Textbook
For events $A$ and $B$ with $P(B)>0$, conditional probability is the probability of their intersection divided by the probability of the condition:
$$P(A\mid B):=\frac{P(A\cap B)}{P(B)}.$$
Event $B$ restricts the outcomes under consideration. Dividing by $P(B)$ renormalizes their weights so that the remaining space has total probability 1. The formula uses original probabilities and also works with unequal outcome weights.
For a fair die, event $A$ is a result greater than 3 and $B$ is an even result. Among $\{2,4,6\}$, only 4 and 6 qualify, so $P(A\mid B)=2/3$. Here $P(A)=1/2$. Also $P(B\mid A)=2/3$ in this particular example; that agreement is incidental, not a general rule.
A condition need not describe a cause. In general, $P(A\mid B)\ne P(B\mid A)$. When $P(B)=0$, this ratio is undefined; conditional distributions in more general settings require a separate definition.
The numerator must satisfy both event and condition: outcomes in $A$ but outside $B$ no longer contribute. Since $A\cap B\subseteq B$, the ratio stays between 0 and 1. In particular, $P(B\mid B)=1$, while incompatible $A$ and $B$ give conditional probability zero. Dividing by the condition’s probability restores total weight one; it does not automatically make the remaining outcomes equally likely.
Definitions and notation: Mathematics for Computer Science; Mathematics for Machine Learning.
Related concepts: Event probability in a finite model.
In plain language
Knowing that a die result is even narrows the discussion to its three even faces. Their original weights are rescaled so that the remaining total becomes 1 again. The shares are equal for a fair die, but filtering does not make unequal weights equal. The condition must have positive probability for this division rule to apply. The frame of the question changes, not the die retroactively; new information is not a face-repair service.
Example
In this constructed scene, a fair die is known to have shown an even number. The chance of a six is computed and compared with the reversed condition.
Conditional probability: reweighting after a condition
Before information
A six has probability 1/6 among all faces.
Step 1 of 4: Before information
Read all steps
Before information
A six has probability 1/6 among all faces.
- Faces
- 123456current
Even faces remain
The condition retains only 2, 4, and 6.
- Condition B
- 1excluded2included3excluded4included5excluded6included
Weight renormalization
The weight 1/6 is divided by the probability 1/2 of an even roll.
- Faces in B
- 246included
- New weight of six
- 1/3current
The reversed condition
If the result is known to be six, evenness is certain.
- Remaining outcome
- 6included
- Probability of evenness
- 1included
After learning “even,” the surviving faces have original total weight $3/6=1/2$. The six occupies $(1/6)/(1/2)=1/3$ of that weight. Reversing the condition leaves only the six, which is entirely even, so the answer is 1. Different denominators explain the different answers: the questions use the same intersection but measure it against different conditioned spaces.
The vertical bar identifies the information already known. Reversing its sides changes the problem.