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Need help with a "creep" concept

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Tue, 08/15/2017 - 22:31
questccg
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Okay so let me explain my design in a nutshell and then I can explain what I need a bit of HELP with! (Hope that sounds reasonable)

My WIP ("SpellMasters") uses Game Tiles to "crawl" a dungeon. The tiles are like corridors the players travel and each one earns a player various Experience Points (XPs).

On each level, there is ONE (1) Monster that must be defeated.

That Monster corresponds to the level's dungeon and is predefined with certain "stats". There aren't too many "stats", just THREE (3):

  • Toughness (makes the Monster harder to attack and can result in counter-attacks).
  • Health (gives the Monster more Health Points, making battles last longer).
  • Damage (gives the Monster a boost in dealing more damage when the rolls fall in its favor).

There has been a TON of "speculation" as to WHERE the "creep"-ing of these "stats" should occur. To explain what I got so far is a HYBRID solution.

  • Each ITEM has the "stats" on the card for THREE (3) level (1-5, 6-10, and 11-15).
  • Players RANDOMLY select three (3) ROOMS per level -- each room defines which "stats" get modified.
  • Player choose to ENTER a subset (1-3) of these ROOMS and DEFINE "How" the "Creep"-ing occurs.

The solution is NOT perfect. Why? Well if a card is 4-1-1, and you choose Toughness, Toughness and Damage... It basically means +5 Toughness to the Monster and a +1 Damage modifier. That +5 is a BITCH and makes it nearly IMPOSSIBLE to defeat the "Monster". Maybe a +4 is again REAL DIFFICULT ... so that card could be a 2-1-1. Again a +3 is STILL very hard to beat.

TL;DR I'll keep it short and leave it at that.

That's MY PROBLEM... How to VARY "creep"-ing such that it is VARIABLE and still PLAUSIBLE in terms of game play.

Thank you.

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Tue, 08/15/2017 - 22:56
#1
questccg
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More on the 4-1-1 example

I agree that 4-1-1 is not for levels 1-5 and 6-10 ... Maybe for 11-15! But still even at those levels (11-15) +5 Toughness is still hard to beat. If you are battling a Monster with "3" Toughness (at start) +5 = 8 Toughness.

That would require a Wizard with "Expert" using an "Oak Staff" for example. That equates to a Attack of 6 Toughness. Your fate dice roll, must be +2 or HIGHER to attack this Monster. Otherwise the Monster attacks BACK each time you FAIL your fate roll.

Which means either A> You do damage or B> The Monster does damage to YOU.

Looking for VARIABILITY and some RANDOMNESS too. It can't be all predictable (and only deterministic).

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Tue, 08/15/2017 - 23:21
#2
X3M
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Hey Kris, When I want to

Hey Kris,

When I want to halve the additions because they are to high in perspective. But I am working with low numbers already. One way to deal with it is actually doubling all other game aspects that are linked directly to the additions.

Is that plosible?

Cheers,

X3M

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Wed, 08/16/2017 - 06:52
#3
let-off studios
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Other Adjustments

You may also be able to make other adjustments to the players or game state, as opposed to just the boss ability scores.

- Force players to discard half their letter tiles
- Re-populate some previously-explored dungeon tiles
- Players must discard 1 loot or suffer damage

...Stuff like that. The main point is to encourage players to face the boss quickly so they don't end up being to weak to face the boss when they come round to it. Mix these in with rare stat boosts and you would have a varied pool of boss-related changes to the game that avoid stat creep in many cases.

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Thu, 08/17/2017 - 18:49
#4
questccg
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Some thinking

questccg wrote:
Looking for VARIABILITY and some RANDOMNESS too. It can't be all predictable (and only deterministic).

And after all of it I have made the following conclusion:

Since the game is cooperative, players should MAXIMIZE the profitability of the Encounters.

What does that mean? Well IF you get a 4-1-1 item on level 11-15, the players (all of them) should consider is a +4 Toughness going to make the battle impossible? If YES, players will only choose a +1 Toughness. If NO, players will choose a +4 Toughness.

I'll let the PLAYERS figure it out.

And why would that have any importance? Well +4 Toughness is also +20 Experience Points (XPs). So each Wizard in the battle would GAIN +20 XP.

Instead of finding some kind of "rational" solution, I have come up with letting the players decide on the finer details. I mean that's what COOP games are all about: figuring out as a team how to beat the game. But at the same time, you want challenges and you want to progress further too.

Thanks for the input.

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Thu, 08/17/2017 - 21:00
#5
questccg
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Another form of Balancing

I guess what we could conclude from my previous comment is that the game should be SELF-BALANCING. If people want an EASY game, people will avoid tougher battles.

But at the same time, IF the battles are CHALLENGING and the DECISIONS interesting, that's what makes a GREAT "cooperative" game!

And even if the game is "co-petive" because each Wizard is battling to score more points than the other players (competitive)... It's has to have a TEAM feel to it -- because it is cooperative also.

So I think this is the best decision for the game: let the players decide how they want to play the game... Will probably make for more meaningful decisions anyhow!

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Sat, 08/19/2017 - 08:50
#6
X3M
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questccg wrote: So I think

questccg wrote:

So I think this is the best decision for the game: let the players decide how they want to play the game... Will probably make for more meaningful decisions anyhow!

I guess so. Even if it is by accident.

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Sat, 08/19/2017 - 18:59
#7
questccg
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31% odds... Am I correct???

X3M wrote:
I guess so. Even if it is by accident.

Well if I can't find a solution to handle all of the cases, the best thing to do is leave it to the players in such a way to "promote" challenging battles... Obviously to earn MORE experience (XP).

By accident? IDK if that 100% accurate. I've thought about it, I asked to see what other designers thought about it... And I've concluded that it's best to dealt in-game, by the players.

