Why doesn't order matter in this

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Discussion Overview

The discussion revolves around the probability of drawing colored balls from a jar, specifically focusing on the independence of order in calculating probabilities for different sequences of draws. Participants explore the implications of this property in combinatorial contexts and seek intuitive explanations and proofs for the general case involving multiple colored elements.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Conceptual clarification

Main Points Raised

  • One participant calculates the probabilities of drawing specific sequences of balls (WWB and WBW) and notes that the probabilities are the same despite the different orders.
  • Another participant highlights that the numerators and denominators in the probability calculations remain consistent across different sequences, suggesting a deeper reason for the independence of order.
  • A participant expresses understanding of the calculations but seeks clarification on the underlying reasoning.
  • Another participant provides a general case for the independence of order, presenting a formula for the probability of drawing a specific sequence of black and white balls and emphasizing the commutative property of multiplication.
  • One participant reiterates the request for intuitive explanations regarding why order does not matter in these calculations.

Areas of Agreement / Disagreement

Participants generally agree on the observation that order does not affect the probability calculations, but there is no consensus on the intuitive understanding or formal proof of this property.

Contextual Notes

The discussion includes various assumptions about the independence of order in probability calculations and relies on the properties of multiplication and fractions without resolving all mathematical steps or definitions.

Who May Find This Useful

Readers interested in probability theory, combinatorics, and intuitive understanding of mathematical principles may find this discussion relevant.

aaaa202
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Say you have two black balls and two white in a jar and you pick 3 at random. When one is picked it is taken out of the jar.
I want to know the probability of getting for instance WWB. That must be:

P(WWB) = 1/2*1/3*1 = 1/6

Now say you want WBW. Then you get:

P(WBW) = 1/2 * 2/3 * 1/2 = 1/6

Amazingly enough the probabilities are independent on the order in which you choose your balls. This result is kind of amazing for me and not really intuitive. So far it stands more or less as a defining property of this kind of combinatorics.
Is it intuitive for you that the order really is independent? What argument would you give for this?
And lastly, perhaps even more interestingly: Can anyone prove that if you general have NA blue elements and NB red elements then the number of ways of getting a blue and b red elements is independent of order. I don't know if it's an established theorem - if so, where can I read more about it?
 
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Just look at those calculations: in the first you have the fractions 2/4, 1/3, and 2/2. In the second, the fractions are 2/4, 2/3, and 1/2. Do you see that, while the fractions are different, the three numerators and three denominators are the same?
Do you see why that is true? Do you see that, because the numerators are the same and the denominators are the same, the products are the same?
 
I see it yes. Why it's true I'm not sure.
 
Ivy, can you please help on the intuition for it?
 
The order does not matter in general case.

Suppose you have [itex]b[/itex] black balls and [itex]w[/itex] white balls. The total amount of balls is [itex]n=b+w[/itex].
Consider a sequence {B,W,B,B,...,B,W}. The probability that this sequence occurs is:
[itex]\frac{b}{n}[/itex][itex]\frac{w}{n-1}[/itex][itex]\frac{b-1}{n-2}[/itex][itex]\frac{b-2}{n-3}[/itex]...[itex]\frac{b-(b-1)}{2}[/itex][itex]\frac{w-(w-1)}{1}[/itex]
Note that the numerator of this sequence is [itex]b(b-1)...(b-(b-2))(b-(b-1))w(w-1)...(w-(w-2))(w-(w-1))[/itex] and the denominator is [itex]n(n-1)...(n-(n-2))(n-(n-1))[/itex].
You should notice that whenever you want to choose [itex]b[/itex] black balls and [itex]w[/itex] white balls, the numerator and the denominator will be the same. The order of the elements of both products will be different, but you know that scalar multiplication is commutative (i.e. order does not matter).
 
aaaa202 said:
Ivy, can you please help on the intuition for it?

The fundamental ideas are the we can multiply fractions by multiplying numerators and denominators separately and that multiplication is commutative. That is, we can rearrange the numerators and denominators as we please.
 

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