Hey, You, Nerd!

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W3H5

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Smart people required. I have a probability/ maths question that is boggling my little brain. Any help would be great. I'm sure the answer is pretty simple I just can't grasp it.

So, 3 dice. Each die face has a single colour; red, blue, green, pink, purple, orange, instead of a number of pips. All 3 dice are rolled to get a colour combination result.
I stacked coloured blocks (digiatally) to work out all possible combinations and came up with the following:
Total combinations: 54
3 of the same: 6
2 of the same: 30
None the same: 18

Can anyone confirm this is correct without going through the labourious block stacking prcoccess I went through, perhaps by using a formula and/ or computer program?

And secondly, what are the percentages on each of these amounts? Is it as simple as finding a normal percentage; 3 of the same: 6 as a percentage of 54 = ~11%?

To take it further, though this might be pushing it a bit, is there a way of working out the percentage of getting a particular combination, say red, green & blue without using anything other than basic percentage mathamatics, that is to say, not using a complex formula? Would it just be 1 in 54?
 
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It's 20 years since I left college so the following could be nonsense.

There are 6^3 = 216 equally likely outcomes. Some of those outcomes look the same to us but can occur in different ways.

Red, Red, Blue and Red, Blue, Red,
look the same as each other but Blue, Blue, Blue
can only happen one way.

6 of the 216 possible outcomes show the same colour 3 times. That's less than 3%.

Your red, green, blue, example can occur in 6 different ways, as you're not bothered (it seems) which die shows which colour.
R, G, B
R, B, G
G, R, B
G, B, R
B, R, G
B, G, R
As with the other example, 2.77...%
 
2.77 for an individual mixed colour outcome? Thanks.
The order of colour is irrelevant for my application so that saved on some further calculations.

E3ECDA35-F483-42F1-9EF8-A1A991FFE1E9.png


I started with this which, as you can see, is far from scientific but it got me some results.

I’m aware that the total combinations equate to 216 from a standard 3D6 but I’m wondering how I only got a total of 54. It seems I might be missing something. Are the 216 combinations from numbered dice different somehow to what I’m getting at here or have I totally messed up my calculations?

Edit: rereading the response from @polysmut am I right in assuming that 216 combinations factors in the order in which the results are displayed?
 
W3HS
Edit: rereading the response from @polysmut am I right in assuming that 216 combinations factors in the order in which the results are displayed?
Click to expand...
Yes.

Some of those are effectively the same result for your purposes. 2 reds and a blue arises from 3 of the 216 outcomes (blue first, blue second or blue third).

Did you count the distinct outcomes as 54 by taking 2 reds and a blue as one possible outcome instead of 3 equally likely possible outcomes which all appear to be the same? (I haven't checked whether that does give 54)
If so, it's a useful number to have but seems to have thrown you and caused you to treat those 54 outcomes as equally likely.

W3HS
2.77 for an individual mixed colour outcome? Thanks.
Click to expand...
If you have specified which 3 colours you want to see, yes. In this case 1 red, 1 blue and 1 green.
If you want to know the probability of seeing any 3 different colours, that's a different calculation.*
W3HS
The order of colour is irrelevant for my application so that saved on some further calculations.
Click to expand...
It can actually make the calculations less simple.

Based on seeing you ask probability questions on GTP before, I think nCr & nPr (as they appear on a calculator) would be good things for you to look at. I vaguely remember the functions being called "choose" and "permutation"


*The first die can be anything, guaranteed winner.
The 2nd must be different, that's a 5/6 chance of success. One face of the die matches the previous die, that's no good to us.
The 3rd must be different again, 4/6 remaining sides are good for us.

Multiplying
6/6 * 5/6 * 4/6 = 120/216 = 55.55..%
 
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polysmut
Did you count the distinct outcomes as 54 by taking 2 reds and a blue as one possible outcome instead of 3 equally likely possible outcomes which all appear to be the same?
Click to expand...

Yeah. Blue, blue, red would be no different from blue, red, blue in the system I want to use, so it wouldn't need to specifiy an order. That's probably how I came up with the number 54 (from the image in the previous post).

Thanks for the help. I'm slowing starting to piece it together in my head.
 
W3HS
Smart people required. I have a probability/ maths question that is boggling my little brain. Any help would be great. I'm sure the answer is pretty simple I just can't grasp it.

So, 3 dice. Each die face has a single colour; red, blue, green, pink, purple, orange, instead of a number of pips. All 3 dice are rolled to get a colour combination result.
I stacked coloured blocks (digiatally) to work out all possible combinations and came up with the following:
Total combinations: 54
3 of the same: 6
2 of the same: 30
None the same: 18

Can anyone confirm this is correct without going through the labourious block stacking prcoccess I went through, perhaps by using a formula and/ or computer program?

And secondly, what are the percentages on each of these amounts? Is it as simple as finding a normal percentage; 3 of the same: 6 as a percentage of 54 = ~11%?

To take it further, though this might be pushing it a bit, is there a way of working out the percentage of getting a particular combination, say red, green & blue without using anything other than basic percentage mathamatics, that is to say, not using a complex formula? Would it just be 1 in 54?
Click to expand...

