**The Legend of Blacksilver****Commodore 64**and Apple II

**Date Started:**7 February 2018

*The Legend of Blacksilver*is the fourth and last of the Charles Dougherty games, which began with

*Questron*(1984) and progressed through

*Legacy of the Ancients*(1987) and

*Questron II*(1988). You may recall that

*Questron*was the first CRPG I ever played--was, in fact, the game responsible for my addiction. I remain fond of it, particularly for the ending, but with 34 years of hindsight, I recognize its weaknesses.

*Blacksilver*is the largest and most polished of the series, but unfortunately it still adheres to some of the odder and less satisfying conventions that Dougherty developed for

*Questron*, some of them based on

*Ultima*(possibly the only CRPG that Dougherty played), and some developed out of a desire to compensate for

*Ultima*'s weaknesses. (For a full discussion, see my interview with Chuck Dougherty.) These include:

- Leveling via plot progression and not by killing monsters.

- A highly original but equally unmemorable monster list, most using two seemingly random words. Entries in the bestiary this time include fetid wheezes, stench creeps, mammoth gulps, and screaming nugs.

Chester battles a "spine quiver." |

- The overall pointlessness of combat, given that monsters deliver no experience and generally cost you more in hit points than you earn in gold.

- Extremely limited character creation and an unchanging character backstory.

- A plot that progresses like a treasure hunt: go to one place, find a key, open a door in another place, find a gem, access a third place, and so forth.

- A magic system that features a limited selection of spells that must be purchased individually.

- Weapon and armor upgrades that appear based on time spent in the game rather than material progress.

- Eminently-cheatable gambling minigames.

- A game world featuring two continents, one of which (the "evil" one) must be discovered during gameplay, this discovery altering many of the fundamental rules established to that point.

One of what I assume will be several promotions during the game. |

*Ultima*innovated and compensated for its early weaknesses throughout the 1980s, Dougherty stuck repeatedly to the same formula. I can't help but lament what might have been if he had achieved the slightest awareness of the conventions being established around him.

*Blacksilver*takes place in the land of Thalen on the planet Bantross, its history well-told in about 14 pages of narrative, letters, and other documents, showing an Origin-esque dedication to world-building. A couple centuries ago, the peaceful people of Thalen would banish criminals to the evil island of Maelbane, sentencing them to hard labor in the mines. Such a fate befell an evil mage named Minon. But while working in the mines, he discovered deposits of Blacksilver, "the source of all magic on Bantross." Using this power, he took over the mining operation, raised Maelbane from the sea until it had grown from a small island to an actual continent, and raised an army to conquer Thalen.

The Prince explains the current threat. |

Like Luke Skywalker, this PC will take any excuse to avoid working an honest day job. |

Character creation consists only of a name. Every character starts with 15 each in strength, dexterity, charisma, endurance, and intelligence, and maximum of 200 hit points. He has 60 food and 20 gold. He is a serf by trade, a theme as common to the Dougherty titles as starting in handcuffs is for

*The Elder Scrolls*series.

"Character creation." |

The player starts in the middle of a swamp and must navigate to civilization. Fortunately, the accompanying game map is relatively detailed. Town names include Iron Forge, Clissold Creek, Crystal Summit, and Bad Axe, and each has a different variety of services: weapon shops, armor shops, magic shops, food shops, jails, banks, and casinos.

A player of previous Dougherty titles will be familiar with the conventions. Banks offer interest and protect your gold from character death. Magic shops sell one spell at a time, although some are unavailable until the character later passes some kind of magical test. Weapon and armor shops start by selling very basic gear--daggers and whips for weapons, leather and studded armor--but slowly add more advanced items as both time and levels progress. This game adds a quality rating to each item, from "shoddy" to "superb," slightly affecting its value.

Weapons for sale late in this session. Giving each store its own name is a nice touch. |

In jails, you can bribe guards to talk to prisoners, then bribe the prisoners for hints and clues. These are the only NPCs of that sort in the game.

This turned out to be a key piece of intelligence. |

Finally, most towns have some kind of trade location where you can work an odd job for a day or get a mini-quest to deliver a package to another town. I believe these were also options in

*Legacy of the Ancients*.

Sure, that seems like a good use of time. |

In addition to the venerable blackjack, there are two new casino games here. "Dragon Wheel" is just a slot machine variant with poor odds. But "heigh-loagh" is devilishly clever. You get dealt a card from a standard deck and then have to guess whether the next card is going to be higher or lower. If you get four guesses right in a row, you win four times your bet (including the original bet amount). I spent longer than I spent playing the game trying to calculate the odds. I knew

*how*to do it, but I couldn't come up with an easy

*formula*to do it. I might actually give this to the students in my statistics class and see how they solve it.

Naturally, you're going to guess "high" on a 7 or lower and "low" on a 9 or higher (aces are always high). An 8 could go either way. You lose on a tie. Thus, in the first round, the best you can hope for is a 2 or an Ace. The only way you can lose is if you get one of the other three cards of the same value, which will naturally happen 3/51 times, or 5.88% of the time, giving you a 94.12% chance of getting it right. An 8 has the worst odds: no matter which way you guess, you only have a 47.06% chance of guessing correctly. Add up all the odds for the individual cards, and you get a 72.40% chance of correctly passing the first round.

