Independence intuition

Solution 1:

How do you think '$4$ on the first die' affects the chances of getting a total of $7$? Do you think it increases it, or decreases it? Let's do the math.

$6$ on the first die: we need $1$ on the second die. One chance in $6$.
$5$ on the first die: we need $2$ on the second die. One chance in $6$.
$4$ on the first die: we need $3$ on the second die. One chance in $6$.
$3$ on the first die: we need $4$ on the second die. One chance in $6$.
$2$ on the first die: we need $5$ on the second die. One chance in $6$.
$1$ on the first die: we need $6$ on the second die. One chance in $6$.

So the chance of getting a total of $7$ is the same regardless of what comes up on the first die. Knowing the first die came up $4$ does not make the chances of $7$ better or worse. Events $A$ and $B$ and independent.

Now what if we want a total of $8$? How is that different?

$6$ on the first die: we need $2$ on the second die. One chance in $6$.
$5$ on the first die: we need $3$ on the second die. One chance in $6$.
$4$ on the first die: we need $4$ on the second die. One chance in $6$.
$3$ on the first die: we need $5$ on the second die. One chance in $6$.
$2$ on the first die: we need $6$ on the second die. One chance in $6$.
$1$ on the first die: we need $7$ on the second die. ZERO CHANCE.

So '$1$ on the first die' lowers the chances of 'total of $8$' (to zero); anything else raises the chances to $1/6$. In particular, events $A$ and $C$ are dependent. The unconditional (i.e. before we know the result for the first die) probability of 'total of $8$' is the average of $\{1/6,\ 1/6,\ 1/6,\ 1/6,\ 1/6,\ 0\}$ which is $5/36.$

Looking at it the other way round, given that the total is $8$, we know that we can't have $1$ on the first die, so the conditional probability of $A$ given $C$ is not $1/6$ but $1/5$.

Solution 2:

Since we are using fair dice, the atomic outcomes have equal probability measure.

$\begin{array}{l} A = \{(4,1), (4,2), (4,3), (4,4), (4,5), (4, 6)\} \\ B = \{(1,6), (2,5), (3,4), \color{blue}{(4,3)}, (5,2), (6,1)\}, & A\cap B=\{(4,3)\} \\ C = \{(2,6), (3,5), \color{blue}{(4,4)}, (5,3), (6,2)\}, & A\cap C = \{(4,4)\} \end{array} \\[2ex]\;\\ \begin{array}{l} \mathsf P( A) = 1/6, &\mathsf P(B) = 1/6, &\mathsf P(A\cap B)=1/36, &\mathsf P(A\mid B) = 1/6 , &\mathsf P(B\mid A) = 1/6 \\ &\mathsf P(C) = 5/36, &\mathsf P(A\cap C)=1/36 , &\mathsf P(A\mid C) = 1/5, &\mathsf P(C\mid A) = 1/6 \end{array} $

The independence of $A$ and $B$ is because the proportion of outcomes for $A$ that occur within the space of $B$ is the same as the proportion of outcomes for $A$ within the total space.   Likewise the proportion of outcomes for $B$ that occur within the space of $A$ is the same as the proportion of outcomes for $B$ within the total space.

However, this is not so for the proportion of outcomes for $A$ which occur within the space of $C$.

NB: if the dice we biased we would have to consider the weight of each atomic outcome, rather than simply counting.

Solution 3:

Here's an analogous example from my previous post:

Pairwise independence can be characterised as:

Let $P(A)\neq0.$

Events $A$ and $B$ are independent iff. the probability of $B$ is unaffected by the knowledge that $A$ occurs.

When considering whether events $A$ and $B$ are independent, the correct intuition is to ask whether $A$ occurring affects the probability of $B$ occurring—not whether $A$ occurring affects which configurations $B$ can occur in.

Consider this experiment: flip two fair coins, letting $H_1$ be the event that the first coin lands on Heads, and $X$ be the event that the coins land on different sides. Then $$ P\left(H_1 \cap X\right)=\frac14=P(H_1)\:P(X);$$ i.e., $H_1$ and $X$ are, by definition, independent events.

Yet, $H_1$ occurring has affected (narrowed down) the possible configurations that $X$ can occur in: one of $X$'s outcomes, $TH,$ is no longer possible.

Solution 4:

$A$ effects the probability of $B$... and leaves it untouched. Getting 4 on the first die excludes many cases of $A$ and $A^c$, but does so evenly, so $P(B|A)=1/6=6/36=P(B)$.