There's no such thing as an overeager vig

[Pie_is_good]

Alright, time for a second vig rant from Pie. The conclusion this time:

Vigilantes should kill every night they get the chance.

There's one intermediate conclusion to reach first, though - I'll also argue that

random deaths are good for the town.

That second conclusion seems a bit counter-intuitive, but its support is all mathematical.

Let A=The number of protown players alive
Let B=The number of proscum players alive
Let C=The average power of a protown player
Let D=The average power of a proscum player

Then AC=Power of the town and BD=Power of the scum. (AC)/(BD) expresses the balance of the game, where a higher number means the town has an advantage and a lower number means scum has the advantage (1 would mean that the game is perfectly balanced).

Now, let's say the mod were to decide to randomly off a player. The game balance after the decision is made (but before the decision is executed) is:

(odds of hitting a protowner)(New Gamebalance after hitting a protowner) + (odds of hitting a proscummer)(New Gamebalance after hitting a proscummer)

[ (A) / (A+B) ] [ ((A-1)(C)) / (BD) ] + [ (B) / (A+B) ] [ (AC) / ((B-1)(D)) ]

[ ((A)(A-1)(C)) / ((A+B)(B)(D)) ] + [ (ABC) / ((A+B)(B-1)(D)) ]

[ ((A)(A-1)(C)(B-1)) / ((A+B)(B)(D)(B-1)) ] + [ (ABsqrdC) / ((A+B)(B-1)(D)(B) ]

((A)(A-1)(C)(B-1) + ABsqrdC) / ((A+B)(B-1)(B)(D))

(AC)((A-1)(B-1)+Bsqrd) / ((BD)(A+B)(B-1)

(AC/BD)((A-1)(B-1)+Bsqrd) / (A+B)(B-1)

(AC/BD)(AB-A-B+1+Bsqrd) / (A+B)(B-1)

(AC/BD)(B(B+A)-1(B+A)+1) / (A+B)(B-1)

(AC/BD)((B-1)(B+A)+1) / (A+B)(B-1)

(AC/BD)(1+(1/(A+B)(B-1)))

...and we know that 1/(A+B)(B-1) + 1 is always a number greater than 1, meaning that the new game balance = the old game balance x some number greater than 1. Therefore, the new game balance is more favorable to the town than the old one.

The obvious next step from this conclusion is that those who have the power to make random deaths happen and are protown should do so. The vig can do exactly that,

and

his kills will be better than random,

and

he knows for sure that he won't be killing himself. His odds of hitting scum over town are significantly better than just a random kill.

(Side note: The whole "random deaths are good for the town" thing is actually a good argument for suicide vigging, too - it shows that gaining a "Free" lynch is, in fact, always a good thing.)

So. There's no such thing as an overeager vig - All vigges should choose to kill every night, anyways.

The most ironic thing about all this is that overeager vigges are more powerful than normal vigges, as the player playing the role is forced to make the correct strategic decision every night.

Pie

[Pie_is_good]

At Kelly - I was referring to Seol's argument made in thread that an extra lynch is not always good. As for the other questions, I'd argue no to the rest of the statements, but those fall outside the scope of this thread.

At Thok - I'd certainly agree that C and D depend on A and B, and also that some roles have synergy with others (2 mafia are less than twice as good as 1 mafia, since the number of nightkills remains the same, but 4 mafia are more than twice as good as 2 mafia, as they can work together in lynches), but for your argument to be valid, you'd have to show that synergy between mafia roles < synergy between townie roles. Until then, because the changing of the C and D variables are just as likely to rise as they are to drop, it's rational to base numerical arguments off their current value. Similarly, the fact that not all townie deaths are the same is another non-issue, given that C represents the

average

value of a protown player. Given only the knowledge that a kill is going to fall on a protowner, C is how much you should expect that kill to hurt the town.

[Pie_is_good]

In terms of making the game more fun, I can see where you're coming from. If only gunning for the win [/badpun], you should always kill. If trying to make the game more fun, I could see where you would nokill, but remember - I'm not actually advocating random vigging. Who to vig is as much of a scumhunt as a lynch is, but only one man participates in it.

[Pie_is_good]

Averaging over the vig outcome and then calculating the change in ratio from that average (which is what you're doing) is not quite the same as calculating the change in ratio from each vig outcome, then taking the average ratio (which is what Pie is doing). In practice the difference will be very small, though. And if you choose to measure game balance with a difference rather than a ratio ("townie power - scum power" rather than "townie power / scum power"), random deaths won't change game balance. Using the expected difference seems more appropriate to my gut than using the expected ratio; I think the latter goes wrong when you try to figure out the effect of multiple days of vigging.[/quote]

I certainly looked at the townie power-minus-scum power formula. The problem with that is that C and D have abstract units attatched. C=1 and D=2 is perfectly equivalent to C=2 and D=4. This throws some obvious problems into calculating by difference that can only be solved by calculating by ratios.