A pretty easy game so far. |

It's the other rounds that are difficult to calculate, because the odds change based on what card you get the first round. [

*Here I digressed into a long analysis of the odds that was both unclear and ultimately incorrect, and the reply comments just pissed me off, so I'm deleting it to try to stave off any more replies.*]

The odds are somewhat important because currency is the lifeblood of the Dougherty titles. Among other things, it's the only reason to engage in combat, since you get no experience, and gaining money by killing enemies (while paying to restore hit points at the same time) is

*slow*. Combat-related payoffs are a bit better here than previous titles, but a lucky trip to the casino can save several hours of grinding.

Despite taking all the time to calculate these odds, I spent most of my time playing blackjack, which also has nearly-even odds. Any game that approaches 50/50 is vulnerable to the martingale betting strategy as long as you're using fictional money. (I'd use it more often at Foxwoods if I could just go outside and kill someone for $120 every time I went broke.) Using my usual rules and strategy, I kept my gold pieces between 300 and 3000 for most of the session.

When this happens in real life, it is among the best feelings in the world. |

Meanwhile, outside, there are new combat rules. In previous games, you could only attack or try to flee by moving away. In

*Blacksilver*, you have four initial options when you encounter an enemy: approach, stalk, wait, or flee. During combat, you now have two attack options: a cautious "battle" and a more reckless "charge." These are still relatively primitive tactics, but better than the previous titles in the series.

The new encounter options. |

This is a conundrum. |

You sure you don't want me to kill your entire garrison just to prove my worth? |

*Questron*, which allowed me to increase my dexterity if I did well. I did okay and got it from 15 to 23. If

*Blacksilver*is like other Dougherty games, some plot development will cause it to increase well beyond what I can accomplish with the minigame, but in the meantime that helps for combat.

The Dougherty titles have always been strong on minigames. |

*Legacy of the Ancients*. Scattered along the archives' walls were a number of exhibits with titles like "Game of Honor," "Etherium," "Vase of Souls," "An Island Retreat," and "Storming Gear." Each exhibit asked for a different type of crystal or gem to view it. (The ones in

*Legacy*had used coins.)

The priest explains how things work. |

Chester thought he had found a valuable artifact, but it turns out that it just bellows, "She'll Be Comin' 'Round the Mountain" incessantly. |

It truly was. |

Beating up blue orcs. |

"Such as, ironically, where to find the Orb." |

I suspect this isn't going to be the vacation that the title suggests. |

The dungeon offers better monster graphics than previous titles in the series. |

The who-ly grail. |

*The Legend of Blacksilver*is a briskly-paced game with a moderate difficulty level. It is likely destined for the high-30s or low 40s on the GIMLET, and will probably be the highest-rated of the Dougherty titles. At the same time, it's a bit disappointing at how little progress has been made since

*Questron*in the areas of NPCs, character development, equipment, and combat tactics, and it thus has the effect of recalling something pleasantly familiar rather than offering something pleasantly new.

*Time so far:**5 hours*

Okay, wait. I'm confused about the "heigh-loagh" game. You say that you lose on a tie. But then a few paragraphs later you say:

ReplyDelete[W]hatever card appeared in the previous round cannot, by definition, appear in the subsequent round. This means that if a 3 appeared in the first round, the odds of any other card appearing in the second round are not 4/51 but rather 4/48: all 3s are excluded from the population.But if that's the case, there can't possibly be a tie. Or am I missing something?

Also, how do you figure that 8 7 8 9 gives you the best chance of getting the last round correct? It seems to me that would be a rather

Deletebadsequence to get; your chances of guessing right on the last round would be only about 52.1% or 55.6% (depending on whether or not ties are possible), not much better than even odds. What about, say, 2 J 9 2? Then your chances of winning on the last round would be 96% if there's still a possibility of a 2 coming up, or 100% if ties are impossible. Or am I completely misunderstanding the rules of the game somehow?This comment has been removed by the author.

Delete"This means that if a 3 appeared in the first round, the odds of any other card appearing in the second round are not 4/51 but rather 4/48: all 3s are excluded from the population."

DeleteNor could a 2 have appeared, because you would have lost in that case as well. ie [3][2] and [3][3] both result in a loss. But concerning yourself with sequencing is only important if you're trying to calculate the win rate by enumerating all winning permutations within the set of all possible permutations.

DeleteThere can be a tie. You lose if there is. But when you're trying to calculate the odds for the second round, you can exclude the possibility of a tie, because you never would have arrived in the second round in the first place if there was a tie. Does that make sense?...No, I'm sorry, but it actually doesn't make sense. No, you couldn't have arrived at the second round if there was a tie in the

firstround, but you're excluding the possibility of there being a tie in thesecondround. If your second card is a three, thefirstcard couldn't have been a three,but your third card can still be a three, resulting in your losing the game with a tie—and that's the possibility you're excluding. By removing all threes from the population, you're not accounting for the fact that there couldn't have been a tielastround; you're excluding the possibility of there being a tiethisround.Also, stopping one card earlier doesn't really help the argument about the 8-7-8/8-9-8. The slight benefit of having eliminated some of the worst cards is more than made up for by your horrible odds on the third guess... your chances of your last two guesses winning the game starting with these three cards are still only about 35.7%, as opposed to 68.5% if your first three cards are, say, 2-9-2. I guess you could say that this sequence gives you a slightly higher contingent chance of guessing correctly on your

fourthguessgiven that you succeed on the third, but that seems like an odd and obscure thing to be concerned with.Actually, I'm curious enough about the odds of the game that I think I'm going to write up a short C program to calculate them... I agree that there doesn't seem to be a simple formula for the overall probability, but writing a program to calculate it shouldn't be too hard.

Okay, actually writing a C++ program to calculate this was easier than I expected. It seems the probability of winning, assuming an optimal guessing strategy, is 393083 out of 1656200, or about 23.734%. So the game is slightly in the house's favor; you get an expected return of 0.95 gold pieces for every gold piece spent.

