This got me thinking on the subject, and I think I've basically figured out how we can make a pretty decent quasi-empirical prediction of what the completion rate of a player solving perfectly should be. To start of it's important to note that the problem consists of two parts, contingent on a few ground rules I set at the outset.Arjádre wrote: Does anyone know of experiments to calculate the optimal percentage of wins on expert, intermediate, and expert (assuming perfect gameplay)? I'd like to compare to the results I'm getting with my solver.

I'm pretty sure I've read someplace that a win percentage of around 25-30% is possible on expert. I just have no idea where that article is!

Thanks!

Let's consider a player who begins a game by guessing randomly, and that the revealed numbers aren't taken into account in calculating the probability of hitting a mine/opening on the next click, then the probabilities associated with the three possible outcomes of the nth click are:

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```
P_failure(n) = p/(A-n),
P_success(n) = N(0)/(A-n), and
P_null(n) = (A-n - p - N(0))/(A-n),
```

Considering an ensemble of games now (all with identical parameters, i.e. identical p, A, and N(0)), it is easy to see that the proportion of successes to failures is simply equal to the ratio p/N(0). Oh, and a success here is defined as hitting an opening, as hitting an opening would allow deductive solving to commence. Most players start this way in my experience, so it should be accurate enough...

Ok, that's the first of the two parts of the calculation of expected completion rates (I know this is scatter-brained, but I didn't sleep and my house almost burned down, so just deal ). The second part is simply taking account of the number of forced guesses the player is going to see on avg. during a game. Then you simply take the two figures and multiply them.

Now, I call this quasi-empirical as the two important numbers in this calculation are at this point only attainable via simulation, namely the number of forced guesses, and the number of 0's on a board (which will of course loosely be a function of 3bv). Fortunately the former has already been touched upon by qqwref in this thread.

An example ballpark would look like this:

For exp qqwref's thread gives around 0.8 50-50 guesses per board, meaning that once we've actually hit an opening our chance of finishing the board is around 57%. Now, lets consider an avg 3bv of about 171 (we're only ballparking, relax). This means that about 210 squares aren't 3bv or mines, or in other words part of openings or opening rims. Lets further assume some totally arbitrary ratio of number of squares on the rim to the number of 0's, something like 9 0's out of every 10 opening squares. This means that we have 189 0's on our fictitious board. Thus the chance of safely hitting an opening is about 52%. Multiply these two numbers (properly, of course), and we get an expected completion rate of 29,6%, or basically 30%.

Ok, granted, half the numbers where pulled out very blatantly from dark places, but the approach, I think is sound. I am of course not a statistics boffin, seeing as I actually failed the subject once, so I might be wrong on any number of points . Anyhow, that's how I approached the problem...

P.S., I'll try and find time somewhere to get a distribution for number of zero's vs 3bv... Should be a nice narrow 2D normalish distribution...