# Variation on the urn problem and frequency distribution

I have $6$ machines each producing different coloured balls. The balls are mixed together in a large vessel. Groups of $6$ balls are extracted at random for packing. Each pack will therefore have a random combination of balls.

Given that for my purposes

1. $a,b,c,d,e,f$ is the same as $f,e,d,c,b,a$ i.e. $6$ all different colours and
2. $a,a,a,a,a,b$ is the same as $e,e,e,e,e,f$ i.e. $5$ of one colour and $1$ of any other colour

I think that that means I have $11$ different cases How do I work out the proportion that will fall into each case ?

Is there a general formula that would work with say packs of $8$ from $10$ different colours? I can work out from $n\choose r$ = $11\choose6$ = 462.

I get from inspection that there are $11$ cases. How do I move on from there to the proportions that fall into each case?

Edit A batch is around 30000 balls, it is therefore reasonable to assume that the proportions stay constant as each pack is removed.

• I gather you aren't replacing the balls. Are there enough balls that you can use an approximation w/ constant proportions, or do you need exact results? Jan 18, 2015 at 15:13
• A batch is approx. 30000, so constant proportions is a valid assumption. thanks Jan 18, 2015 at 19:07

To calculate the number of possible combinations of 6 different colors selected for each outcome type, take Comb(6, # of different colors selected for that outcome type)

6 balls all one color = Comb(6,1 color selected) = 6

5 balls one color, 1 ball another color = Comb(6,2) = 15

4 balls one color, 2 balls another color = Comb(6,2) = 15

4 balls one color, 1 ball another color, 1 ball another color = Comb(6,3) = 20

3, 3 = Comb(6,2) = 15

3, 2, 1 = Comb(6,3) = 20

3, 1, 1, 1 = Comb(6,4) = 15

2, 2, 2 = Comb(6,3) = 20

2, 2, 1, 1 = Comb(6,4) = 15

2, 1, 1, 1, 1 = Comb(6,5) = 6

1, 1, 1, 1, 1, 1 = Comb(6,6) = 1

However, the combinations above only give you the number of ways color combinations can be selected for the given outcome type (i.e. 4 balls one color, 2 balls another color can cover two different colors in 15 different ways when there are 6 colors). It does not factor in the number of balls selected for each respective color (i.e. colors blue and green can be covered two ways: one with 4 blue and 2 green and the other with 2 blue and 4 green). Therefore, you need to also account for the ordering of the selection by multiplying the Combinations above by the number of possible orders. The result is the number of possible outcomes for each outcome type.

6 balls all one color = Comb(6,1)*1 possible order = 6

5, 1 = Comb(6,2)*2 possible orders = 30

4, 2 = Comb(6,2)*2 possible orders = 30

4, 1, 1 = Comb(6,3)*3 possible orders = 60

3, 3 = Comb(6,2)*1 possible order = 15

3, 2, 1 = comb(6,3)*6 possible orders = 120

3, 1, 1, 1 = Comb(6,4)*4 possible orders = 60

2, 2, 2 = =Comb(6,3)*1 possible order = 20

2, 2, 1, 1 = Comb(6,4)*6 possible orders = 90

2, 1, 1, 1, 1 = Comb(6,5)*5 possible orders = 30

1, 1, 1, 1, 1, 1 = Comb(6,6)*1 possible order = 1

There are 462 total possible outcomes in this example, as you mentioned. Since all specific outcomes are equally likely, you can simply divide the number of possible outcomes for a given outcome type by 462 to get the respective probability of occurrence for that outcome type.