DeleteOf course, if Commentman is right and the draws aren't randomized, then this is all moot anyway...

Oops... just realized I made a slight mistake in the program, though it doesn't affect the odds much. The corrected winning probability is 147167 out of 621075, or about 23.696%; the expected return is about 0.948 gold pieces for every gold piece spent.

DeleteHere's the program if someone wants to check my work, though I admit I didn't write it with readability in mind so some of the code may be a little arcane. Though, again, if the draws aren't randomized this is all moot.

...and I just realized I made another mistake. Dagnabbit. Okay, corrected winning probability 588668 out of 2436525, or about 24.16%; corrected expected return 0.966 gold pieces for every gold piece spent. Corrected program here.

DeleteAt this point I don't think I can necessarily guarantee there isn't another slight mistake somewhere, but I'm confident the probability should be pretty close to that.

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DeleteNo, I understand you're calculating one round a time, but you're still making a mistake. If you exclude all the threes, you absolutely

Deleteareexcluding the possibility of a tie in the current round. But yeah, it's not worth arguing the point further; I'm okay with dropping the matter here.This comment has been removed by the author.

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DeleteOkay, so if I understand you correctly you're assuming that you

Deletesucceededon the third guess, and that you're waiting for the fourth card to come up. Okay... then that falls into what I was saying about the "contingent chance of guessing correctly on your fourth guess given that you succeed on the third", which doesn't seem to me like a particularly meaningful thing to worry about (and actually, you get exactly the same probability for 8-5-8 or 8-10-8 or whatever; that second card doesn't matter for this result, since if you're guessing optimally your third guess would always have been on the opposite side of that second card). But again, it's probably not worth arguing further; I'll drop the matter here.Hm, that's odd... I got an e-mail notification of another comment from you here, but I don't see it. Did you post a comment and then delete it, or is it taking a while to show up?

DeleteIn any case, I think I may owe you an apology. I'm sorry if the words "wrong" and "mistake" seemed harsh, but mathematics is one of those places where there generally

isone right answer. But, more importantly, I'm sorry for not giving more consideration to the possibility that we were just taking different approaches. The possibility did occur to me, but I couldn't think of any approach to the problem where it would make sense to exclude the 3s from the population. But thinking further about the matter, yes, there is a way to approach the problem where that would make sense, and presumably that's the way you were doing it, which means I was wrong about your being wrong. (I think you're calculating thecontingentprobabilities of winning each round given that you won the previous, and then multiplying those together, is that right? That's a different approach than I'd take, and not one that would necessarily have immediately occurred to me, but yes, of course it would give the right answer, and if you're doing it that way then what you're doing with excluding the 3s from the population makes sense.)I apologize for getting so argumentative here. I don't think normally I'm an argumentative person (though I'm a lot more argumentative online than I am face to face, which is probably true of most people), but I guess there are certain topics that get me going, and mathematical mistakes (or apparent mistakes) is one of them. But I should have given you the benefit of the doubt and given more consideration to the possibility that you were just using a different approach; I'm sorry. And I'll try not to be so confrontational if like matters come up in the future.

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Delete...Now you've got me wanting to apologize for apologizing.

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DeleteYeah, it's definitely a tough problem.

DeleteFor what it's worth, I just wrote up a Monte Carlo program to try to get a rough probability another way and check my answer (program here)... it's just using the standard C++ rand() pseudorandom number generator, but I figured that might be good enough for a quick and dirty estimate. Out of 10,000,000 trials it got 2432904 wins and 7567096 losses, so about 24.33%, which is in the ballpark of previous answers.

No question it's a tough problem, though. But an interesting one.

I brute forced it in R; the number of combinations of 5 different cards (disregarding color) is not that high, even if my program is not optimal at all. I get a result of 24.28% of winning. Program follows.

Deletex <- c(1:13) # cards as numerical values

# includes all sequences of five cards, even impossible ones (five aces)

combinations <- expand.grid(x,x,x,x,x)

n <- nrow(combinations)

total.win <- 0

for (i in 1:n) {

# is sequence a winner

winner <- TRUE

bet.high <- ifelse(combinations[i, 1:4] < 7, TRUE, FALSE)

differences <- diff(as.numeric(combinations[i,]))

if (any(differences == 0)) winner <- FALSE # ties = loss

is.higher <- differences > 0

was.correct <- bet.high == is.higher

if (!all(was.correct)) winner <- FALSE

if (winner) {

# calculate probability of combination

# will be zero for impossible combinations

p <- 1

each.card <- rep(4, length(x)) # four copies of each cards (no color)

for (j in 1:5) {

p <- p * each.card[combinations[i,j]] / (52 + 1 - j)

# remove card from deck

each.card[combinations[i,j]] <- each.card[combinations[i,j]] - 1

}

total.win <- total.win + p

}

}

Maybe in not getting it, but why would 8 7 8 or 8 9 8 give best odds? Wouldn't 8 7 A or 8 9 2 be best? I get that by removing two 8s you have a lower probability of a tie, but with any 8 you stool habe 50/50 of being high low.

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DeleteEh. I'm still dubious; I tried it two different ways and both times got under 25%. But it's not worth arguing further.

DeleteOkay, wait a second... aargh, i really don't want to get into an argument again, but... mrfl.

