If you have other strategy points to discuss, emailing me is fine.

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