In the last post I looked at the “hole-in-one” game, where the aim was to get an empty column before being forced to deal another row of 10 cards from the stock. I asked the following questions:
What are the chances of winning, assuming perfect play and a “proper” random number generator?
How would you go about estimating it?
The first question was intentionally difficult. It is not trivial to find a perfect strategy, even for this simplified game where the player doesn’t have to worry about what happens after 10 cards are added to the tableau. Even I don’t have a definitive answer. Instead I wanna focus on the second question.
If you know anything about research, you might think the win rate can be estimated by playing a large number of games, tallying up your wins and losses and then doing some basic math to arrive at a winning percentage between 0 and 100. Or if you want to spice things up, you might guess a winning percentage before playing and see how accurate your guess was 😊
In this case, I estimated my win rate should be 30%. I played 30 games (to give myself the chance of an exact guess if I happened to win exactly nine games). Alas I won ten games out of 30, one too many – so my estimated win rate should be 33%.
my sample size of 30 games is small. Ergo, 33% could be an overestimate or underestimate.
Even the noble Spider GM may not have played the perfect strategy, so the correct win rate has been underestimated.
A more subtle point is optimal strategy may change when we only focus on empty columns. In a real game nobody would trash their game in every way possible just for the sake of increasing their miserly chances of getting that hole-in-one from 2% to 3%. In other words, 33% could be an over-estimate of our chances of getting that hole-in-one.
The real point of the exercise is to give the student a feel of how the simplified game works. If you have played 30 (or more) games of hole-in-one you may have come across some interesting decisions. Here is an example:
There are a number of options to consider:
Two of Clubs onto the Three of Hearts, turning over a new card in the only column containing only three face-down cards.
Three of Hearts onto the Four of Clubs, ensuring the previous option won’t decrease our minimum guaranteed turnovers.
Two of Hearts onto the Three of Hearts, turning over a new card in a column containing four face-down cards but building in-suit
Ace of Clubs onto the Two of Clubs, keeping the above options open.
Although a win-rate of 33% is to be taken with a pinch of NaCl we could still conclude that:
If you do obtain an empty column before dealing from the stock then you are probably in good shape
If you don’t get the empty column then you are not necessarily in bad shape. I know from previous experience I can clear all eight suits about half the time sans ROT13(haqb). If you were close to an empty column but didn’t quite make it, you are probably doing okay.
This is a simplified 4-suit game designed for beginners. The aim is simply to obtain one empty column. It doesn’t matter if you expose five Aces and seven Kings in the process. Any empty column means victory. The bad news is you’re not allowed to deal any cards from the stock. As usual ROT13(haqb) is not allowed.
The purpose of this game is to allow the beginner to actually win something at Four-Suit Spider Solitaire. More seriously, I think this will help the beginner improve his game by focussing on one concept at a time. As a chess analogy, suppose you wanna teach a child how to play for the first time – unless he, she or it is exceptionally gifted chances are it will be overwhelmed by the sheer complexity of the game. So we could consider simpler games e.g.
Pawns only: you win if a pawn reaches the 8th rank, or opponent has no legal moves.
Kings Rooks and Pawns only
Kings Bishops and Pawns only
To win, capture 5 enemy pawns or pieces.
In this way, it will most likely figure out interesting concepts by itself, e.g. zugzwang for the “pawns-only” game or the fact friendly bishops and pawns can protect each other (but rooks and pawns cannot). Although I haven’t tested this method myself (I haven’t even googled), I won’t be surprised if this method actually works.
Admittedly Hole in One ain’t the greatest pun, but “hole” refers to empty column and “one” means no help from the stock – if you need to deal the stock once before getting an empty column that’s a hole-in-two and so on. This means, for example, if an exposed Nine of Hearts is covered by another card, that other card will always be an Eight of any suit.
Here is an example game, with start and end game state shown. This game is lost because there are no legal moves left (apart from dealing another row) and we don’t have an empty column.
Some questions for the reader:
What are the chances of winning, assuming perfect play and a “proper” random number generator?
How would you go about estimating said winning chances?
Apologies for the long break, but it’s been rather busy both onside and outside the workplace.
Today I got an email from a fan, and (no surprise) this means a lot to me particularly in what’s been a difficult year for all of us. This fan is clearly keen to improve his game, and I already like his engaging writing style.
He mentions “Version 1607” of Windows 10 Spider Solitaire. I am not aware of this version. I always thought Windows 10 SS meant the game that is part of the Microsoft Solitaire Collection and unfortunately, I am not computer-literate enough to work out what version number this is. All I can do is post a screen-dump and ask if someone else can verify if this is indeed Version 1607 (and if not, then where do I get it?)
