So I'm planning on having d6's with the following break down:
X, A, A, A/B, B, C
I don't know enough about probability and multiple dice, but I can't just add the percentages up together, can I? Something tells me it's a lot more complicated then that
I would very much appreciate a crash course, or a point in the right direction. And you'll have my unending thanks. :-)
using D6, the chance of getting any number is 1 in 6, represented by 1/6
the probability of getting a specific result from 2 D6s is calculated by multiplying 1/6 by 1/6 to derive at 1/36
with a third dice, you just get 1/36 and multiply it by another 1/6 to get 1/216
with every additional dice, just multiply the probability by the previous probability and you will get your answer. this works even when you want to put other types of dice into the mix, like if you want a D10 to be the next dice, just multiply the last probability by 1/10, if you want a D20 just multiply the last probability by 1/20
this is assuming you are looking for a specific result on the dice, you will need to define what result "A" is to appraise if your own calculations are accurate
Www.anydice.com will be your bestest friend when it comes to dice probability :)
I always suggest a software called "small roller" to calculate probability without knowing any math. But from what I can see, this app seem to be hard to google. So I decided to upload it on my website until the next moon. It can be found here:
http://lariennalibrary.com/extern/temp/SmallRoller.zip
Enjoy!
http://lariennalibrary.com/extern/temp/SmallRoller.zip
Enjoy!
Thanks. :-)
I would use Excel to create a table with all possible values - there will only be 216 for 3 dice.
For two dice, it's pretty easy
1) Rolling two A's - out of 36 possibilities, you will get two A's on 9 of them (3x3). That makes your probability 1/4.
2) Question: does A/B mean A *OR* B, or does it count for both A and B? That will alter your probability.
For three dice - can you get MORE than the requirements?
For two dice, it's pretty easy
1) Rolling two A's - out of 36 possibilities, you will get two A's on 9 of them (3x3). That makes your probability 1/4.
2) Question: does A/B mean A *OR* B, or does it count for both A and B? That will alter your probability.
For three dice - can you get MORE than the requirements?
A/B is A or B.
Multiplying the fractions seems to work for what I need. I planned on difficulty being set as the number of symbols required and need a way to group difficulties.
If the difficulty is 2 A's, you'd have to assign at least two dice to the roll. More would be better.
What the fractions and die rollers don't help me with is rerolls or rolling extra dice.
you really will have to state what it is you intend to achieve
from what you have said so far, it seems you are trying to go for the probability of success of an action based on increase the amount of dice used for that action
the scale in achievability scales so exponentially, it might make the action avoided altogether.
if you did not know how to change the probability to percentage, assuming every side of the D6 is a different result:
a 1/6 probability with one D6 is a 16% chance
a 1/36 probability with two D6s is a 2% chance
a 1/216 probability with three D6s is a 0.4% chance
as said, you will need to explain what result "A", "B", "C", and "X" is, right now these are your terminologies, not general terminologies everyone knows about.
if you have the same result on several sides of the dice, this changes the probability dramatically.
I was thinking of some a little like Elder Sign. You would roll dice and try to match symbols on cards.
Assuming the die faces are such:
Miss, Punch, Punch, Punch/Kick, Kick, Headbutt
Abbreviated [M] [P] [P] [P/K] [K] [H]
(P/K can be either/or, not both)
We have three faces with Punch, two faces with Kick, and one each with Miss and Headbutt.
Say you have two cards in you hand:
And you have three "Fights" in front of you:
So Winston v Jacob is a 50% chance to beat, 66% if you roll an [H].
Churchill v Jacob... He gets two 50% chances, but I don't how much more likely you are to win.
Winston v Hamm: 17% if you roll an [H] (1/6)
Churchill v Hamm: 25% to win (3/6 x 3/6)
Churchill v Pete: 17% to win (3/6 x 2/6)
Winston+Churchill v Pete: ... no clue, or 17% if you roll an [H]
I wanted to separate challenges in to easy, medium, hard and figured I could base that off the minimum number of dice needed.
just note that anything i state that sounds obvious is for my own clarification, not an insult to your intelligence
okay using the faces on your dice, ill calculate the number of sides one result has:
1M, 3Ps, 2Ks, 1H
this is based on the assumption that the P/K face is a face that can switch between the 2 results when it is revealed.
so the chance of M is 1/6
the chance of P is 3/6 or 1/2
the chance of K is 2/6 or 1/3
the chance of H is 1/6
now calculating the probability of each card's success:
Winston, 1/6, or 16%
Glass Jaw Jacob, 1/2, or 50%
Mitch Hamm, 1/2*1/2, or 1/4, or 25%
Prize Pete, 1/2*1/3, or 1/6, or 16%
i dont understand what churchhill can do so i didnt analyse it, but your last example does bring up an issue, what is the math behind "Winston+Churchill"? if you do not know what the math is, youll have to state your intention behind that situation
In my example, Winston and Churchill are "creature" cards you play from your hand. Jacob, Hamm, and Pete are "missions" that you would play Winston or Churchill onto in order to defeat and "win the mission".
