Probability of duplicate colours on 2 gos (4 drawn total)

[njmiller]

A single go consists of picking 2 balls from a randomly ordered set of coloured balls (4 distinct colours).

Possible distinct colours = n = 4

Balls chosen per go = r = 2

They're spaced out in such a way that you cannot get the same coloured ball twice in a single go, but you can get the same colour across multiple go's.

There are 20 green (Gn), 18 blue (Bn), 17 red (Rn), and 15 yellow (Yn) balls = 70 total balls.

What is the probability of getting the same coloured ball across 2 go's

>>> I need help with this problem. So far, I have been able to map out the n!/(n-r)! possibilities = 4!/2! = 12 possibilities for a single go (no repetition). I have deduced that across the 2 go's there are 12^12 possibilities = 144, and i think 12 of those do not contain duplicates. I need to find an equation to calculate the probability of getting a duplicate colour for any n, any r, and any counts for each colour (Gn, Rn, Bn, Yn).

Any help would be greatly appreciated.

Thanks

 

[blamocur]

Is this a part of your homework? Do you need an exact answer or an estimate?

 

[njmiller]

Is this a part of your homework? Do you need an exact answer or an estimate?

It's a small part of a bigger project. An estimate would be helpful if not the exact answer or even ideas on how this should best be approached, my statistical skills aren't the strongest! Thanks

 

[Dr.Peterson]

Probability of duplicate colours on 2 gos (4 drawn total)

A single go consists of picking 2 balls from a randomly ordered set of coloured balls (4 distinct colours).
Possible distinct colours = n = 4
Balls chosen per go = r = 2
They're spaced out in such a way that you cannot get the same coloured ball twice in a single go, but you can get the same colour across multiple go's.
There are 20 green (Gn), 18 blue (Bn), 17 red (Rn), and 15 yellow (Yn) balls = 70 total balls.
What is the probability of getting the same coloured ball across 2 go's?
>>> I need help with this problem. So far, I have been able to map out the n!/(n-r)! possibilities = 4!/2! = 12 possibilities for a single go (no repetition). I have deduced that across the 2 go's there are 12^12 possibilities = 144, and i think 12 of those do not contain duplicates. I need to find an equation to calculate the probability of getting a duplicate colour for any n, any r, and any counts for each colour (Gn, Rn, Bn, Yn).
Any help would be greatly appreciated.

I don't understand how you can be choosing randomly, from a random collection, and yet be unable to take two of the same color at a "go". This claim makes it hard to judge what the probabilities would be.

Also, it is not clear whether balls are replaced after one "go".

In any case, you can't calculate probability by counting only distinct outcomes; they will not be equally probable. You need to take into account the different numbers of each color.

It's a small part of a bigger project. An estimate would be helpful if not the exact answer or even ideas on how this should best be approached, my statistical skills aren't the strongest!

It may be very important for you to tell us the larger problem, as it is quite possible that you have already made a mistake in arriving at this formulation.

 

[njmiller]

It's a system that issues 2 balls each time without repetition - never the same colour per each go/2 balls drawn. Other than this they are randomly ordered.

I want to understand the probability of getting 1 or more duplicate colour across 2 go's (4 balls drawn), based on the number in each pot, so that we can alter the number of balls in each pot to reduce the chance of duplicates. No replacement. I see what you are saying, still not sure the best way to approach

 

[blamocur]

and i think 12 of those do not contain duplicates.

While I agree with the total of 144 with non-duplicated colors in single go's, the number without any duplicates seems wrong.

 

[blamocur]

I don't understand how you can be choosing randomly, from a random collection, and yet be unable to take two of the same color at a "go".

One can imagine a "semi-random" ordering of all balls which do not allow same colors inside pairs, i.e. balls 1 and 2 must be different, so must 3 and 4, and so on. But, for example, 2 and 3 can have the same color.

 

[njmiller]

Hey thanks for the info!! Ive revised this question based off what ive learned today, also because i think the figures were wrong previously:

There is a pot of balls of 4 distinct colours.

The balls are ordered in a semi random way which does not allow the same colors inside pairs so 1 and 2 must be different, so must 3 and 4, and so on, but 2 and 3 for example can have the same color.

For a single go, you chose one pair (2 balls no repitition). The number of possibe arrangements is then C(n,k) = C(4,2) = 6 (no reptition on a single go and order doesn't matter).

This would leave me to believe that if there was an equal number of each colored balls, the probability of getting a duplicate colour across 2 gos would be 30 possibilies for duplicate/36 total possibilties (6*6 for 2 goes).