DeleteOkay, I don't know if this affects your calculation for the case of the full game, but I'm pretty sure there's an issue with your simplified two-round, four-number illustration. Basically, Anonymous above is right. Sure, you can conflate the 3-4-2 and the 3-4-1 as "effectively the same hand", but then if you go on to assume that each of the 22 resulting hands has a probability of 1/22, you're assuming all those hands are

equally likely, which they're not. Sure, if you stop a losing game after the first round, you get 22 total combinations, but not 22equally probablecombinations.Consider this: any number is equally likely for the first number, agreed? And since the game doesn't stop till after the second number, any remaining number is equally likely for the second number, yes? So there's a 1/12 chance of getting a 2-1 for the first two numbers, and there's a 1/12 chance of getting a 1-2. The first combination immediately loses the game; the second doesn't. So in the second case, we draw a third number. There are two equally likely possibilities for that third number: it's either a 3 or a 4. Since there was a 1/12 chance of getting a 1-2 for the first two numbers, there's a 1/24 chance of getting a 1-2-3 for the full game, and a 1/24 chance of getting a 1-2-4 -- as opposed to a 1/12 chance of losing the game in the first round to a 2-1. The same calculation holds for any combination in this simplified two-round, four-number game: all the combinations that lose in the first round have probability 1/12, while all the combinations that make it to the second round have probability 1/24. So yes, there are 22 possible outcomes, but they're not all equally likely. (It's easy to verify that the probabilities add to 1, as of course they must: there are 16 possible winning combinations with a probability of 1/24; 4 combinations that lose in the second round with a probability of with a probability of 1/24; and 2 combinations that lose in the first round with a probability of 1/12: 16*(1/24)+4*(1/24)+2*(1/12)=1.)

As a check, this game is simple enough that it's easy to write a Monte Carlo program that'll give an estimate as to the results. I'll do that, and I predict it'll win about 67% of the time, not 72.7%. I'll post the results after I write the program.

Okay, here's the Monte Carlo program, and this time I did take the time to thoroughly comment it and make it easy to understand, so anyone can verify that it does what it's supposed to. (And incidentally, it does immediately chalk up a loss and start a new game if it loses on the first round.)

DeleteIt ended up with 6,669,859 wins and 3,330,141 losses out of 10,000,000 trials. As expected, the win probability for this simplified version of the game is indeed 67%.

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DeleteI agree that Chet's reasoning in the simplified game is flawed, and thus suspicious of its application in the large game.

DeleteFor what it's worth, my program that exhaustively searched came up with a winning probability of 65948/270725 ~= 24.36-%. This disagrees with every other calculation that I've seen reported here, but is at least the closest so far to Jalen's Monte Carlo approach.

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DeleteArgh, apologies for contributing to your annoyance, Chet -- your comment suggesting that this all be dropped arrived while I was writing my previous one, so I did not see it in time.

Delete(And further apologies if this comment was received twice; had some strange issue posting it the first time.)

Well this is embarrassing. I was a few sheets to the wind last night, got annoyed at my e-mails, and decided the thing to do was just delete all my comments. I apologize. That was unreasonably petulant of me and somewhat unethical besides. I'll try to avoid doing that again in the future.

DeleteFriendly reminder to all fellow nerds out there- arguing for paragraphs about something pretty inconsequential, even if you're right, starts to get rude after a while. I get it! But sometimes it's best to say "one of us is right, one of us is wrong, I know what I think is correct, let's forget about it."

DeleteThis is fascinating. I loved following along and wrapping my mind around the math. Thank all of you for going down the rabbit hole!

Delete"However, the calculations are somewhat important..."

ReplyDeleteNope, they're not. The game uses a simple but devious method to ensure that players don't win too easily. I don't want to spoil it (and don't remember the details anyway), but your approach won't work.

I find these Dougherty games strangely appealing, even though I have no nostalgia for them. I think I played LoB or the first time in 2001 or so. once i realized that combat is almost pointless and character progression mostly fixed, I stopped playing. But the games managge to create a special kind of atmosphere and it's hard to pinpoint how they do so exactly. It's probably the attention to little details like named shops, weird enemies and also the graphical style.

Also, I forgot to mention the intro tune is awesomely cheesy: https://www.youtube.com/watch?v=BeIaKMVQdIE

DeleteAre you talking about the fact that the guards attack you if you win too much? Or are you saying that the draws aren't truly randomized?

DeleteThe draws aren't randomized.

DeleteYou intrigued me, so I just played 50 hands and recorded the cards. Among the 50 hands, there were 191 cards played. In an even distribution, each would have come up 14.7 times. The real distribution ranged from 8 (2) to 20 (6), but the chi-squared test on the distribution gives me p=0.66, which is about as randomized as you can get.

DeleteI DID win more than I expected--18 times vs. my expected 14--but even that's not enough to conclude bias (p=0.21).

So as far as I can tell, the draws are randomized, but perhaps there are platform differences or I'm missing something. Either way, thanks for sending me on the exercise. I went from 1,484 gold to 1,904.

Yeah, the game uses a simple way to make you feel like it's truly random, that's what I meant when I said it's "devious". Here's how it works (from A. Schultz' FAQ):

Delete1) gurer ner bayl svir qvssrerag unaqf lbh pna fgneg bhg jvgu rnpu

gvzr nsgre lbh fcrnx gb gur Urvtu-Ybntu qrnyre.

2) gurl ner nyy jvaanoyr.

3) gur tnzr nsgre gurz serdhragyl unf gjb pneqf va n ebj gung ner gur

fnzr, fb lbh znl bayl jnag gb cynl bapr.

This worked for me and I only ever played the C64 Version.

I've played this game a lot and taken notes. I think the hands are randomized to the best of whatever random number generator is used in the game, because the total set of hands is larger than I think anyone would have bothered to manually enter into a data table.

