# Intuition on independence of two events.

The following question is taken from a textbook: Stanley H. Chan: Introduction to Probability for Data Science, 2021. (pp. 87-88)

Please note that this is not a question on how to calculate independence, I am fully aware of this. Instead, the intuition is what I am looking for in the below confusing question.

Consider the experiment of throwing a die twice. One should be clear from the context that the outcomes are in the form of a tuple $$(\textbf{dice_1}, \textbf{dice_2})$$ and the sample space is:

$$S = \left\{(1, 1), (1, 2), \ldots, (6, 6)\right\}$$

Define the three events below:

$$A = \{\textbf{1st dice is 3}\} \quad B = \{\textbf{sum of two die is 7}\} \quad C = \{\textbf{sum of two die is 8}\}$$

We want to find out if events $$A$$ and $$B$$ are independent? How about $$A$$ and $$C$$?

Now the answer is easy to get if you just use the formula and show that if $$P(A \cap B) = P(A)P(B)$$ and in a similar vein for event $$A$$ and $$C$$. However, I want to understand it more intuitively as my intuition failed me immediately, when I saw the question I thought that both should have similar answer since they are asked similarly.

We focus on the independence of $$A$$ and $$C$$ first. The author said that intuitively, given that event $$C$$ has happened, will this affect the probability of $$A$$ happening? I assume that this means we do have to know the probability of event $$A$$ without $$C$$ first.

We can enumerate and see that event $$A$$ has the following set representation:

$$A = \{(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)\}$$

which amounts to $$P(A) = \frac{6}{36} = \frac{1}{6}$$. Now if $$C$$ happened, we know that the two rolls have a sum of $$8$$, and we cannot construct a sum of $$8$$ with a roll of $$1$$. To me, I immediately know that event $$A$$ cannot have the outcome that has a $$1$$ in the second roll, and thus the outcomes should only be limited to $$5$$ instead of $$6$$ and hence dependence is established.

However, I believe somewhere my intuition is flawed, the author mentioned that:

If you like a more intuitive argument, you can imagine that C has happened, i.e., the sum is 8. Then the probability for the first die to be 1 is 0 because there is no way to construct 8 when the first die is 1. As a result, we have eliminated one choice for the first die, leaving only five options. Therefore, since C has influenced the probability of A, they are dependent.

I think I cannot understand why the author mentioned about "first die" when in event $$A$$, the first die is already a $$3$$. If we follow this line of logic, does this mean we do not actually need to know how event $$A$$ is defined? Is my interpretation wrong?

• P(A) = 1/6. Now to compute P(A|C), we dont have 6 choices for A as the choice 1 is now irrelevant. So of the five options, you want just one option (face 3). So P(A|C) = 1/5. You can also do this using Bayes, P(A|C) = P(C|A) P(A)/P(C) = 1/36/(5/36) = 1/5. The author is pointing to the fact that C occurring influences the total options of A.
– sku
Mar 15, 2022 at 7:35

Define the three events below: $$A = \{\textbf{1st dice is 3}\} \quad B = \{\textbf{sum of two die is 7}\} \quad C = \{\textbf{sum of two die is 8}\}$$

Neither your explanation nor the author's is correct.

• Let $$S$$ be the event that the first roll is smaller than $$10.$$

Then, by the author's reasoning:

event $$C$$ occurring affects the options for the first roll and consequently event $$S,$$ so events $$C$$ and $$S$$ are dependent.

This is of course false, since $$S$$ is a certain event and is thus independent of every other event associated with the experiment.

event $$B$$ occurring immediately eliminates the outcome $$(3,6),$$ so limits how event $$A$$ can eventuate, so events $$B$$ and $$A$$ are dependent.

This is false, since events $$P(B\cap A)=P(B)P(A).$$

When considering whether events $$X$$ and $$Y$$ are independent, the correct intuition is to ask whether knowing that $$X$$ happens changes the probability of $$Y$$—not whether it restricts how $$Y$$ can occur.

I wrote more here.

• Thanks for this, I will read in more details before accepting the answer! I've upvoted, however, for the clarity in the (counter) example you provided.
– nan
Mar 15, 2022 at 11:14
• When I ask whether C occurring affects the probability of A occurring, the answer is yes since the denominator (sample space) has been reduced from 36 choices to 30 choices, and thus affecting the probability of A. But if I think of it this way, then I am still narrowing down the configurations of A. It would be helpful if you can provide your intuitive take on why A and C are dependent.
– nan
Mar 15, 2022 at 11:37
• @nan $C$ occurring affects the configurations—and the number of ways—in which $A$ can occur, and also the probability of $A$ occurring (changing the latter from $\frac6{36}$ to $\frac15);$ it is this last point by which we conclude that events $A$ and $C$ are dependent on each other. Mar 15, 2022 at 11:44
• BTW, notice that when the author wrote "we have eliminated one choice for the first die, leaving only five options. Therefore, since C has influenced the probability of A", (I) what they meant to say was that $C$ has reduced the number of ways in which $A$ can occur, (II) they are reasoning wrongly, because, as I've pointed out, the dependence is indicated by the probability of $A$ being altered rather than by the possible configurations of $A$ being narrowed down. Mar 15, 2022 at 11:46
• @nan Oh, in case you don't mind more reading, the bottom section of this supplementary piece (Visualising independence of events) that I wrote gives a clearer picture of how reducing the sample space may or may not alter the probability of an event (i.e., its original probability may or may not equal its conditional probability). Mar 15, 2022 at 15:05