This is probably not a good idea, but I wanna start a “new game with a difference”. Standard Four-suit Spider Solitaire, sans rot13(haqb) and there is a second win condition which will become apparent in due course. Not gonna spoil this obviously, but I can mention that (i) some basic cultural literacy is assumed and (ii) if you are familiar with the usual quirky humour in this blog, you will not be disappointed 😊
The game shall be played on Microsoft Windows at the “Four-Suit Random” setting. Here is our start position:
We have four turnovers, two of which are in-suit. But we also have an Ace and King showing. If given the choice, would you take a Mulligan or play the hand?
If you take the Mulligan do you want Master difficulty (between expert and GM) or random?
Edifying Thoughts of a Spider Solitaire Addict 
The quick answers: “yes”, and “random”.
One key question is what our goal is. Let’s assume it is to maximize the chance of winning. I would think so unless we’ve said something else in advance, like “Produce an interesting hand”. I don’t think even GM claims to have a win record of 50% or above. Let me say “40%” as a guess. Mine is worse and I think SA’s is too. There is also the question of whether the outcome here will be improved by committee work or made worse. Given how infrequent and partial our communications are, I would say worse, and GM seems to take our input only rarely. So let’s say the chances of winning are GM’s 40%. That I think implies that you would take a Mulligan on an initial deal that is only very slightly above average. You might think that we would only keep an initial position that is in the top 40% given what we know at the time, but the “real” math is undoubtedly not that simple.
I could put together a computer simulation to classify opening deals by various criteria combined into a single score, so we could at least be guided by that. “# guaranteed turnovers +(0.3 * in-suit turnovers) – (0.5 * number aces) – (0.25 * number kings)” would be an example. Another component might be “number of ranks that will produce a known new turnover after known turnovers are done”. In our case it would be King, Jack, NOT 10 because something out-of-suit was put on the 9, 9, 7, 6, 3, and 2. That’s 7 of 13. Is that good? We could even weight them by how many of each rank are left in the deck. The program then lets you evaluate any given deal against the score and tell you the chances that redealing will give a better score. But the weights on the components of our formula will just be guesses, so this way lies a lot of work for nebulous gain.
The one tweak I would make to GM’s evaluation is that one of the 2 in-suit turnovers is a queen on a king, which is less valuable than other in-suit turnovers in that it does not help with future turnovers. It’s very close, but I judge this is exactly average, so I would take the Mulligan.
As for question 2…
“Would you choose random or master?” is interesting… it sort of presupposes we WILL be taking the Mulligan, or GM wouldn’t ask the question. But that doesn’t affect our own judgment, right? We are independent thinkers, right?
I honestly don’t know what the characteristics of the different levels are. However, the system gives you 8,000 points if you win at Master level and only 6,000 if you win at Random. Assuming there’s a good reason for that, and that it applies to “nosy key” hands even if designed for “zeeky” hands, I should think we would take “Random”.
I could make shorter posts, honest, but so far this blog hasn’t been plagued by too much text in replies.
LikeLike
Hi Bart, thanks for your reply. I agree that in the last few hands I often refused your suggestions – and perhaps this was my fault. This time, the reader will be a lot more involved. The format of this hand will be several suggestions are made and then you (or Schistocerca Americana) have the final decision. The “extra win condition” will further test your decision making skills but more on that later 🙂
I always enjoy reading your posts no matter how short or long. Keep up the good work!
LikeLike
Play, always play.
LikeLike
So… the simulation with just one variable was easy to do and meaningful. How many turnovers will we see when we look at a fresh set of 10 cards?