DeleteHowever the seeding of the random number function seems quite poor, so that the set of initial hands is around 5 for each session, and once you have them mapped out, you can just bet 333, win 999, and leave and restart.

In a way I found it satisfying to solve the game by taking careful notes, which is an RPG tradition. But at the same time it was kind of silly to defeat the software instead of the game.

That looks like the font from the Gold Box D&D games.

ReplyDeleteI can't vouch for how this game handles it, but I like the idea of leveling more through plot progression instead of killing things. There's nothing wrong with the latter approach for combat focused games, and maybe that's why it's a mismatch for this game.

ReplyDeleteBut for other games it bugs me that I'm missing out on rewards if I figure out a way to evade an enemy, or cause them to flee, or solve a problem with diplomacy or a bribe. A goal oriented reward system allows for more ways to approach problems, and even quests that don't involve violence at all.

And the leveling-for-killing often encourages contrived game play like grinding that I don't particularly enjoy. Even worse is when only the character that delivers the killing blow gets the experience, which forces all kinds of bizarre micromanagement in combat.

It still needs to be implemented well, but for games that have more going on than just combat, I suspect an achievement-based reward system can bring out a more natural and deeper gameplay.

I quite like that idea, too, but it's a bit pointless if you still have to combat-grind for money. At least it more or less assures that you're not over/underleveled for the next part of the story.

DeleteDon't get me wrong: I certainly don't object to alternate methods of earning experience. If you can level up by sneaking past enemies, picking locks, passing dialogue checks, and turning in quests, I'm all for it. I think a system like

DeleteThe Elder Scrolls, where you level up via skill development and most of those skills are non-combat, is fantastic.What I object to is the idea of leveling up at fixed intervals, and thus having every character and every player arriving at each stage of the game in the exact same condition. It makes it feel like character development is illusory and your own effort or playing style doesn't matter.

That's fair. An objective-oriented reward system would work best in a non-linear game with lots of options. Even better, rewards could vary with the quality of the solution.

DeleteBut implemented right, I think it's potentially a better approach than just tying everything to body count.

It's another conversation, but I actually don't like the Elder Scrolls approach that much. It sounds good in theory, but in my experience it leads to artificial, contrived gameplay where you repetitively perform actions you wouldn't otherwise to power up a particular skill. Skill-grinding, I guess. Sure training is realistic, but it should be abstracted out the same as needing to eat food.

I just finished Shadowrun: Dragonfall. One of my favorite rpgs now. Their system is really simple, but still manages to make tactically interesting, challenging scenarios. It's pretty linear, and all level ups come from achieving goals, none from combat.

DeleteIn Dragonfall the order of the runs in the middle section is pretty much your choice, so you can make runs easier by taking them later (making the runs you take earlier instead harder, of course).

DeleteGiving "experience points" (karma) for runs and special achievements is standard in Shadowrun. Combat is rather undesireable, after all, when you work from the shadows (and those High Threat Response teams are nasty).

Dragonfall is one of my favourites, too (The Aztechnology run alone). Played through it three times, SR: Hong Kong is next for a third spin.

True- not entirely linear, I meant more in the sense that you can't go traveling everywhere, but I did appreciate being able to choose mission order. I also really loved the Aztechnology level- I was disappointed that I didn't have high enough stats to do anything other than a frontal assault, but the result was so tense and epic I quickly accepted :) Playing Hong Kong now!

DeleteI never could get myself to try any of the Dougherty games myself. Even back in the days, with a limited of amounts of games and lots of time, they never seemed exciting enough to be worth trying

ReplyDeleteAnd 30 years later they seem exceedingly boring compared to comtemporaries like Wizardy, M&M, Phantasie, Bard's Tale and the Gold Box games.

So they remain the only (I think)professional CRPGs of the '80s I never even tried.

And this blog entry does not make it sound any more tempting.

Blacksilver sounds like a CRPG made for gamblers. So maybe that was the appeal of the games; gambling? After all they must have been somewhat succesful for Dougherty to be able to publish four games.

I wouldn't call them primarily gambling games. But they do all feature gambling, and with the exception of

DeleteQuestron II, carefully-balanced odds. This means it's hard to get rich unless you cheat, but the games offer an alternative form of "grinding" than combat grinding.The title screen looks like the face of a disappointed father look at his son

ReplyDelete"Look at you. Half naked and running around waving that toy sword of yours with that ridiculous Viking helmet. Why can't you sit still and study magic to succeed my megalomaniacal ambitions?"

DeleteA 'spine quiver' makes me think of a little spine-flinging porcupine, like the ones in the fields everywhere at the start of Diablo 2. The ones in this game don't look like them though.

ReplyDeleteI doubt the game has a simple closed-form solution. A few quick calculations suggest that the odds rarely drift far from 70%, and if you assume that figure always, you will expect to win a little less that 1/4 of the time. Which, it seems, is about right.

The fastest way to get an exact solution is probably to program the calculation. A language like Python (or APL, if anyone still uses that) would be ideal, but anything you are familiar with would work.

As for not having experience from killing monsters, the motivation is very logical - it means you can in principle play as your character should. A thief doesn't get punished for sneaking past the guards without killing them. But in *practice*, players just don't like it as much. Secretly, and in defiance of all ludonarrative theory, we're there for the grind :)

I think you're right, and the easiest way to get exact probabilities, if you need them and can't settle for simulated ones, is probably by dynamic programming.

