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Strong dice mechanic

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questccg
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I'd like to know what the other Game Designers think about using or having a STRONG "dice mechanic"?!

Each time you conquer a unit, that unit "drops Loot" which can be in five (5) different types. So far, all of that is working well and strategically it is also okay.

Now comes to the "mechanic" by which players MAY collect "Loot":

Rolling 3d6s: 2 White and 1 Black.

The concept is a bit like "dice pooling" but simplified since there are only three (3) dice.

How it works is that EACH "Race" has a way of "matching" the dice rolled with the conquered cards in the area of play.

For example:

5 Black or "A Pair" is the Warlord ("Sir Calahan").

Each turn you have THREE (3) Dice rolls. And what die (or dice) you re-roll is your option. So IF you roll a "5 Black" and there is a "Citadel" with Loot, you may collect ONE (1) Loot of your choice. But that leaves the two (2) WHITE dice...! You can re-roll the two of them twice (2x) more... And see if you can meet the dice requirements of another card!

So you see, that the game relies on DICE ROLLING to figure out what "Loot" that you may collect each turn...

Do you see this as a big turn-off???

Jay103
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It could be a fun thing, if

It could be a fun thing, if it's quick and obvious and doesn't feel just like, well, if I need 5 black and you need 3 black, it's just 1 in 6 no matter what.. is there any logical connection between what rolls are needed and who that character is?

questccg
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Based on a designer's suggestion

Well I created this mechanic based on a suggestion by Stephen (@let-off studios) which said to look at "buildings in Kingsburg". And when I watched the Tom Vasel Review video (of Kingsburg)... I really thought that the DICE ROLLING mechanic was NEAT. So I opted to design something inspired by Kingsburg.

Each Character Class has it's OWN "building" and races have their own pre-set rules to what is available to them.

For example:

Humans are "Black or White" and could be "Fighter: 1 Black OR 7+ White" to "Loot" the "Sparring Circle".

So it means that you have two (2) ways to Loot from the "Sparring Circle".

High Elves are "White only" and could be "Zealot: 6- White" to "Loot" the "Chapel".

So it means that you have only one (1) way to Loot from the "Chapel".

In any event the issue remains the SAME.

Note #1: It's a STRONG "dice rolling" mechanic and subject to much luck... I guess you could say: "So is craps..." and most gambling games (rely on luck...) So maybe it's not so bad of a thing TBH.

Note #2: The dice are also CUSTOM: 1-5 and a Star (*). The "Star" can be ANY value you choose. Plus remember you have three (3) rolls to get as much as you can (in terms of "Loot").

The way it works is that you LOCK dice... For example suppose BOTH the "Warlord" (Citadel) and "Zealot" (Chapel) were available to loot. You roll a "5 Black" and "10 White". You can NOW "LOCK" the Black die and ONLY re-roll the 2 White dice ... to try to get less or equal to 6 (6- White). If this fails, you can re-roll TWICE (+1 initial roll = 3).

Jay103
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The good news is, dice that

The good news is, dice that are 1-5 and a symbol are cheaper than dice that are completely custom.

I don't see a problem with added randomness, as long as it doesn't overwhelm what is (i assume) an otherwise less luck-dependent game.

I mean, Magic The Gathering is all luck, because you're drawing cards from a random deck. But of course, it isn't, because the game is about anticipating the possible outcomes and maximizing the results you get.

questccg
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I agree with you 100%

Jay103 wrote:
The good news is, dice that are 1-5 and a symbol are cheaper than dice that are completely custom.

Yes I only pay for one-sided mould (or $60).

Jay103 wrote:
I don't see a problem with added randomness, as long as it doesn't overwhelm what is (i assume) an otherwise less luck-dependent game.

The rest is all about "Area Control" and playing the units into the Keep. Again this is all strategic and not luck-based. You usually have all kinds of options whereby you can play offensively or defensively too.

Jay103 wrote:
I mean, Magic The Gathering is all luck, because you're drawing cards from a random deck. But of course, it isn't, because the game is about anticipating the possible outcomes and maximizing the results you get.

Same with my game, you draw cards from YOUR own deck. And you can CUSTOMIZE your deck before playing a game... Because it's a "Customizable" Card Game (CCG).

questccg
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Although...

The "Loot" drops are ALSO "Random" too... So you're never certain of what the "Loot" is going to be after conquering a unit/card/tile. You then pick out of the defeated pouch the corresponding number of "Loot". And it's RANDOM.

How you collect "Loot" is determined by the DICE. But picking the "Loot" as you play is according to your own "strategy". Like you may opt to collect the "less" abundant of a type of a "Loot" or collect the "Loot" you need the most, etc.