The issue is that the sizes of the pots are different for each colour, and they are not replaced.

What's the best approach to give either a probability of receiving a duplicate across 2 gos, or the average number of duplicates you can expect across 2 gos? I want the final probability/average to compare different set ups for number of colors in each pot, and find the one that is best at reducing duplicates across multiple gos. I'm starting to think I'd be better looking for the average.

Let me know your thoughts, thanks!!

 

[Dr.Peterson]

The balls are ordered in a semi random way which does not allow the same colors inside pairs so 1 and 2 must be different, so must 3 and 4, and so on, but 2 and 3 for example can have the same color.

You've borrowed this statement from @blamocur, but I don't think it helps. You seem to suppose that the balls have been placed into one row, and all you are doing in "choosing" is taking the next two balls in order. If this ordering is "semi-random", then we need to know exactly what that means! The probability will come from the way in which someone ordered them, which we don't know. This just pushes the problem back further.

Suppose they randomly took balls from four pots of different colors, following the rule that no pair (odd-even) can have the same color, I can imagine that he reaches the end and has no choice that fits the rule; what does he do then? Throw out the whole arrangement and start over?

But in any case, this is where you would have to work out the probabilities.

For a single go, you chose one pair (2 balls no repitition). The number of possible arrangements is then C(n,k) = C(4,2) = 6 (no reptition on a single go and order doesn't matter).

No; there are not 4 equally-likely colors to choose from, so you can't just count arrangements of colors. There are distinct balls, and the number of each matters. You have to count arrangements of balls.

This would leave me to believe that if there was an equal number of each colored balls, the probability of getting a duplicate colour across 2 gos would be 30 possibilies for duplicate/36 total possibilties (6*6 for 2 goes).

The issue is that the sizes of the pots are different for each colour, and they are not replaced.

Yes, this is the issue. You have to consider how the balls are put in the given order. That is what is missing.

What's the best approach to give either a probability of receiving a duplicate across 2 gos, or the average number of duplicates you can expect across 2 gos? I want the final probability/average to compare different set ups for number of colors in each pot, and find the one that is best at reducing duplicates across multiple gos. I'm starting to think I'd be better looking for the average.

I'm confused there. I thought each pot contained balls of a single color? Or did you never actually say what the pots are? At one point you indicated there is one pot:

There is a pot of balls of 4 distinct colours.

 

[blamocur]

If this ordering is "semi-random", then we need to know exactly what that means!

My interpretation would be: take all possible sequences without "single-go duplications", then pick randomly one of them with equal probability.

 

[blamocur]

What's the best approach to give either a probability of receiving a duplicate across 2 gos, or the average number of duplicates you can expect across 2 gos?

I cannot think of any analytical way to solve this. My quick-and-dirty script showed something like 0.825 probability of finding duplicates in the first 2 go's.

 

[njmiller]

You've borrowed this statement from @blamocur, but I don't think it helps. You seem to suppose that the balls have been placed into one row, and all you are doing in "choosing" is taking the next two balls in order. If this ordering is "semi-random", then we need to know exactly what that means! The probability will come from the way in which someone ordered them, which we don't know. This just pushes the problem back further.

Suppose they randomly took balls from four pots of different colors, following the rule that no pair (odd-even) can have the same color, I can imagine that he reaches the end and has no choice that fits the rule; what does he do then? Throw out the whole arrangement and start over?

But in any case, this is where you would have to work out the probabilities.


No; there are not 4 equally-likely colors to choose from, so you can't just count arrangements of colors. There are distinct balls, and the number of each matters. You have to count arrangements of balls.

Yes, this is the issue. You have to consider how the balls are put in the given order. That is what is missing.

I'm confused there. I thought each pot contained balls of a single color? Or did you never actually say what the pots are? At one point you indicated there is one pot:

This is really helpful thank you. It is one single pot, containing 4 different colours (it's the number of each different color that is different, but from the same pot) sorry for my wording!

Imagine it more as a computer system that has pre arranged these colours so that they can't be consecutive but other than that, the ordering is random - they are randomly inserted into the pot where there is at least 1 different color between same colors.

I understand now the probability depends on the exact ordering, which I won't know before I need to make this calculation.

Instead of a probability problem, knowing the approx. average number someone will get with 2 go's (4 draw, no repitition for 2 at a time) of duplicates would be just as useful. There must be a way to mathematically do this based on the number of distinct colours (4 in my example), and the total number of each colour (will vary for each set up), and the number of colours the user will select for a single go (2 in my example). This is a problem I have that I have generalised to a coloured balls problem - but its really a system chosing items - the problem is the same.