DeleteSuppose you've guessed three cards right and you're guessing the final card. The state of the game is the cards that have been drawn, plus which card was the last (fourth) one. There are (52 choose 48) * 4 possible scenarios (a few less given the tie restrictions, but those can all be brute-force handled by setting their win probabilities to 0%), and each of those has a win probability and an optimal guess (high or low). That's around a million different scenarios, so doable by computer.

Suppose you've guessed two cards right. There are (52 choose 49) * 3 scenarios and the win probability for each is maximum of the win probability given you choose higher and you choose lower. The win probability if you choose higher is the mean of: 0% for those scenarios where the card ends up lower or a tie, and when the card ends up higher, the win probability for the three-guess scenario that it leads to.

And so on... I think. The memoization would solve the game as a side effect.

Posted this in another thread above, but since it relates to this one: I wrote a C++ program to calculate the probability--not that C++ was necessarily the best language for the job; it just happens to be one I'm comfortable with. (You can see the program here, if you're curious, though it's not necessarily written for easy comprehensibility, and I used a different method from the one Xavier suggests.) I got a probability of 147167 out of 621075, or about 23.696%, so Gerry's right; you'd win a little less than 1/4 of the time.

DeleteOops... found a slight mistake in the program. Corrected winning probability 588668 out of 2436525, or about 24.16%; corrected expected return 0.966 gold pieces for every gold piece spent. Corrected program here. Still matches up with Gerry's prediction of winning a little less than 1/4 of the time, of course.

DeleteIf you want an analytic solution, you could look at a continuous problem in which a real or rational number between 0 and 1 is selected, and you guess whether a number that is uniformly distributed between 0 and 1 will be lower or higher. This is the limit of the card game with infinitely many card values, and infinitely many cards of each value.

DeleteThe odds are a bit better, because having discrete values punishes the player. The continuous game has no ties.

Your chance of winning a game if the previous card was 0.0, 1.0, or (interestingly) 0.5 is 75%. If it had another value, your chances are a little less, though.

The distribution of chances is not differentiable but maybe there's some means of hacking it away and getting an analytic solution of the chance of winning the next round after winning infinitely many rounds. Wouldn't surprise me if it was something simple like Sqrt(2)/2.

Hi, I get 65948/270725 ~= 24.36% for winning a single hand. Looking at Jalen's program, it appears to me that the g and h values are being incorrectly accumulated though iterations of the inner loops. (These affect the high/low decision when playing from an eight.)

DeleteI'm spending entirely too much brainpower trying to work out why the game is called "heigh-loagh". Obviously it's just a funny spelling of "high-low", and I have to wonder why (within the universe's internal logic) it's not just called that.

ReplyDeleteGoing to assume that it's published by the universe's equivalent of Milton Bradley and they decided "high-low" wasn't trademarkable.

In earlier Dougherty games, there was a game called "hi-lo" where you just guessed which of two values the cursor would stop on. (You theoretically had 50/50 odds, but a fast player could control the outcome.) I suspect he called it something goofy to differentiate from the earlier game.

DeleteI wish I had played these games as a kid, even now they sound interesting enough to waste a few hours while watching TV.

ReplyDelete"Natural blackjack pays double!"

ReplyDeleteDoes the version of blackjack in this game pay 2:1 on a blackjack instead of the conventional 3:2 (or, for suckers, 6:5)? That would certainly tip the balance in the player's favor, even if it didn't take away the entire house edge.

Just curious... on the side jobs like making candy, is there any actual player involvement there? Is there anything you have to

ReplyDeletedoor interact with to make candy, or is it just a matter of time passing and you getting money?More importantly, does the candy machine actually go faster and faster that you end up having to eat them and stuff them in your armor and helm?

DeleteYou know, you still owe me a screenshot of the "jelly nymph" from Questron II. :) If they exist in LoBS be sure to get one!

ReplyDeleteI spent way too much time thinking about the card game. Anyway...

ReplyDeleteThe basic strategy will be to consider all draws of 5 cards, then partition based on whether a given card in the draw is high or low. This generates 32 partitions to consider. Initially, we ignore 8's (we will correct for this afterwards).

Notes:

1) A permutation of cards is called winning if the optimal strategy (default high) would lead to a win.

2) Permutations will be partitioned in low-high cases, represented by a sequence of 5 L or H (e.g. LHHLL is the partition where the first card is low, second is high, third is high, fourth and fifth are low).

3) Let T(n) = 6Cn * 4^n.

* Conceptually, this is the number of ways a sequence of n L's or H's can be made winning assuming this is the first run to be filled. As expected, T(1) is 24.

** This is because if considering e.g. LLL, a winning sequence must be strictly increasing in value (otherwise there is a tie or a guess would be wrong). This means the possible permutations of values in the example is just 6C3, and there are 4 choices (suits) per value, hence 6C3 * 4^3.

4) Let T(n0, n1, n2 ... ) = T(n0) * ( T(n1) - n0 ) * ( T(n2) - n1 - n0 ) * ...

* Conceptually this is the ways to fill segregated run lengths of a given type, e.g. T(1,1,1) is a way to fill L / L / L (with none of the L's adjacent to each other, i.e. in this case LHLHL). For this example it'd be 24 * 23 * 22.

** This _only_ works if only n0 is greater than 1. In particular, T(2,2) does not actually compute LL/LL; this comes up later.