And there are the Chaos Gems: when a player collects 2 of them, the game ends and he/she is declared a winner (Alternate Victory Condition).

X3M
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Check for chances

See, how big the chances are, with the suggested locking dice. I bet, it is high.

Also, if the loot is random. Perhaps you could go for a single loot each time. But if a player may collect 2. It has to choose the one it likes?

questccg
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That's also why I have a wildcard on the 6 of each die

X3M wrote:
See, how big the chances are, with the suggested locking dice. I bet, it is high...

Well that sort of the whole point of the dice: to collect "Loot" in order to win the game. Locking gives you the option to collect two (2) "Loot" per turn. Why is this important??? Because on turns 1-3 usually there is no conquering, that means nine (9) rounds left to score points. And that means that it's on average ONE (1) "Loot" per turn (to collect and win with 9 VPs). Having the option to collect MORE than one (1) per turn makes the outcome more realistic ... since you can say "Average one (1) per turn for nine (9) rounds...

X3M wrote:
Also, if the loot is random. Perhaps you could go for a single loot each time. But if a player may collect 2. It has to choose the one it likes?

The player on the LOSING side of a conquest RANDOMLY choose the "Loot" from his pouch and places them on the vacant building. Players roll dice and try to land the right numbers to collect a "Loot". Of course, each player may choose the "Loot" he/she prefer from each location.

So it's already like that... With the exception that players can collect up to two (2) "Loot" per turn.

X3M
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I kinda calculated the loot for...

A 5 on the black die and the pair of eyes on the white.

The chances are very low to reach 0 loot. But 1 or 2 loot seem to be relatively equal in chance.

There is roughly a 70% chance that the black die succeeds within 3 rolls.
The white where more complicated. But they too come close to that number.

Either way. I think it would be wise to have the exact numbers for the loot. Seeing as how you wanted different strengths for loot.

A 'black and white' OR 'black or white' is a big difference too!
If you have both of these, I think it would be wise to warn the player about the importance of the AND or OR.

Knowing you, you already did(?)
:)

questccg
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70% odds ... That's GOOD!

X3M wrote:
A 'black and white' OR 'black or white' is a big difference too! If you have both of these, I think it would be wise to warn the player about the importance of the AND or OR...

I only used "or" when the possibility is for either Black OR White. With a success rate of 70% that's pretty good. Because I don't want it to feel like a "Dice Game" where you roll and get NOTHING. That was my biggest concern. 70% one (1) or two (2) "Loot" is GOOD IMHO. Because it means players are ACCUMULATING "Loot" towards an eventual "Victory".

Sure the "Chaos Gems" could change the outcome... This too is GOOD because that's why they are called "CHAOS" Gems because they "screw" with the plans of the players and can alter the outcome of a game.

There is another FACTOR that influences the dice rolling. And that is IF you roll the dice and can collect a "Resource" but it was not the one that you had planned to get... This introduces some kind of Push-Your-Luck and try to re-roll ALL the dice to get the "Resource" you wanted versus the one that you lucked out in rolling.

That too is IMPORTANT. Because it plays into the dice psychology. Do you take the risk to roll again and get NOTHING. Or do you "LOCK" what you already have and try only for more... Could be reasonable game table thinking and strategy.

Note #1: Victory Conditions all start with 2 of each "Resource" and ADD an additional +3 "Resources" (which vary from Lord/Lady). 6 + 3 = 9 (nine) a re-occurring value in my game (e.g. 9 races, 9 Victory Points, etc.)

questccg
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Do you think you can HELP me with the odds???

I'd like to really know the "correct" odds... Can you HELP me???

1. Rolling a # on the Black die

Three (3) chances of rolling a "5" given that the "6" is a Wildcard (can be any value you want).

1/5 + 1/5 = 2/5 or 4/10 (40%) for only 1 roll.

2. Rolling an amount on the White dice

Three (3) chances of rolling greater than a "7". Again "6" is a Wildcard BUT it's values go from 1 to 5...

Or three (3) chances of rolling less than or equal to a "6". Again "6" is a Wildcard BUT same as before, the values go from 1 to 5...

Have no idea how to compute these two...

Do you think you could EXPLAIN to me the MATH (for the ODDS computation)???

X3M
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erm, I am bad at explaining...

The black die is the easy one.

A die has 6 sides, right?

So, the die has the following options: 1,2,3,4,5,*
Where * can be any number we want.

We want a 5? * is now also a 5.
The die acts as a 1,2,3,4,5,5

Rolling a 5?
Success is 2 out of 6. Is 33%
Fail is the remaining 4 out of 6. Is 67%

However, you allow re-rolling.
And this occurs when you fail your first time (and second)
So, we only need to calculate the fails and subtract the answer from the 100%, if we want these success rates.