Winston can roll 1 die to try and defeat one of the missions (and has a special ability), Churchill rolls two dice.
So knowing that Churchill rolls two dice, what's the percentage chance that he will defeat Glass Jaw Jacob?
I googled this and apparently the answer is 75%.
If Winston and Churchill go up against Prize Pete (together, they're 3d6), what's the chance that they'll defeat Pete?
Couldn't find an answer to this one.
i think im finally beginning to understand your game a little more
so the dice probabilities you mentioned for jacob, mitch, and pete, were requirements to defeat them
and the dice effects you mentioned on winston and churchhill were their special abilities
from what you said, it seems like winston can only use one dice while churchhill can use 2 dice
correct me if any of the above imentioned is wrong
heres what i find weird, since Mitch and Pete need 2 dice to defeat them, does winston still fight with only one dice?
and when a fighter(winston or churchill) wins a challenger(jacob, mitch, and pete) based on the challenger's criteria of defeat, is the fighter discarded? cause your effect on winston states that you can discard him to win if you have a H result, is the discard to win exclusive to winston's effect or is it a default action?
so far this game seems to have alot of potential mix and matches, you could include multiple rolls as and ability, or reconsidering a result as another result (eg for this card, K is regarded as K/P)
That's how I read it first pass through. Winston can use a headbutt, so his die wins on an H independent of the target's defeat requirement. To meet the target, they roll 3 dice and hope to get at least one P and one K amongst the dice.
I should hone up on my combinatorics, but to do this with semi-brute force I'd split the calculation into the 5 die results Winston could have. Then put in the probability from there.
StagC, one quick note on that 75% probability for you. You get there by saying that my chance of getting at least one P on the dice is 100% minus the chance of getting no P. Going backwards you can multiply the "missed" probabilities, but you can't multiply the hit probabilities. A lot of probabilities are calculated that way, but the combinatorics version might look prettier.
If you split your start set, you can multiply the size of the group times its success probability and re-add. In this case, we have Winston's 1/6 of an H times the chance of that winning (100%) and have our first element: 16.7%
Now the P/K die (1/6), needs one P or K amongst C's two dice. Since 4/6 of the die is a P or K, the chance of not getting one is 2/6=1/3. That's the first of C's dice. Same with the second 1/3 of missing. 1/3 * 1/3 = 1/9 of missing. And so 8/9 of at least one P or K, times the 1/6th for this die is 8/54 or 14.8%
etc.
At least in this case, seems simpler to just generate the 216 results in Excel and make a formula to check. All summed up, I get about a 2/3 chance of the two of them winning.
I haven't messed with that dice sim, but I might. Seems useful once you realize how messy branching and multi-role dice can be. I do wonder if it is rough for the designer, how is it for the player... Are they happy enough on guy instinct?
and the dice effects you mentioned on winston and churchhill were their special abilities
from what you said, it seems like winston can only use one dice while churchhill can use 2 dice
If you split your start set, you can multiply the size of the group times its success probability and re-add. In this case, we have Winston's 1/6 of an H times the chance of that winning (100%) and have our first element: 16.7%
Now the P/K die (1/6), needs one P or K amongst C's two dice. Since 4/6 of the die is a P or K, the chance of not getting one is 2/6=1/3. That's the first of C's dice. Same with the second 1/3 of missing. 1/3 * 1/3 = 1/9 of missing. And so 8/9 of at least one P or K, times the 1/6th for this die is 8/54 or 14.8%
Here's what I would do in Excel:
Create columns of possible die rolls. So you label the columns d1, d2, and d3, for instance.
We'll go with 3d6 example because that's the most complex.
You are going to need a total of 216 (6^3) rows, plus that header row.
In the d1 column, put in H, P, P, P/K, K, and M - each in its own row. (so 6 rows) Repeat that sequence 36 times. (Copy and paste it)
In the d2 column, put in 6 H's, then 6 P's, then 6 P's, then 6 P/K's, then 6 Ks, then 6 Ms - each in its own row (36 total rows). Copy and past that 6 times.
In the d3 column, put in 36 H's, 36 P's, 36 P's, 36 P/K's, 36 K's and finally 36 M's, That's it (that's 6x36 = 216)
You now have all possible combinations for your dice.
The weird parts you will encounter are because of the P/K and the 3rd dice, which may or may not be necessary. (I assume even if you roll 3 P's, that you still win in an encounter that only requires 2)
If you only want to look at 3 dice, it might be easiest to "brute force" it rather than figure out a formula. What I mean by that is to simply do a manual check on what the 3 rolls are and then figure out if it's a "win" or "loss". You'd set up different columns for each scenario that you want. Then just put W or L in it. When you find them all, just add them up (SUMIF) and then divide by 216 for the probability.
Hopefully, I don't think I need to Brute Force it. I just wanted the probabilities for establishing rewards; anything factoring extra dice and rerolls should fall on the player side and they'll most likely go with their gut anyway.
Thanks a lot everyone. I think I have enough knowledge to continue working. :-)
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That was very helpful, thanks. :-)
I'll have to work with that website a little more too. Thank you. :-)