Although it also seemed counter-intuitive to me (I originally did the math setting out to prove that game odds were the

same

after a random kill, then checked my math about 7 times before deciding I was right), the following scenario made more sense in my mind:

Imagine a game of A=12, B=4, C=1, D=3. The game balance is perfectly even right now - each side has a 50% chance of winning. Now we start introducing random kills until A=3 and B=1. With more random kills introduced, balance is no longer even - If you were to finish out the game by killing randomly, the town would have a 75% chance of winning. Hence, as you get closer and closer to the end of the game, the random deaths help the town more and more (which is reflected in the formula - the number that the original ratio is multiplied by gets higher and higher as the numbers left alive go down).

[Pie_is_good]

C and D could certainly be distractions, but I wanted the math to cover all scenarios - not just the scenario where the game happened to be perfectly balanced.

To clarify my attempt at explaining the weirdness about the town gaining an advantage through random deaths...

Let's say 9 town and 3 scum. It's perfectly reasonable to assume that this is a balanced game - 50/50 chance of winning on either side.

Now, let's say the mod decides to make 11 random kills. Suddenly, the odds of the town winning are 75%. Conclusion being that each of those random kills helped the town 1/11 of the 25% gain the town saw.

[Pie_is_good]

blah blah blah Since the C and D didn't change in these two scenarios blah blah blah

I disagree - you explained yourself why C and D each change. C and D certainly have the power to change over the course of the game - a cop, for example, is more valuable late-game when he has information, so his C goes up as the game goes on.

C and D certainly have the power to be dynamic variables. I would argue that that fact doesn't change the math attatched, because there's no reason to suppose that C and D don't go up proportionally to each other.

Pie, in a mountainous game, what is the "power" of the town and scum? Please come up for a formula in terms of the # of town and # of scum at the beginning with a day start.

I have no idea. Personally, I'm not big on mountainous theory. I'd guess the ideal equation would use exponents, as the wiki numbers seem to suggest that higher numbers in the beginning favor town (a trend that can easily be balanced out by mods).

Your mathematical model is inadequate, as it's based on the naive concept of random lynchings. It disregards

information

, and therefore cannot possibly account for my counter-contention that the desirable strategy is not the one that maximises the

number

of kills that are town-controlled, but the one that maximises the

quality

of the kills (with random town kills being superior to scum kills, but inferior to informed kills). You can't prove

anything

about Mafia strategy from first principles.

It also assumes that balance is linear, which is not the case (for example, four mafia is more than 33% more powerful than three mafia).

I certainly agree that balance is not linear, although I believe the equation does reflect that in the changing of C and D (again, the only way to invalidate the equation based on the fluctuating C and D variables is to show that power of the scum goes up as numbers get lower).

Which brings me to the second argument. Let's say each vig kill yields 1/3 of the information that a lynch would (certainly up for contention, but not entirely unreasonable - I'd say more than half the information of a lynch comes from the information gained from the player's death). In that case, overeager vigging would be the play if two or more kills are happening per night (really not that uncommon).

[Pie_is_good]

i agree with pie

that is all

:goodposting:

[Pie_is_good]

Does it yield 1/3 of the information? That's a tricky one - there's no voting record or arguing for or against, you don't learn anything from anyone apart from the target. How do you quantify information - I sure can't. Does it yield the information you

need?

Sometimes - quite often, in fact - it is the play, but you can't say that "in general scenario X, vigs should all be overeager".

I honestly have no idea if it yields 1/3 of the information; I pulled that number out of nowhere. It seemed to be erring on the low side to me, but I don't know for sure. I don't claim to be able to conceretely quantify information - just as I can't concretely come up with formulas, units to put C and D in, or anything of the like. That doesn't mean those concepts don't exist, though, and it's not futile to make reasonable guesses.

Sometimes, if you're having two nightkills per night, you need as many opportunities to take out scum as possible. Sometimes you need to slow the game down - after all, it's not just vigs that have town nightchoices.

It should be noted that this whole string of "vig kills are worth 1/3 the information" was not an argument unto itself - it was intended as a refutation of the "Vig kills shorten the game" line. I was pointing out that a) they really don't shorten the game all that much, and b) the information that they give is sometimes worth shortening the game for.

It should also be noted that vigkills can overlap with scum kills as well as scum/scum overlaps.

I should also make it clear that I don't claim that

vigges should kill every night no matter what.

I'm just saying that making kills should be the default, and a vig needs a good reason

not

to.

Maybe only .999... percent of the time.

[Pie_is_good]

But if C and D change in consequence of the vig's kill, then the conclusion you've drawn about the new balance necessarily being greater than the old balance isn't necessarily true.