Each roll goes down by putting the fail ratio to the number of times rolled.

Roll 1: Fail (4/6)^1=67%, Success 33%
Roll 2: Fail (4/6)^2=44%, Success 56%
Roll 3: Fail (4/6)^3=30%, Success 70%

***

The white are more complicated. However, if 1 white die acts the same as the black die. Then the chances here are much, much higher.
Either way, instead of a pair, you want a roll that is equal or higher than a 7?
Doesn't that mean that the * is always a 5? Just to make sure you get above 7?
There is only one roll with one * that will not reach 7 or more. And that would be rolling a 1 with the other die. But then you only need to re-roll that one a couple of times.

I can't do the math here. This is one of the cases, where I rather get my paper, and write it all out.

The first roll gives the following options for the fails, where some are more often to occur:
1+1, 2x(1+2), 2x(1+3), 2x(1+4), 1+5, 1+*
2+2, 2x(2+3), 2x(2+4)
3+3
That is 15 fails out of 36.
42% fail, 58% succeed.

This first roll already looks easier than that of the black die.
But more so, because you can simply hold on to the 5 or *. The 4 is questionable and 3 already gives me doubts.

Now, if you have to roll both whites for the re-roll. It would be simple:
Roll 1: Fail (15/36)^1=42%, Success 58%
Roll 2: Fail (15/36)^2=17%, Success 83%
Roll 3: Fail (15/36)^3= 7%, Success 93%

But as you have said before, any number of dice can be hold. So players will hold on that one 5 or *. Only rolling the other die. Which decreases the chances to fail a lot.

I strongly suggest, forcing players to roll both white dice if they need to re-roll. Because holding 1, will increase the succeed number to roughly 97 to 99%. Which isn't even a worthy roll.
If so, the loot is distributed as following:

0 loot: Bfail x Bfail = 2.1%
1 loot: Bsucceed x Wfail + Bfail x Wsucceed = 32.6%
2 loot: Bsucceed x Wsucceed = 65.3%

0 loot: 2.1%
1 loot:32.6%
2 loot:65.3%

PS. I take my words back at loot 1 and 2 being equal. It looks like there is a 2/3th chance to get 2 loot.
You might as well say that a player only needs to roll once to get that +1 loot on the one loot that it already gets.

questccg
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Let me clarify a bit...

Okay ... so there seems to be a "question" about LOCKING. When and what can you LOCK?

1. Either you can LOCK the "Black" die.

2. Or you can LOCK BOTH "White" dice.

You can't just lock one (1) "White" dice because you rolled a Star (*) for example... So that explains your two (2) choices...

Does this affect the odds/probabilities (Probably yes... I think by your remarks)?

Do you need to do some "re-calculations" with these constraints??

BTW your explaining is real good. I understood the "Black" dice right away... Your two (2) White dice explanation is good... But I am unsure with the above mentioned constraint (roll BOTH White dice) that the math will be affected.

Please let me know and thank you for your effort!

questccg
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Just because I re-read your message... and

X3M wrote:
...This first roll already looks easier than that of the black die...

Now, if you have to roll both whites for the re-roll. It would be simple:

Roll 1: Fail (15/36)^1=42%, Success 58%
Roll 2: Fail (15/36)^2=17%, Success 83%
Roll 3: Fail (15/36)^3= 7%, Success 93%
...

Is this ACCURATE??? Both White dice have a 93% probability of successfully rolling 7+??? Really? Wow that's some serious odds! Am I correct in these assumption?

X3M
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To clarify

Well, if only one of the 2 white dice where rolled, while the other one was locked. The math would become very hard.

But having them both locked or re-rolled is much better.

A mistake that I made was that I had both the black one and the 2 white ones re-rolled.

But if it is either one or the other. That means a double fail would mean re-rolling only one of the 2 colours.

In which case, white would be re-rolled, because the success rate is much higher.

The chances will get a bit lower. I need to re-calculate.

***

I need the following data:
Bf = 2/3
Bs = 1/3
Wf = 5/12
Ws = 7/12

The first roll can contain:
Bf x Wf = 2/3 x 5/12 = 10/36, a re-roll on white
special note: black remains a fail here
Bf x Ws = 2/3 x 7/12 = 14/36, a re-roll on black
Bs x Wf = 1/3 x 5/12 = 5/36, a re-roll on white
Bs x Ws = 1/3 x 7/12 = 7/36, no re-roll

If you add up the results, you get 36/36.

We have 3 starting points now...
Which will divide the chance into lower parts.
All victories will be added up at the end though.
That is why I will also include the no re-roll in each step.
Just to keep a good overview.