All this lets us take advantage of a lot of symmetry in the problem. So the zero-8 cases (32 of them):

(first value is the number of repetitions)

2 * T(5) [ HHHHH, LLLLL ]

4 * T(4) * T(1) [ HHHHL, LLLLH, HLLLL, LHHHH ]

4 * T(3) * T(2) [ HHHLL, LLHHH, LLLHH, HHLLL ]

2 * T(3) * T(1,1) [ HLLLH, LHHHL ]

4 * T(3,1) * T(1) [ HHHLH, HLHHH, LHLLL, LLLHL ]

2 * T(2) * (6 * 51) [ HHLHH, LLHLL ] - special cases, covered below

8 * T(2,1) * T(1,1) [ HHLHL, HLHHL, LHHLH, LHLHH, LLHLH, LHLLH, HLLHL, HLHLL ]

4 * T(2,1) * T(2) [ HHLLH, HLLHH, LLHHL, LHHLL ]

2 * T(1,1,1) * T(1,1) [ HLHLH, LHLHL ]

which are 32 cases, covering all the possible winning permutations without an 8.

For HHLHH (and inverse):

The first HH can be covered by T(2) as normal. The second set has two cases: a) at least one value is shared with the first HH, or b) no value is shared with the first HH.

a) One value is shared (6 choices) and the other has to be a different value (22 - 3 = 19 choices). The specific choices determine the requisite ordering, so it is a unique pairing (6 * 19 cases)

b) No value is shared (16 * 12)

=> 6*19 + 16*12 = 6 * 51.

Aaand character limits (I have a truly remarkable proof for the 8-cases, but not enough space in the margins to record it...). 8-cases are broken up into number of 8's in the winning sequence (can only be 0 - covered, 1, 2 or 3).

*** FWIW I'm not completely confident about the 8-valued sequences. I'm not sure if I miscounted somewhere (there's a LOT of special cases) or if I just misentered a figure somewhere. The net result I get is 71,417,280 / 52P5 ~= 22.9%, which seems low. I'm planning on writing something to recheck my arithmetic. ****

DeleteNow for the 8's. * represents a possible H/L (but not 8)

a) Exactly one 8 (all multiplied by 4 for the choice of 8)

Allowable combinations are

8H*** (no 8L*** due to default guess), H***8, L***8

T(4) [ 8HHHH ]

2*T(3) * T(1) [ 8HHHL, 8HLLL ]

2*T(2,1) * T(1) [ 8HHLH, 8HLHH ]

T(2) * T(2) [ 8HHLL ]

T(2) * T(1,1) [ 8HLLH ]

T(1,1) * T(1,1 ) [ 8HLHL ]

(3x copies)

H8L**,

T(3) * T(1) [ H8LLL ]

2*T(2) * T(1,1) [ H8LLH, H8LHH ]

T(1,1) * T(1,1) [ H8LHL ]

L8H** - note this is a copy of H8L** (x2)

HHH8L

HHL8L

HLH8L

LHH8L

These 4 are an exact analogue of H8L** (x3)

LLL8H

LLH8H

LHL8H

HLL8H

These 4 are also an exact analogue of H8L** (x4)

HH8L*

T(2) * T(2) [ HH8LL ]

T(2,1) * T(1) [ HH8LH ]

LL8H* - note this is a mirror of previous (x2)

HL8H*

T(1,1) * T(1,1) [ HL8HL ]

T(2,1) * T(1) [ HL8HH ]

b) Exactly two 8 (all multiplied by 4 * 3 for the choices of 8's)

There are 6 possible positions for the 8's, and 8 possible H/L permutations per positions (48 total) but many can be ruled out, or are identical in nature.

Disallowed by strategy:

8L*** (12 possible)

8H8H* ( 2 possible)

8HL8L

8HH8H

L8L** ( 4 possible)

H8H** ( 4 possible)

L8H8L

H8L8L

LL8L8

LH8L8

HL8L8

HH8H8

leaving (48 - 30) = 18 viable choices.

8 * T(2) * T(1) [ 8H8LL, 8HH8L, 8HLL8, 8HHL8, L8HH8, H8LL8, LL8H8, HH8L8 ]

9 * T(1,1) * T(1) [ 8H8LH, 8HL8H, 8LH8H, L8H8H, H8L8H, L8HL8, H8LH8, LH8H8, HL8H8 ]

T(3) [ 8HHH8 ]

c) Exactly three 8 (multiplied by 4 * 3 * 2)

T(1) * T(1) [ 8H8L8 ]

All others are ruled out.

I was looking at the games list, and I'm wondering why Final Fantasy V, VI, IX, and X aren't on there. While the PC versions are recent, they didn't make massive changes to the games that would make them closer to modern games than ones from back then. Beyond shiny new graphics, they're the same as they were originally.

ReplyDeleteI believe it was the fact that they are emulated, and Chet didn't want to open the floodgates of emulators as well.

DeleteThe PC versions of FF III, IV, V, VI, VII, VIII, IX, X, X-2, and XII are all proper ports, not the sort of packaged emulators that the Addict has rejected. The primary difficulty the Addict mentioned last time he weighed in on the subject was what year to place them in.

DeleteAs far as I recall, VII and VIII are both solidly in their original release year (because the original ports were contemporaneous to the console release, so there was no problem), III and IV won't be played until the blog reaches the time of their DS releases (because the PC version is a port of the greatly modified DS games rather than the originals), and IX, X, X-2, and XII are too far in the future to worry about.

That means that the only games that need a decision are V and VI, and even VI was released late enough not to be a concern. The Addict will need to make a decision on V soon, as BY 1992 is the earliest valid place to put it. He has an entire Blog Year to think about it, so there's not a lot of pressure.

Well Gnoman I apologize, should have checked that more before commenting, I didn't realize they were proper ports. Makes me want to try FF IX again since my playstation os long gone and I loved that game.