What I do know is I have another reason to revive this blog, so readers can expect to see regular updates – and there might be an odd video or two on Youtube as well 😊
If you are keen to improve your game, don’t be afraid to drop a comment in one of my posts.
It turns out we are able to clear Diamonds (this is an exercise for the reader). Note that only one Ace is currently exposed and this gives us some decent flexibility. Perhaps my strategy of refusing to complete a suit for fear of exposing three new Aces has paid off. We can then clear Spades. With three empty columns and two suits cleared, things are looking up!
And the lucky last card is a …
drumroll … drldrldrldrldrldrldrldrldrldrldr …
Ten of Diamonds!!!!!!
Winning this game is left as an exercise for the reader 😊
In summary: this game started well and if we were to assume random shuffling, then I would estimate our chances of winning are heavy odds-on. But given this was a “master” level hand (the middle-level difficulty of Four Suit spider), I anticipated there would be difficulties in the middlegame and I was “not disappointed”. But I got through in the end. I hope you enjoyed this game as much as I did.
Ninja Monkey did not enjoy this game however. The poor thing only managed to win 1 game in 50 even with its improved random move algorithm.
We expose a card in Column 4 with some trepidation, knowing the game is not headed in the right direction. It’s a Seven of Hearts and we must deal another row.
Now we have a problem: it is not possible to shift the 9-8-7-6-5-4-3 in Column g onto one of the tens despite an empty column. This is a critical stage of the game and margins are extremely thin. We might be on the wrong side of the ledger.
We turn over a card in column ‘b’ and pray for luck. We get a useless 10 of clubs and must deal the last row of cards.
Last week I added my third video – this time about Empty Columns. And I have a <sarcasm> whopping </sarcasm> 13 views at last count.
Of course the real motivation for this vid is revenge on one of my Facebook friends for a really lame prank. This kind of thing should have been stamped out many years ago, but I can’t give too much away. You just have to watch the video here and judge for yourself.
Unfortunately there’s a lot going on at the moment so I won’t have time to update the blog as much as I used to. Recently I solved an extremely difficult Sudoku puzzle on Puzzling Stack Exchange and I’m also into inventing my own Sudoku puzzles. And I still have a full time job that has nothing to do with mathematical games or puzzles …
Recall that we can remove a full set of Diamonds. There is a hidden catch I didn’t mention from last week. Removing Diamonds would imply we uncover no less than three Aces. There are exposed Aces in columns d,g,j. Any experienced player knows that too many exposed Aces can kill a game (perhaps even quicker than Kings) because nothing can play onto an Ace. Too many Aces exposed means a restricted set of legal moves at every stage of the game and the only way to fix this is removing a complete suit or dealing a new row of cards. One of these is usually not desirable and the other is difficult to achieve. I’ll let you guess which is which 😊 In any case we know that there will always be a full set of Diamonds exposed no matter what turns up on the next deal of cards. I decided to gamble by turning over a card column c even though I am no longer certain to remove Diamonds. I exposed a Two of Spades.
Next is the Five of Diamonds. We shift that onto a Six and reveal the Eight of Hearts. We are forced to deal another row of cards:
Well that was awkward. We have three Fours and no Fives. Perhaps I should have removed Diamonds while I had the chance, but on the other hand chances are I would not have liked those ten cards no matter what I did.
This raises an interesting point: if you were paying attention you might have noticed I am playing “Spider Master” instead of “Spider Random”. Since Master is the “intermediate level” of all Four-Suit games (the levels are Expert/Master/GrandMaster) we do not expect an easy ride. We started with some good luck in the beginning and therefore we were due for some bad luck. Under normal circumstances, this reeks of “Gambler’s Fallacy”. But given that I did not choose “Spider Expert” or “Spider Random” I will stand by this judgment. Needless to say, Whinging About The Injustice Of It All is not a recognised strategy by the experts, so how would you continue?
This is a reasonable set of 10 cards. We can obtain no less than three empty columns – if we so choose. But what do we do for an encore? Recall that an action ends when we expose at least one more card, so getting an empty column doesn’t cut it for an advanced player. Column ‘b’ sounds like the best source of new turnovers since any other column requires us to spend at least two empty columns. So we have three guaranteed turnovers since there are three empty columns and at least three face-down cards in column ‘b’.
