# Probability Computation

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questccg
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Joined: 04/16/2011

So I'd like to know how to "compute" the Probability of the following:

Choose 3, values must add up to 7.

1, 1, 5
1, 5, 1
5, 1, 1
1, 2, 4
1, 4, 2
4, 1, 2
4, 2, 1
2, 4, 1
2, 1, 4

etc.

My problem is how to compute for the TOTAL... If you can explain it too. That would be great. Because I am unsure about the TOTAL. It was 9 and that may be too high and 6 was too low... 7 seems to be a bit more than the 6 ... But I'd like to know how many possibilities exist.

Is this even possible to compute, IDK?!

One of you PRO mathematicians (I know we've got a few of them around!), my great gratitude if you can help me with the formula.

Also is there anywhere where I can get ALL the possibilities??? Some kind of website, maybe?!

Cheers.

Fri
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Joined: 09/06/2017
Anydice.com

Anydice.com

3 d6s is their example. It includes the probability of getting a seven.

nand
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Joined: 07/27/2008
This line in nanDECK creates

This line in nanDECK creates a sequence (named "dice") of all the 216 permutations with repetitions of three elements taken from a set of six:

pr[dice]3=1|2|3|4|5|6

And this line saves that sequence into a file (named "res.txt"):

[out]=savelabel(res.txt,dice)

And this line creates a sequence (named "seven") taking only the elements that sum 7 from that sequence:

[seven]=filter(+[dice],7)

i.e.:

115
124
133
142
151
214
223
232
241
313
322
331
412
421
511

X3M
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Joined: 10/28/2013
Fri beat me to it

Anydice...

Ridethewave
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Joined: 06/04/2020
Mind spinner

I don’t know if this helps you but.....
There is a one in six chance of getting 7 with the first two die...

Therefore there is a 2 1/2 in 6 chance(5 in 12) chance that you rolled under 7 with the first two dice.... with 5 ways of making it up to 7(however 6 is far more likely than two)with the last dice.
Looking at it this way definitely gives analysis paralysis lol

John

Posted this a couple hours after writing it... and now Having seen the answer I understand what the question was about... I quite like screwing my mind up with stats and probabilities so I’ll leave it up for a laugh for any similar likeminded math wannabes lol

questccg
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Joined: 04/16/2011
I actually found a way to compute the series

Let's say my total is "9":

 1, 1, 7 2, 1, 6 3, 1, 5 4, 1, 4 5, 1, 3 6, 1, 2 7, 1, 1 1, 2, 6 2, 2, 5 3, 2, 4 4, 2, 3 5, 2, 2 6, 2, 1 1, 3, 5 2, 3, 4 3, 3, 3 4, 3, 2 5, 3, 1 1, 4, 4 2, 4, 3 3, 4, 2 4, 4, 1 1, 5, 3 2, 5, 2 3, 5, 1 1, 6, 2 2, 6, 1 1, 7, 1

And this has allowed my to compute for 8, 7 or 6 also. That's how you do it "MANUALLY"... Anyhow 9 was tops and has the most amount of choices.

X3M
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Joined: 10/28/2013
You are using d7?

Where is Lewisher to kick your butt?

Ridethewave
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Joined: 06/04/2020

You seem to have a 7 on your d6.
Average on 3xd6 is 9.5 so I’m guessing 10 and 11 are equally good

questccg
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Joined: 04/16/2011
It's NOT dice probabilities ... but STAT distribution!

X3M wrote:
Where is Lewisher to kick your butt?

I think everyone misunderstood that they are NOT dice probabilities but STAT distributions. I am trying to figure out if the TOTAL number of points should be 9, 8, 7 or 6.

And there are 3 resources. Ergo. Choose 3, totaling "?".

There is no SEVEN (7) die. It means that a stat has 7 resources. Of course, I have removed those and stuck with only 5 or lower in the distribution. What I am saying is that AT MOST any STAT can be 5 or less.

Sorry. I was trying to figure out how ANYDICE was going to help. I wanted something more like a formula but have no idea how to compute for a TOTAL. But I did manage to find a MANUAL way of doing the results of the computation...

questccg
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Joined: 04/16/2011
Specifically this had to do with end-game scoring

I was a bit concerned for the end-game scoring which resulted in a "stalemate". So one Designer proposed lowering the Victory Points (think resources) by a point or two. So that led me to wanting to have outcomes for 6, 7, 8 and 9 VPs (or resources). See in the game player must collect "Crystals" which add to their Lord/Lady's "fervor". The three (3) in-game vices are Hunger, Greed and Bloodlust.

So I wanted AT MOST 5 of any given vice. Just because getting five (5) of the vices is HARD ENOUGH. Players may opt for more BALANCED Lord/Lady's "fervor": like 4, 3, 2 or 3, 2, 4. Anyways there are 10 possibilities and only nine (9) Lords/Ladies. And I've removed 3, 3, 3... because that's just too balanced and is the easiest of the STATS to win with. That leaves now 9 possibilities and just enough Lords/Ladies.

I'm sorry for the confusion... My mistake to use the incorrect terminology in the OP. But thanks for your input, Nand actually did a pretty good job listing the outcomes (for 7).

