# Is the probability of a chain of dependent events, independent of the order in which they occur?

So I have a bag of 6 balls: 3 red, 2 blue and 1 green.

Events:

A: I draw a red ball

B: I draw a blue ball

C: I draw a green ball

I do not replace the balls (thus, resulting in conditional probability).

Is the probability that I remove balls in the order: R, B, G, the same as the probability that I remove them in the order B, G, R? Or any order?

Intuitively, the probability of order R-B-G is $\frac{3}{6}.\frac{2}{5}.\frac{1}{4}$ and B-G-R gives $\frac{2}{6}.\frac{1}{5}.\frac{3}{4}$, both of which are $\frac{1}{20}$.

Thus, is p(A and-then B and-then C) = p(A and-then-C and-then B)? If so, why?

• "Intuitively.." I would rather call that a mathematical confirmation of intuition. Commented May 30, 2015 at 12:25

1. Order does not matter. You are right.

2. I think the rule for which you are looking is called the chain rule (of probability).

http://en.wikipedia.org/wiki/Chain_rule_%28probability%29

If you want to compute

$P(A_1, A_2, A_3)$, you can do so like this:

$$P(A_1, A_2, A_3) = P(A_3 | A_1, A_2) P(A_2 | A_1) P(A_1)$$

But since $P(A_1, A_2, A_3) = P(A_1, A_3, A_2)$, we can do:

$$P(A_1, A_2, A_3) = P(A_2 | A_1, A_3) P(A_1 | A_3) P(A_3)$$

Oh, when I say $P(A_1, A_2, A_3)$, I mean $P(A_1, A_2, A_3) = P(A_1 \cap A_2 \cap A_3)$

Edit: Um, $A_1$ can be removing the green ball, $A_2$ for red and $A_3$ for blue.

• Thank you, this is the answer that I was looking for :) Commented May 30, 2015 at 13:16
• @abhidivekar Here's a related question if you are interested math.stackexchange.com/questions/1283804/…
– BCLC
Commented May 30, 2015 at 13:20
• This answer is not really dealing with the order in wich the balls are removed. Commented May 30, 2015 at 13:29
• I am not talking about the question (yes, that is about the order in which the balls are removed), but about the answer. Commented May 30, 2015 at 14:59
• Yeah I got it thanks :) the chain rule link was very helpful. Commented May 30, 2015 at 16:41

Let $E$ denote the event that the first $3$ balls that are drawn have distinct color.

Do you understand that $P(BGR\mid E)=P(RBG\mid E)$?

(Here $BGR$ stands for the event that the first ball removed is blue, the second is green and the third is red.)

This because working under condition $E$ is the same as drawing $3$ balls from a bag containing exactly $3$ balls that have distinct color.

But here we have also $BGR\cap E=BGR$ and $RBG\cap E=RBG$ which leads to:$$P(BGR)=P(BGR\cap E)=P(BGR\mid E)P(E)=P(RBG\mid E)P(E)=P(RGB\cap E)=P(RBG)$$

I think your own explanation is already a mathematical proof that both probabilities are indeed equal. Your products of probabilities are correct and hence the result follows. Intuitively, this is due to the fact that you are only drawing $1$ ball per color. Things would be more complicated if you would consider the sequence RRB for example.

• Aside from the fact that the probabilities change, how would it become more complicated? It does not change the result if the order was swapped ; p(RRB) = (3/6)(2/5)(2/4) = 1/10; p(RBR) = (3/6)(2/5)(2/4) = 1/10 ; p(BRR) = (2/6).(3/5)(2/4) = 1/10; Commented May 30, 2015 at 12:28
• @abhidivekar I only meant that the computation would be somewhat more complicated: you have to add that the ordering of the red balls does not matter, which will give you a factor of two (for all three cases). I did not mean to say that the result would change. Commented May 30, 2015 at 15:21

Yes, for the mentioned problem the result is independent on the order. Let's say you have $n$ balls and their distribution by colors: $(n_1, n_2, n_3 = n-n_1-n_2)$. Then you extract $i$ balls and their distribution by colors: $(i_1, i_2, i_3 = i-i_1-i_2)$, then the answer:

$$\frac{i!}{n!}\cdot \frac{n_1!}{(n_1-i_1)!}\cdot \frac{n_2!}{(n_2-i_2)!}\cdot \frac{n_3!}{(n_3-i_3)!},$$ regardless of the order.

• But does it work for other problems? Is it a general result? Commented May 30, 2015 at 12:23

After a few years and learning more probability theory, I think the answer I was looking for is twofold:

1. If you assume three events happen together, then the probability is $$p(A, B, C)$$, which just means $$p(A \text{ and } B \text{ and } C)$$. Since "and" (conjunction) is independent of order, $$p(A \text{ and } B \text{ and } C) = p(B \text{ and } C \text{ and } A)$$, or any other permutation.
2. Another way to look at three events is in terms of information: what information does the occurrence of event A have on the probability of B? Subsequently, what does the occurrence of events (A and B) have on the probability of event C? If we assume that the chronological order of events was A, B, C, we can instead look at the conditional probability, i.e. $$p(A)p(B|A)p(C|B,A)$$. This view assumes that event A occurs first (causing us to update our probability that B occurs, given that A has occurred), then B occurs (causing us to update our probability that C occurs, given that A and B have occurred).

The chain rule just equates the above probabilities. It says that you can view a set of events in both ways: as occurring all at once, or one after the other, and you end up with the same probability value. This is quite convenient, as in certain problems, we might have different probability values available.

Another thing worth noting about the chain rule is that it is actually n! rules in one: since $$p(A \text{ and } B \text{ and } C) = p(B \text{ and } C \text{ and } A) = \dots$$, you are free to choose any permutation of events in the chain rule's RHS. You should pick the one which allows you to make use of the probability values available to you in your problem.

• Yes. And mathematically, not only does the order of a chain of dependent trials not affect the joint probability of their successes, neither does it matter whether the trials occur concurrently or in succession. Commented Sep 11, 2021 at 6:07