Explain dependency between probability event

Question

Say I have 7 cards that are numbered from 1 to 7.

$D = \{1,2,3,4,5,6,7\}$

and events A and B:

$A =:$ getting an even number $\rightarrow \{2,4,6\}$

$B =:$ getting a number bigger than $4 \rightarrow \{5,6,7\}$

Then it follows:

$P(A) = 3/7 \hspace{0.5cm} \text{and} \hspace{0.5cm} P(B) = 3/7$

$P(A\cap B) = 1/7 \hspace{0.5cm} (\text{number 6})$

$P(A) \cdot P(B) = 9/49$

One of the rules for checking events dependency is

$P(A\cap B)= P(A)\cdot P(B) \rightarrow$ independent events

$P(A\cap B)\neq P(A)\cdot P(B) \rightarrow$ dependent events

From the example above

$P(A\cap B)\neq P(A)\cdot P(B)$

meaning that we are dealing with dependent events. However, I can not conceptually understand how these events are actually dependent, i.e. we can get an even number without throwing getting a number higher that 4. Can somebody help explaining?

This example expands from a 6-sided dice scenario. Events A and B are the same, but in this case $P(A\cap B) = P(A)\cdot P(B)$

Answer 1

Two events are independent if information that one has happened tells you nothing new about the probability of the other.

In this case, if you know that the number is greater than $4$ the probability that is is even changes from $3/7$ to $1/3$, so those two events are not independent.

(For an ordinary die even has probability $1/2$ whether or not you know that the number is greater than $4$.)

The argument is symmetrical. If you know that the number is even then the probability that it is greater than $4$ changes from $3/7$ to $1/3$.

The fact that "even" and "greater than $4$" are equally likely is a coincidence that has nothing to do with independence.

Answer 2

As to a more mathematical exploration, start with the Bayes theorem result:

$P(AB) = P(A|B)P(B)$

Or, equivalently:

$P(AB) = P(B|A)P(A)$

If you multiply both equations together:

$P(AB)P(AB) = P(A|B)P(B)P(B|A)P(A)$

Now, if the case of independent events, it is known that:

$P(AB) = P(A)P(B)$

If we substitute this result in the equation above and cancel terms:

$P(AB) = P(A|B)P(B|A)$ which equals $P(A)P(B)$

which implies for independent events that:

$P(A) = P(A|B)$ and $P(B) = P(B|A)$

Or, in other words, for ascribed independent events A and B, they do not present any new information content on each other.

This clearly not the case in this problem.