Ridethewave
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Joined: 06/04/2020
Ah that explains it better

The mention of dice threw me. It is of course a matter of counting the permutations that are allowed to make the final answer.
Perhaps like playing cards you could have higher ranking(classes) of vices to break tiebreaks. Or could you be given secret cards at the beginning of the game which give a target/bonus for a particular set. Therefore also increasing replayability.
My apologies as I am new and have not yet read through your game.

John

larienna
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Joined: 07/28/2008
From a mathematical point of

From a mathematical point of view, the probability of anything is always the number of permutation desired, over the total number of permutations.

For exampple: roll 10+ on 2D6 is
desired permutation: 3 + 2 + 1 = 6 permutation
total permuation: 6x6 = 36

desired/total = 6/36 or 1/6.

If the number of permutation is high (3d or more) you generally use combinations and arrangements to know the number of permutation in any category (desired or total)

Or if you are lazy, you can use rolling softwares. Or you can code your own random generator and count dices if you know how to program. I actually did this for one of my game where decisions had to be made between rolls.

X3M
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Joined: 10/28/2013
I see, no dice

In that case, you got a very good method already.
But I noticed something.

It starts at 3:
1,1,1

Then you get 4:
1,1,2 2,1,1
1,2,1

I could continue with 5 etc., but I see a patern here.
In your case, you have a triangular number.

Combinations=0.5*((Total-2)^2+(Total-2))
Max=Total-2
I just realized, could have done this the other way around.

Total:Max:Combinations
3:1:1
4:2:3
5:3:6
6:4:10
7:5:15
8:6:21
9:7:28
10:8:36
11:9:45

How far you wanna know?

questccg
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Joined: 04/16/2011
You could easily be a "Professor" of some kind of mathematics

X3M wrote:
In that case, you got a very good method already. But I noticed something... But I see a pattern here. In your case, you have a triangular number.

Yeah I see the "triangular pattern" too.

X3M wrote:
Combinations = 0.5*((Total-2)^2+(Total-2)). Max = Total-2

Total:Max:Combinations
3:1:1
4:2:3
5:3:6
6:4:10
7:5:15
8:6:21
9:7:28
10:8:36
11:9:45...

Yes this seems to be accurate. How you derived that FORMULA ... IDK. Man you rock! You are like a Wizard with numbers and math. With your mind, I'm sure you could easily become a "Professor" in the event you get tired of your day job. Obviously in some field of Mathematics or research...

X3M wrote:
How far you wanna know?

I only needed 6, 7, 8, and 9. Thanks Ramon... Doing it with the Triangular data worked for what I needed ... But your equation and the combinations seems accurate to me! Cheers.

questccg
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Joined: 04/16/2011
I think X3M's formula computes the total number of combinations

Indeed ... With that formula and "Total:Max:Combinations" that explicitly explains and computes all the values that I need. Like I said, I only needed it for 6, 7, 8, and 9. The main reason that 9 could end the game in a "stalemate" or draw... So I wanted to see IF the another lower value could be used.

The main problem is winning the game TOO EARLY. The later rounds can be critical like 9, 10, 11, and 12. It's a THIRD of the game (potentially) and it is important to experience the game as much as possible. a value of "9" does stretch the game further... It just means that the "possibility" of a draw is more likely.

But I'd rather have a game that is more complete (playing the later turns) than one that finishes too prematurely. Like in the 8th or 9th round...

I will do more playtesting ... Anyhow I have the Lord/Lady cards to re-design anyhow. I re-ordered them to be more according to a sorted order. And there are NINE (9) of them.

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Joined: 02/11/2015
They were right, you can use anydice

For choose 3, total 7
So on any dice, type "OUTPUT 3d7". It produces a probability list for each total. Multiply the % value (4.37317784257%) of total = 7 (in decimal form 0.0437317784257) by the total number of permutations (in this case 7^3=343) and you get 15.
Replace the 7s for 8s and get 21. (In other words, match X3M's results.)

Not as quick or clean as X3M's method, but doable.

Masacroso
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Joined: 05/05/2014
questccg wrote:So I'd like to

questccg wrote:
So I'd like to know how to "compute" the Probability of the following:

Choose 3, values must add up to 7.

1, 1, 5 1, 5, 1 5, 1, 1 1, 2, 4 1, 4, 2 4, 1, 2 4, 2, 1 2, 4, 1 2, 1, 4

etc.

My problem is how to compute for the TOTAL... If you can explain it too. That would be great. Because I am unsure about the TOTAL. It was 9 and that may be too high and 6 was too low... 7 seems to be a bit more than the 6 ... But I'd like to know how many possibilities exist.

Is this even possible to compute, IDK?!

One of you PRO mathematicians (I know we've got a few of them around!), my great gratitude if you can help me with the formula.

Also is there anywhere where I can get ALL the possibilities??? Some kind of website, maybe?!

Cheers.

There is a general formula for this kind of problems, that is, to find the probability that throwing n fair dices of D sides (with values from one to D) you get a sum of S.

Now, if you wish, you can implement this formula in any programming language or mathematical software to get tables, probability mass function graphs or whatever you like, but probably the fastest way to find what you want is using anydice.com