DeleteOne of the neat things about living in a rural area is all the wildlife in my yard. Every time I look out my window, I can see at least one or two rabbits, and I know there are dozens of them in the area. Even when there's snow on the ground, I can see them foraging around the bushes and huddling together for warmth. They're so adorable.

DeleteI mention this, because from now on, any time someone encourages me to play a

Final Fantasygame, I'm going to open my window and shoot one of those rabbits. I won't enjoy it. It will make me sick to my stomach. But I'll do it if it makes my commenters think twice about starting this discussion again.Rifle's locked and loaded. Continue this thread at some poor little innocent bunny rabbit's peril.

You should totally add this to the FAQ. :) Maybe with a rabbit counter if people don't get the hint...

DeleteThis comment has been removed by the author.

DeleteDeleted my previous reply here because, on reread, it came off a lot nastier than I intended it to. Humor is hard to convey in text, which is (quite appropriately) what I was trying to rib the Addict for.

DeleteIf you lived in Australia that threat would have the opposite effect Chet, damn pests.

Delete(reposting in correct thread with typos fixed...)

DeleteWhy not just make a rule that you're not playing any game or series that originated on consoles? Your current rules seem to oblige you to play Final Fantasy games sooner or later, you don't want to -- but one imagines you feel this whole project is cheapened if you flout your own rules -- so change the rules, boom, problem solved...

...I'm not sure if I'm kidding or not.

Either way I'd never "encourage" you to play a FF game; they're not that great (some are outright hraka), embody many things you dislike, etc.

Not sure if other commenters are trying to encourage you either, so much as sound out the scope of your rules and figure out which games will/won't get the badge of honor for a playthrough.

("Lendri? We don't need no embleer lendri." Mel Brooks -- er, I mean, Weird Al version, of course.)

If you request Shin Megami Tensei, Chet's rifle is upgraded to an M249.

DeleteLOL

Delete(Not speaking to Chet here or encouraging him to do anything). For those of us who have played FF1, this thread feels especially hilarious. FF1 would easily fit in with the Questrons and Ultima clones on this list, as a polished top down grindy adventure. The sequels would have too much plot and anime, of course, haha.

Delete@PK that rule would actually eliminate Mass Effect, which is.. interesting. Off the top of my head, that's probably the most noteworthy western RPG series that would be affected.

DeleteFable just popped into my head has well. Though they aren't western, this would also eliminate Dark Souls and its brethren.

DeleteYeah, I wouldn't eliminate games originally published on a console because so many games are INTENDED for both console and PC and just happen to be published on the console first. When the delay is 20 years, on the other hand, I don't know.

DeleteI'm flattered that people like my writing and look forward to my coverage of particular games, but I don't like the pressure when multiple people won't shut up about the same game. I know that whatever I write isn't going to satisfy them. PARTICULARLY when the games occupy a genre that simply isn't part of my history. I was born too soon to be part of the generation that fetishizes Japanese entertainment. I don't like anime. I think a Nintendo and its games look like toys for children. I think Pokemon looks like something babies should be playing with. The graphics of early JRPGs are goofy and weird and off-putting. The plots of later ones are confusing and blend technology and fantasy in ways that are uncomfortable to my tastes and experiences.

So given all that, you can see why constant prodding to play

Final Fantasygames is annoying. I think they depend on developing a taste, like coffee or beer. You developed it when you were 6, so you don't remember what it's like not to have it, but I never did. I'm the guy spitting out a lager after one sip saying, "People LIKE this?!" and you're sitting over there saying,"How can you not like beer?"The issue here is that, unless you've been playing off on your own, you aren't that guy. You're the guy sitting over in the Middle East saying "Beer? I saw a picture of that in a magazine once. Looked like urine. I know beyond a shadow of a doubt that nobody sane could like it."

DeleteOr, perhaps, the German sitting around insisting that all American beer is practically water, as if the massive craft brewing industry doesn't exist.

That's why the FF PC ports are getting pushed. You'd have to rewrite your rules for the specific purpose of excluding them, and even if you weren't happy with them, at least you'd be basing your comments on actual first-hand experience rather than -at best- reading somebody else's blog about them.

I always took your opinion on JRPGs as just a difference in taste, but I never thought about the fact that some of my earliest tv shows were anime and I am inclined towards it, dragon ball and Gundam Wing still are classics to me. But I get it, and I think you'll get a good enough taste of JRPGs even if you do exclude most of a series which has been reported to death anyway. Of course I'd love to get your view but hate for you to suffer through games you know you won't like.

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DeleteI am absolutely appalled at what I read about rabbits.

DeleteMr. Bolingbroke, are you telling us you have rabbits jumping around just outside your window & you don't have a stew at least once a week?

I certainly wouldn't need people to tell me to play Final Fantasy to get the little suckers.

As fluffy as they look, a rabbit always looks best amidst potatoes and other assorted vegetables.

I think the best reason for skipping them is they are overly discussed already, and the conversation is almost never deep.

DeleteThere are some japanese console RPG games that you can more or less tell are still carrying a torch of sorts of Wizardry. Some of them are old and some of them are not, but I'm not going to recommend any of them.

This comment has been removed by the author.

ReplyDeleteThis comment has been removed by the author.

ReplyDeleteFun fact: Bad Axe is a real town. https://en.wikipedia.org/wiki/Bad_Axe,_Michigan

ReplyDelete"

ReplyDeleteBut "heigh-loagh" is devilishly clever."This reminded me of the game show Card Sharks. I used to enjoy that one.

a sample: https://www.youtube.com/watch?v=VrlYsRJV5n0