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?

  • $\begingroup$ 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. $\endgroup$
    – sku
    Mar 15, 2022 at 7:35

1 Answer 1


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.

  • By your reasoning:

     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.

  • $\begingroup$ 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. $\endgroup$
    – nan
    Mar 15, 2022 at 11:14
  • $\begingroup$ 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. $\endgroup$
    – nan
    Mar 15, 2022 at 11:37
  • 1
    $\begingroup$ @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. $\endgroup$
    – ryang
    Mar 15, 2022 at 11:44
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    $\begingroup$ 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. $\endgroup$
    – ryang
    Mar 15, 2022 at 11:46
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    $\begingroup$ @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). $\endgroup$
    – ryang
    Mar 15, 2022 at 15:05

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