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Subject: Custom dice probabilities rss

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Jonathan Warren
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I am working on a design which might use four custom d6:

A on three of the d6 faces
1 on two of the d6 faces
X on the remaining face

I am wanting to calculate the probabilities of rolling A, 1, X when rolling 1 die, 2 dice, 3 dice and finally all 4 dice (all identical custom d6).

Could someone with a better understanding than I have help me out please If you could provide me with a calculation that I can use in future, that would be great.

Thanks for any help in advance.
 
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aahs jpw
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When you say rolling A, 1, X I assume you mean rolling at least one A, 1, X on the given roll of # dice. This is important.

What helps me think of those odds and how they work is that:
5/6 faces aren't X
4/6 faces aren't 1
and 3/6 faces aren't A

So each time you fail to roll the desired face on a single roll, the odds against also failing on the next roll compound at those rates. This is the same with rerolls or multiple same dice.

As a formula resulting in a percent chance of success, you get:
(% chance of at least 1 success) = 100% - ((100% - odds against success)^(# of rolls))

As a table of number of success faces (counting 1-6 faces on a d6) and number of rolls (up to four d6s as specified) you get a success chance of:

16.67% 30.56% 42.13% 51.77% (X)
33.33% 55.56% 70.37% 80.25% (1)
50.00% 75.00% 87.50% 93.75% (A)
66.67% 88.89% 96.30% 98.77%
83.33% 97.22% 99.54% 99.92%
100.00% 100.00% 100.00% 100.00%

The math gets more complicated when the faces are not exclusively success/fail but are additive like trying to roll higher than a certain number on actual d6.

Hope that helps!

aahsjpw




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Johannes Hihn
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I'm not an expert on using anydice.com so I'm not sure if you can simulate dice with other faces than number values. But probably it's worth checking out.

http://anydice.com/program/8f19

This is obviously not what you're looking for since you cannot see what a certain value consists of but maybe there's a way to replace the numbers with strings I couldn't find on the fly.
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Ranko M
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I'm guessing you don't know of Any Dice?

1dCustom
2dCustom
3dCustom
4dCustom
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aahs jpw
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I've worked with anydice in prototyping. It has what you describe.
The functions you're looking for look like:

http://anydice.com/program/8f1e

Cheers!

aahsjpw
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Michał
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These are result count tables for different amount of dice:
1 die
2 1
3 A
1 X

2 dice
4 11
12 1A
4 1X
9 AA
6 AX
1 XX

3 dice
8 111
36 11A
12 11X
54 1AA
36 1AX
6 1XX
27 AAA
27 AAX
9 AXX
1 XXX
4 dice
16 1111
96 111A
32 111X
216 11AA
144 11AX
24 11XX
216 1AAA
216 1AAX
72 1AXX
8 1XXX
81 AAAA
108 AAAX
54 AAXX
12 AXXX
1 XXXX

Now to get probability of any event find the rows that match that event. For example at least 2 A-s in 4 dice matches these:
216 11AA
216 1AAA
216 1AAX
81 AAAA
108 AAAX
54 AAXX

Add all the counts and divide by 6 to the power of the number of dice.
In the example case: 891/1296 = 68.75%.
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Jeremy Lennert
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If you want to use anydice, then instead of calling your sides 1, 2, and 3, it's probably better to call them 1, 10, and 100. That way you will be able to see how many of each result was rolled in the results table (e.g. a result of "201" means 2 of the result called "100" and 1 of the result called "1".

http://anydice.com/program/8f28
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Jonathan Warren
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aahsjpw wrote:
When you say rolling A, 1, X I assume you mean rolling at least one A, 1, X on the given roll of # dice. This is important.

I probably haven't explained myself very well. Let's assume my d6 are marked: O, O, O, B, B, X (this notation makes it easier for me to understand ).

If I roll 1 die, I want to know the probability of rolling either O or B or X. Turns out that this is quite an easy calculation:

O = 50%
B = 33.33%
X = 16.67%

Now, I want to understand the probability percentage when rolling two of the same dice dice on all permutations:

OO
OB
OX
BB
BX
XX

I also want to understand the probability when rolling three dice on all permutations:

OOO
OOB
OOX
OBB
OBX
OXX
BBB
BBX
BXX
XXX

And finally, the same with using all four dice:

OOOO
OOOB
OOOX
OOBB
OOBX
OOXX
OBBB
OBBX
OBXX
OXXX
BBBB
BBBX
BBXX
BXXX
XXXX

So, from the above permutations, I want to understand which is the easiest to roll (obviously the more O's the easier the roll - the higher the percentage, the easier it is to achieve the result). I am not worrying about mitigating the dice results by being able to re-roll dice for now, I'm just interested in the percentage for each permutation. It's probably really simple to calculate, but I'm having some trouble.
 
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Jeremy Lennert
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OK, so either Michał's table or mine (previous 2 replies) should tell you everything you wanted to know.
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Jonathan Warren
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Antistone wrote:
OK, so either Michał's table or mine (previous 2 replies) should tell you everything you wanted to know.

Thanks Jeremy. I think I've got it. Please bear with me for my lack of understanding while I ask yet another question. Here is a chart that I put together, based on the results of your link:

BBB 3 3.7
XBB 12 5.56
XXB 21 2.78
XXX 30 0.46
OBB 102 16.67
OXB 111 16.67
OXX 120 4.17
OOB 201 25
OOX 210 12.5
OOO 300 12.5

There must be some sort of theory behind this, but my brain cannot work it out! There are three faces showing O, two faces showing B and one face showing X. You would suppose that sides with more faces have a better chance of being rolled than results with less faces. Picking one result at random: How come that rolling XBB (5.56%) has a greater chance of being rolled that BBB (3.7%)? Same happens in two dice and four dice rolls, such as with two dice rolling BB has exactly the same chance as BX being rolled - this doesn't make sense to me.

BB 2 11.11
BX 11 11.11
XX 20 2.78
OB 101 33.33
OX 110 16.67
OO 200 25


BBBB 4 1.23
BBBX 13 2.47
BBXX 22 1.85
BXXX 31 0.62
XXXX 40 0.08
OBBB 103 7.41
OXBB 112 11.11
OXXB 121 5.56
OXXX 130 0.93
OOBB 202 16.67
OOXB 211 16.67
OOXX 220 4.17
OOOB 301 16.67
OOOX 310 8.33
OOOO 400 6.25

Red selection shows another - surely getting the three O results should have more chance than getting some of the other results (the 16.67 results or the OOOX)? Please help me to understand this
 
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Jeremy Lennert
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JoffW wrote:
Picking one result at random: How come that rolling XBB (5.56%) has a greater chance of being rolled that BBB (3.7%)?

Because "XBB" is actually three different combinations (XBB, BXB, and BBX), whereas BBB is only one. So although "B" is two times as common as "X", "XBB" stands for three times as many combinations as "BBB", so it's more common overall (because 3 > 2).
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