That conclusion is not by accident.

It's the best solution that I could come up with. Maybe if someone finds a better solution, well then I'm all "ears". But so far, to avoid impossible odds on the certain occasion (Rare or not - I cannot compute the odds).

My math is bad (3/20 + 2/19 + 1/18 = 31%) Is this even correct??? There are a total of 20 modifier/creep cards, 3 are for Toughness (the most difficult variable to control)...

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Sat, 08/19/2017 - 21:01
#8
X3M
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With accident, I meant that

With accident, I meant that little fiasco that I and my cousin let loose on Tradewars Homeworld. I am sure you remember that we had it all wrong back then.

Regarding that 31%. Are you looking for the chance to have at least 1 right card out of a deck of 20? Where the player has 3 right cards in the deck. And is allowed to pick 3 cards?

***

If we assume that we are not going to pick the right card at all. The chance on this is 3/20 * 3/19 * 3/18 = 4080/6840 = 59,6%
100 - 59,6% = 40,4%.
There is a chance of picking at least 1 card with strength, when allowed to pick 3 cards.

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Sat, 08/19/2017 - 21:03
#9
questccg
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Fiasco = Fun!!! LOL

X3M wrote:
With accident, I meant that little fiasco that I and my cousin let loose on Tradewars Homeworld. I am sure you remember that we had it all wrong back then.

No worries - we've all had things that were not 100% clear by the rules. Joe pointed out to me that when attacking a Player with 2 Starships versus 1 Starship, IF you ENGAGE the starship with only 1 (of 2), then the 2nd starship could ATTACK DIRECTLY the player's Homeworld.

Never even thought of that. And it's great because it lowers the game's length too... We've added a new Battle Front scenario which also attempts to try to "correct" the game time... Speeds up the game so that it's more won in 60 minutes instead of lingering upwards of 90-120 minutes.

X3M wrote:
Regarding that 31%. Are you looking for the chance to have at least 1 right card out of a deck of 20? Where the player has 3 right cards in the deck. And is allowed to pick 3 cards?

Yes there are three (3) "Toughness" increase cards out of a deck of 20 cards. And the odds are trying to select ALL THREE (irrespective of order) but in one picking of the three (3) cards... Are my calculations correct???

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Sun, 08/20/2017 - 05:58
#10
X3M
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If a player is allowed to

If a player is allowed to pick 3 cards.
And there are 3 cards of 1 type in a deck of 20.
And all 3 cards picked have to be of that same type.

Chances are very, very low.

Instead of adding up, you had to multiply those odds.
3/20 * 2/19 * 1/18 = 6/6840 = <0,1%.

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Sun, 08/20/2017 - 15:09
#11
questccg
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X3M wrote:Chances are very,

X3M wrote:
Chances are very, very low.

Instead of adding up, you had to multiply those odds.
3/20 * 2/19 * 1/18 = 6/6840 = <0,1%.

Okay @X3M, thanks for the correction. I'm very surprised that the odds are so very LOW! That's good to know. But it also may be BAD too...

I guess my question should be: "What if I want to know the odds of ONE (1) of those cards being chosen out of the three (3) cards?"

I want to make sure picking ONE (1) cards is reasonably LIKELY... Or odds-wise probable...

Update: My math is telling me that it should be 3/20 or 15%... Am I correct???

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Sun, 08/20/2017 - 15:32
#12
X3M
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You are correct (sorry for

You are correct (sorry for the late response between posts).

First pick is 15% indeed.

Things get more complicated if you calculate the "at least 1" for multiple picks.

Picking 2 cards with at least 1 card being strength. The chance is 28,4%. But wanting to have 2 cards both being strength, the chance is merely 1,6%.

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Sun, 08/20/2017 - 15:41
#13
questccg
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Okay so I'm partially correct...

X3M wrote:
Things get more complicated if you calculate the "at least 1" for multiple picks.

Picking 2 cards with at least 1 card being strength. The chance is 28,4%. But wanting to have 2 cards both being strength, the chance is merely 1,6%.

How do you do the math for Picking 3 cards with at least 1 card being Toughness???

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Sun, 08/20/2017 - 16:03
#14
FrankM
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questccg wrote:X3M

questccg wrote:
X3M wrote:
Things get more complicated if you calculate the "at least 1" for multiple picks.

Picking 2 cards with at least 1 card being strength. The chance is 28,4%. But wanting to have 2 cards both being strength, the chance is merely 1,6%.

How do you do the math for Picking 3 cards with at least 1 card being Toughness???


Sometimes it's easier to find out how likely it is to NOT happen, then subtract from 100%.

There are 3 strength cards, so there is a 17/20 chance of not picking one for the first card.
There is a 16/19 chance of not picking one the second time.
There is a 15/18 chance of not picking one the third time.

17/20 * 16/19 * 15/18 = 59.65%

So the probability of picking at least one strength card (that is, failing to perform at least one of the steps above) is 100% - 59.65% = 40.35%.

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Sun, 08/20/2017 - 16:21
#15
questccg
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Wow ... Awesome!

Well looking at the numerical "tendancy", I predicted 40%. Said it would be like 28% + 12% = 40%...

But thanks for that great explanation. 40% is "satisfactory". That means that is "probable" that a Toughness card is picked. And I want this to be TRUE, because it helps players EARN more experience... The biggest "creep" is Toughness and it gives +5 XP bonus per level.

So if my Toughness = +3, then it's +15 XP per Wizard that is part of the battle.

Thanks again!

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Sun, 08/20/2017 - 17:03
#16
X3M
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Great that you have reached

Great that you have reached the same percentage (see post 8 :) )

If only I knew how to calculate these kind of effects in anydice.com in a simple matter. I could blue print them. I suspect, we need looping for that.

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