3 equal and 2 equal faces is easy. There is one unique combination that gives 3 equals, so that's a total of 6 for all the faces.
For 2 equal faces the last dice can have 5 different faces, so that is 5 combinations per face, so a total of 30.

For all faces unique it gets tricky, since we're only counting unique solutions. Let's call the faces A, B, C, D, E, F.

The easiest way to do this is simply to list all the combinations. We can do this by exhausting the faces one at a time:

AB+(C to F) = 4
AC+(D to F) = 3
AD +(E to F) = 2
AE+F = 1

BC+(D to F) = 3
BD+(E to F) = 2
BE+F = 1

CD+(E to F) = 2
CE+F = 1

DE+F = 1

And we run out of unique solutions. In total we have 20 unique solutions for all faces being unique.

So to summarize we have 56 unique combinations. 6 for three equal faces, 30 for 2 equal faces and 20 for all faces being unique.

The percentages depends on how you are planning to generate these combinations. If it's by random dice throw then it's:

3 equal faces: 1*(1/6)*(1/6) = 1/36, or around 2.78%
This is because the first dice can be anything (1), while the other two needs to be the same as the first one, so (1/6)*(1/6).

2 equal faces: 1*(1/6)*(5/6)*3 = 15/36, or around 41.67%
One dice can be anything (1). One dice must be the same value as one other dice (1/6 probability). One dice must be a different value from the two others (5/6 probability). The order of the dice doesn't matter, so there are three combinations through which you can achieve this: aab, aba or baa. That means we multiply by 3.

All unique faces: 1*(5/6)*(4/6) = 20/36, or around 55.56%
The first dice can be anything (1), the second dice must be a different value from the first (5/6) and the third dice must be a different value from the two others (4/6). In this case there is only one combination, because the probabilities are determined by the order of the dice.

If it's not by random dice throw, but let's say that you are storing all the different unique combinations in a list and draw a combination at random from there, then the probability is simply the number of items of that kind in the list / the length of the list. So to get three faces of the same colour would have a probability of 6/56 = 11%. For two faces we get 30/56 = 54%, and for all faces unique we get 20/56 = 36%.
 
W3HS
I’m only counting 54, seems I’m missing two...
Click to expand...

Each colour should appear in ten combinations, but you only have nine per colour. I think the combinations missing are those that combine column 1,3, 5 and 2, 4, 6.

Edit: Turns out there is an "easy" (if you know it) way to calculate the number of unique combinations where all colours are different. It's the binomial coefficient, "six choose three".

binom.png


I also found that Python has a function for generating all the combinations, included repeated values (i.e. a number can occur more than once in a combination). I used these colour codes: R = red, Y = yello, G = green, C = cyan, B = blue, M = magenta. And there are indeed 56 different combinations in total.

Code: 
import itertools

def printCombinations(i):
    counter = itertools.count()
    for a, b, c in i:
        print(next(counter), ":", a, b, c)

colours = ['R', 'Y', 'G', 'C', 'B', 'M']

combinations = itertools.combinations_with_replacement(colours, 3)

printCombinations(combinations)

0 : R R R
1 : R R Y
2 : R R G
3 : R R C
4 : R R B
5 : R R M
6 : R Y Y
7 : R Y G
8 : R Y C
9 : R Y B
10 : R Y M
11 : R G G
12 : R G C
13 : R G B
14 : R G M
15 : R C C
16 : R C B
17 : R C M
18 : R B B
19 : R B M
20 : R M M
21 : Y Y Y
22 : Y Y G
23 : Y Y C
24 : Y Y B
25 : Y Y M
26 : Y G G
27 : Y G C
28 : Y G B
29 : Y G M
30 : Y C C
31 : Y C B
32 : Y C M
33 : Y B B
34 : Y B M
35 : Y M M
36 : G G G
37 : G G C
38 : G G B
39 : G G M
40 : G C C
41 : G C B
42 : G C M
43 : G B B
44 : G B M
45 : G M M
46 : C C C
47 : C C B
48 : C C M
49 : C B B
50 : C B M
51 : C M M
52 : B B B
53 : B B M
54 : B M M
55 : M M M
 
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Yes, this is a combinations problem, specifically with repetition allowed (or replacement).

The combinations with repetition equation is:
combinations-repeat.gif

where n is the number of choices and r is the times you choose.

This is slightly different than the equation that @eran0004 posted, which is for combinations where repetitions are not allowed.

In your case, you have 6 colors (choices) per dice and you are choosing 3 colors (1 per die), so the equation becomes (6+3-1)!/((6-1)!*3!) = 8!/(5!*3!) = 40320/(120*6) = 56 combinations of colors, which @eran0004 showed with the Python script that allowed replacement.
 
I bow before your mighty arithmetic knowledge. These answers are really helpful guys, thanks for your time. Although some of it went over my head, I must admit.

The basic function of this system is to introduce a game to my students using probability (something I plan to teach them) attached to colours (something they know well). The odds will help me to set prizes for certain outcomes based on likelihood.

You can all feel good in the knowledge that you’ve assisted a bunch of 5-6 year olds become more proficient in English and maths!
 

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