Key: Num turnovers, Proportion, Cumulative Proportion, raw count
0 turnovers: 0.00593, 1.00000 (11864)
1 turnovers: 0.03422, 0.99407 (68445)
2 turnovers: 0.10995, 0.95985 (219905)
3 turnovers: 0.21659, 0.84989 (433177)
4 turnovers: 0.27594, 0.63330 (551872)
5 turnovers: 0.22018, 0.35737 (440368)
6 turnovers: 0.10574, 0.13718 (211484)
7 turnovers: 0.02778, 0.03144 (55562)
8 turnovers: 0.00350, 0.00366 (7007)
9 turnovers: 0.00016, 0.00016 (316)
The “cumulative” column is most interesting, I think. It is roughly, “How likely am I to see this number of turnovers or more when I look at 10 cards?” So if you see 4 turnovers, chances are 63% you will see 4 or more if you Mulligan. If you see 5 turnovers, chances are 36% you will see 5 or more if you Mulligan. Four is slightly worse than average and five is slightly better.
To compute turnovers, I first compute the lesser of the count of any two adjacent ranks. Add this up for all 12 such adjacent pairings, and that’s the guaranteed turnovers, I think. Do you think?
It’s going on 9 years since anyone has paid me to write programs. Beware.
In the case given, since our 4 turnovers is slightly worse than average, I would say that is a “pro-Mulligan” argument.
LikeLike
Hi Bart, this is a very interesting analysis!
I remember doing something similar many years ago when I wanted to verify the stats quoted by Steve N Brown in his book Winning Spider Solitaire Strategies. At first I thought you goofed because 4 was supposed to be slightly better than average, but I found the true idiot in the mirror because I confused mean with median.
It turns out 4 turnovers is better than average if you argue the “expected number” of turnovers is 3.97. In layman’s terms this means if you deal 1000 hands you can expect the number of total turnovers is close to 3970.
In other words, 4 turnovers is good (bad) if you’re looking at medians (means). But the difference is so small I would be more swayed by other factors such as how many builds are in-suit or the number of Aces and Kings showing.
LikeLike
I had forgotten such a number was in that book, which is totally par for my forgetting these days. But I do get the same number of 3.97 for mean turnovers when I tweak my program to report that, so that’s a good sanity check.
You are kind to say my different conclusion has to do with means and medians, but it is more simply a mistake on my part to say, “Four is slightly worse than average and five is slightly better.” Five is much better than average. I was misinterpreting my own table. Part of my confusion arises because there are so many ties. Here’s another breakdown:
Under 4: 0.3666
Exactly 4: 0.2759
Above 4: 0.3574
So if you take your Mulligan and get something other than that same 4 again, chances are slightly better of things getting worse. That is consistent with 3.97 for the mean.
(As an aside, I think mean is more relevant than median in this case. We care a lot how much better than average a given good result is, and how much worse than average a given bad result is).
“I would be more swayed by other factors such as how many builds are in-suit or the number of Aces and Kings showing.” This is surely true when we eliminate my erroneous idea that 4 was worse than expected.
I still think our answer can change in accord with our expected base rate of wins.
To take an extreme example of this, suppose our goal is to get 3 6s showing on 3 dice. We first get to roll, then after keeping any 6s we get, reroll the other dice. Now suppose on our initial roll we get one 6, and then we are offered a Mulligan. Do we take it? Clearly getting a single 6 is an above the average result of 0.5 6s. However, we do better to take the Mulligan.
Without it, chance of winning is 1/36, rolling for two 6s, or .0278.
With the Mulligan, the math I get is:
(125*1 + 75*6 + 15*36 + 1*216) / (216^2)
or .0285, which is bigger than .0278.
(Terms above are chances of getting 0, 1, 2, 3 6s on base roll, multiplied by chances out of 216 of getting remaining needed 6s).
Although our first roll was above average, our goal is so difficult to achieve that we do better to try again in our “Hail Mary” situation.
(And what are the chances I’ve done this right? Groan…)
Now getting an initial result that is twice as good as chance is a big effect, and we need a very difficult problem to make it worth giving up that big advantage. But if our chances of winning Spider was (say) 1 in 4, then we might take a Mulligan on an initial result that is slightly above average.
LikeLike