We should also start thinking about removing suits. There is no suit with all 13 cards visible, but we can “threaten” to remove a suit or two. The Six of Spades is a nice card since it enables us to complete a run from King to Three. We can also guarantee a run from King to Five in Diamonds. With the 3-2-A in sight we have a “twelve-suit” with only the Four missing. With 30 cards in the stock and 15 in the tableau it is heavy odds on the Four of Diamonds will be found in the stock (recall there are two decks so the odds are better than 2:1). Unfortunately, building these partial runs will imply a reduction in our minimum guaranteed turnovers. I guess one could compromise, e.g. turn over one card and still retain options of building partial runs. So there is plenty to think about and it is difficult for even an expert to find a “definitive best play”.
Another consideration is that we have had a relatively easy run so far, so we might expect some “bad luck” given the hand is not random (we chose Master level). The initial row had 5 turnovers, and both rounds 1 and 2 have at least 4 turnovers (pretending they are the start of a new hand). So I won’t be surprised if something unpleasant happens on the next round of 10 cards. I guess if the next round is e.g. all odd cards then we probably can’t do much about it anyway. So we will cross that bridge when we come to it, if you pardon the terrible cliché.
At this stage of the game I will not give a sequence of moves but simply give the resulting diagram and leave it as an exercise for the reader to verify this diagram can indeed be obtained from the initial position.
The Jack of Hearts is a good card giving us an easy move that even my Dad can’t 70,85,67,75 up.
That’s a nice card. Shifting the 5s onto the 6s gives us a suited build, an empty column (Q of Hearts onto K of Diamonds) and a turnover (8c onto 9s). Nice!
Note the use of procrastination. We can always get our empty column back.
Aha! We drew a Nine so we retrieve our empty column without having to commit to <ed>. We might choose to empty column ‘j’ – this allows us to connect the 32A in Diamonds and recall that we have a twelve-suit in Diamonds with only the Four missing – no wait up, we just turned a Four of Diamonds two moves ago. Things are looking good!
Exercise for the reader:
Can we complete a full suit of Diamonds?
If yes can we procrastinate? i.e. can we turnover a new card (not necessarily column ‘b’) and still retain the option of removing the Diamonds?
I have now started a new project: Spider Solitaire videos on YouTube with the aim of teaching other players how to play well at Four-Suit Spider Solitaire sans boop. So far I have twovids, an introduction to the rules and a lesson on guaranteed minimum turnovers.
This is also an opportunity for me to “upskill” e.g. how to use OBS Studio or how to prepare material for lessons etc. I’ve learnt quite a bit after two videos – for instance listening to my own voice just downright 83,85,67,75,83 etc. With the pandemic not going away any time soon, I thought it was just as well to learn something new – who knows, my mad skillz might come in handy one day.
Unfortunately this means the silly stories you all know and love will be put on hold temporarily, but hopefully I can compensate with even cornier jokes. I can’t claim to be raking in the views just yet, so my modest goal is to achieve at least the square root of the number of videos for Aad Van Der Wetering’s Pi Sudoku – apparently you can write the first 20 decimals of pi in an approximate circle inside a 9×9 grid, add in a King constraint (no two equal digits diagonally adjacent) and voila – there is a unique solution! This video is sitting pretty at 301,405 views so I need 549.0036429751628 views (at the time of writing) and I’m only up to 45 views for both videos combined. So if you have plenty of spare time on your hands, haven’t seen my vids yet and are desperately keen to improve your Spider Solitaire game then you know what to do 🙂
Continuing from last week, we had the following position
We have only one guaranteed turnover. Let’s hope its a good one.
< gc, ab, jb, ef, if, hf, dj> As
Note that I deliberately broke the 7-6 of Clubs in column ‘i’. This is because if we expose an Eight then we get an empty column. The down-side is of course if we don’t draw an Eight then we are stuck with one less in-suit build then we “deserve”. This type of trade-off is typical in the middlegame. It would be nice if we could ensure (1) any Eight wins back an empty column and (2) we don’t lose any in-suit builds. Unfortunately I can’t see any way to achieve this.
Another reason for breaking the 7-6 of Clubs is I can arrange ten out of thirteen cards in Spades in a single column. From experience I find this increases the chances of completing Spades when the missing cards finally do come out. This type of “advanced” consideration can easily outweigh “basic statistics” such as the number of in-suit builds.
Finally we have to decide which of the four left-most columns to turn-over. I chose column ‘d’. I turn over an Ace of Spades and must deal a new row.
As usual, whenever a new row is dealt it is wise to study the game state in some detail before making any moves.
Do you think this is a good set of 10 cards under the circumstances?
How many guaranteed turnovers do we have?
Can we complete any full suits? Yes, No or Not Even Close?