Group Bf x Wf = 10/36
Wf = 10/36 x 5/12 = 50/432, a re-roll on white
special note: black remains a fail here
Ws = 10/36 x 7/12 = 70/432, a re-roll on black

Group Bf x Ws = 14/36
Bf = 14/36 x 2/3 (x 4/4) = 112/432, a re-roll on black
Bs = 14/36 x 1/3 (x 4/4) = 56/432, no re-roll

Group Bs x Wf = 5/36
Wf = 5/36 x 5/12 = 25/432, a re-roll on white
Ws = 5/36 x 7/12 = 35/432, no re-roll

Group Bs x Ws = 7/36
While not a re-roll is needed. We want to see how this one fits in with the rest.
7/36 (x12/12) = 84/432

We get 7 groups now. But this system allows us to collapse the doubles. By simply adding up the scores. It doesn't matter how we got to this point. Only the results belonging to a group matter.
(I do this all the time for my wargame)

Bf x Wf = 50/432, a re-roll on white
special note: black remains a fail here
Bf x Ws = (70+112)/432 = 182/432, a re-roll on black
Bs x Wf = 25/432, a re-roll on white
Bs x Ws = (56+35+84)/432 = 175/432, no re-roll

If you add up the results, you get 432/432.
All I need to do now, is repeat the process, but enter the second roll value's. We will be getting the 7 groups again, which we can collapse to the 4 main groups. Or even the 3 loot groups. I am going to do the latter.

Group Bf x Wf = 50/432
Wf = 50/432 x 5/12 = 250/5184; Bf x Wf; 0 loot
Ws = 50/432 x 7/12 = 350/5184; Bf x Ws; 1 loot

Group Bf x Ws = 182/432
Bf = 182/432 x 2/3 (x 4/4) = 1456/5184; Bf x Ws; 1 loot
Bs = 182/432 x 1/3 (x 4/4) = 728/5184; Bs x Ws; 2 loot

Group Bs x Wf = 25/432
Wf = 25/432 x 5/12 = 125/5184; Bs x Wf; 1 loot
Ws = 25/432 x 7/12 = 175/5184; Bs x Ws; 2 loot

Group Bs x Ws = 175/432
While not a re-roll is needed. We want to see how this one fits in with the rest.
175/432 (x12/12) = 2100/5184; Bs x Ws; 2 loot

If you add up the results, you get 5184/5184.
Yaaay! That means it worked.

Now to collapse:
0 Loot: 250/5184 = 4.8%
1 Loot: (350+1456+125)/5184 = 1931/5184 = 37.2%
2 Loot: (728+175+2100)/5184 = 3003/5184 = 57.9%

Here ya go lad:
0 Loot: 4.8%
1 Loot:37.2%
2 Loot:57.9%

X3M
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Indeed

questccg wrote:
X3M wrote:
...This first roll already looks easier than that of the black die...

Now, if you have to roll both whites for the re-roll. It would be simple:

Roll 1: Fail (15/36)^1=42%, Success 58%
Roll 2: Fail (15/36)^2=17%, Success 83%
Roll 3: Fail (15/36)^3= 7%, Success 93%
...

Is this ACCURATE??? Both White dice have a 93% probability of successfully rolling 7+??? Really? Wow that's some serious odds! Am I correct in these assumption?

Yes, that was on white only.
Henceforth, in the calculation above, I only consider rolling the whites, before the black ones are re-rolled.

X3M
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It is possible

To have an excel sheet with the rules.
Calculating all chances.

Seeing as how it can be done on paper. The excel sheet is simply a fast way to plow through the different conditions.

The black die can have:
2 or more, 3 or more, 4 or more, 5.
With * as any number you like.
That is 4 options.

The white dice can have:
3 or more, all the way to 10.
With * as any number you like.
That is 8 options.

Seeing as how a re-roll on one of the 2 makes the chances ridiculous good to get a loot.
A limit on re-rolls is also of influence.
Rolling, once, twice or trice means 3 options.

Being allowed to re-roll only one colour or both colours. Will need to change the sheet a bit. But this too will offer 2 options.

In total, there will be 4x8x3x2=192 chance distributions for the 0, 1 or 2 loot.

Of course, the single colour can easily be included.
Only black or only white. Which would mean 4x3=12 and 8x3=24 more chance distributions.
A total of 228. I can also include the average loot.

***

Speaking of average loot.
The last calculation that I did:

1 Loot: 1x (350+1456+125)/5184 = 1931/5184 = 0.372
2 Loot: 2x (728+175+2100)/5184 = 3003/5184 = 1.159
0.372+1.159=1.531

Your loot is 1.531 each turn.

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