Trueish (the statement is a bit ambiguous). I agree that it's not necessarily true in the sense that

there's a possibility the kill could hurt the town

. But that doesn't matter, because

C and D are mostly unpredictable, as are the overall results of the random vig kill.

So sure, it's not necessarily going to wind up in your favor, in the same way that getting 3:1 odds that a coin will come up heads isn't neccesarily going to wind up in your favor.

If your final term increased by, say, 10% and C fell by 20% while D remained unchanged (I believe that's a plausible set of assumptions for when the vig kills a strong pro-town role), then the overall balance would be worse for town.

True, and certainly plausible. The coin you bet on could come up tails, but that doesn't mean you made a bad decision.

And, I repeat for the second or third time, leave C and D out of it entirely by making all pro-town roles the same and all the scum roles the same and construct some game scenarios where the A/B ratio at the beginning of the game is the same.

I've already explained why this isn't an accurate refutation. You can show me all you want that C and D change with the number of players left alive; I'm with you there. The point I'm making is that it's near-impossible to predict beforehand the change of C and D, so it's rational to make plays assuming the average change (0, unless you'd like to show otherwise).

I promise you that the games don't have the same balance. I gave the example of 5:1 and 10:2 from the Wiki numbers, as well as an easily analyzed scenario of 2:1 and 4:2. In neither case is the expected win probability for town the same in each case. I note however, that you haven't tried to refute either of those arguments.

The reason I haven't directly refuted them is because I agree with the conclusions they yield (although they don't yield the conclusions you think they do). What those models show is that C and D change based on A and B. I have already agreed to this. But C and D could change either way.

[Pie_is_good]

You're the one arguing the unconvetional point of view. It's your duty to convince us.

I certainly agree, and that's what I spent the first post doing mathematically.

Now a refutation was made, saying that C and D

do

change. I made the claim that, while C and D do change, average change in C and average change in D both equal 0. If that statement is true, the mathematical model is still accurate. If someone would like to claim that that statement is not true, they're welcome to, but they now hold the unconventional point of view, and it is now their duty to convince me.

[Pie_is_good]

a. Your initial mathematical model is horribly naïve. Mafia can not be modelled with a function like that.

...begging the question of

why?

Some parts of mafia certainly can be modelled mathematically, so I'm not sure what you're basing that claim on.

b. Even if you could come up with functions C and D to make P = AB/CD hold (which is doubtful), at the very least they would clearly depend on A and B, as pointed out by Thok right at the start, and so the whole construction is dubious as you treat them as independent variables.

ci. Even ignoring a. and b., claiming that they average change of either C or D is 0 is ridiculous when you don't even know what C and D are.

These can both be answered by defending the

average change = 0

thing.

While I certainly accept that C and D depend on A and B, they are still effectively independent until a correlation is established. It's like this:

I choose a random integer 1-20, then add another random integer between -3 and 3. I call the second random integer A and the end result C. While A and C are certainly not independent variables, because I could not tell whether A was more likely to make C bigger or smaller, the average of C is the same as it would be without introducing A in the first place.

If you want to convince me that my randomizer is broken and A is more likely to subtract from C, you can certainly do that - but the onus is on you to show me how my randomizer is broken.

cii. Further, the onus is still very much on you. You have claimed to have a case backed by mathematical arguments. People have pointed out these mathematical arguments are flawed. To then turn around and make a completely unsupported claim to fill the hole and claim it is somehow now our duty to convince you you're wrong is stupid.

The All-Knowing Razor states that the person making the positive claim has the burden of proof. I initially had the burden of proof, which I fulfilled. The refutation of my argument took the form of a positive claim, though (

as A goes down, C goes down

, whereas

No notable correlation between A and C

is the default). I don't need support of my assumption that A and C have no correlation; you need support that A and C

do

have some correlation.

I wish I could get away with stuff like that all the time, it would make my work a lot easier. It's one of those things that can't be argued against

I believe that's called "being right"

because the claim is based on an entirely misguided foundation - the most I can say about C and D is that I don't believe they exist in any useful form.

C and D are abstract representations of town and scum power. The concepts of town and scum power

do

exist. They certainly don't have measurable units (the town is not 3 meters more powerful than the scum), but that's immaterial as they express a ratio rather than individual value.

d. Even if everything else is fine, which it isn't, you still can't treat C and D as constants just because on average they don't change. For a very simple example, say P = 1/D, and D is initially 2. Say a coin flip happens and D either goes up by 1 or down by 1. Clearly the average value of D is still 2, but now the average value of P is (1/2)(1/1 + 1/3) = 2/3.

...except that the whole point of my argument is that as long as we don't have a simple P=1/D formula to plug in, we have to assume P its average. I'm saying that we don't have a P=1/D formula to use (the applicable formula is just as likely P=D^1/2, in which case the coinflip will, on average, lower the value of P)