What “given” means
A conditional probability is the chance of an event once you know something else has happened. It is written P(A | B) and read “the probability of A given B”. The vertical bar is the word “given”.
The idea is that new information shrinks the list of possibilities. Instead of asking about every outcome in the sample space, you ask only about the outcomes consistent with what you know, and see what fraction of those satisfy A.
The standard formula is:
P(A | B) = P(A and B) / P(B)
It looks abstract, but in practice it simply says: of the cases where B is true, how many also have A?
A dice example
Roll two standard dice. What is the chance that the total is 10 or more?
There are 36 equally likely ordered outcomes. The totals 10, 11 and 12 can occur in 3, 2 and 1 ways, giving 6 outcomes, so the chance is 6/36 = 1/6.
Now suppose you are told the first die is a 6. Only 6 outcomes remain: (6,1) through (6,6). A total of 10 or more needs the second die to show 4, 5 or 6, which is 3 of those 6 outcomes. So P(total ≥ 10 | first die is 6) = 3/6 = 1/2.
Check with the formula: P(first die is 6 and total ≥ 10) = 3/36, and P(first die is 6) = 6/36, so 3/36 ÷ 6/36 = 1/2. The two routes agree.
The information changed the answer from 1/6 to 1/2 because the first die is part of the total.
When information does not matter
Now ask a different question: what is the chance the second die shows a 6, given the first showed a 6? It is still 1/6. The dice do not influence each other, so knowing one result tells you nothing about the other. In symbols, P(A | B) = P(A), which is the definition of independence.
This is exactly why believing a number is “due” after a streak is the gambler’s fallacy. The conditional probability of a fresh roll, given any history of earlier rolls, is unchanged.
Cards change when cards leave
Cards drawn from a deck and set aside are different. The deck’s composition changes, so later probabilities are conditional on what has already come out.
Illustration: a standard 52-card deck contains 4 aces. Draw one card.
- Chance the first card is an ace: 4/52 = 1/13.
- Given the first card was an ace, the chance the second is also an ace: 3/51, about 5.9%, because only 3 aces are left among 51 cards.
- Given the first card was not an ace, the chance the second is an ace: 4/51, about 7.8%, because all 4 aces remain among 51 cards.
The chance both are aces is 4/52 × 3/51 = 12/2652 = 1/221, about 0.45%.
A natural puzzle follows: if the second card’s chance depends on the first, what is the chance of an ace on the second card when you know nothing about the first? Combine the two cases, weighting each by how likely it is:
- First is an ace (4/52), then ace: 4/52 × 3/51 = 12/2652.
- First is not an ace (48/52), then ace: 48/52 × 4/51 = 192/2652.
- Total: 204/2652 = 1/13.
The answer is 1/13, the same as the first card. Without information, every position in a shuffled deck is equally likely to hold any given card. Conditions only change the picture once you see something.
Removing cards to shift odds
Seeing cards leave a deck is how information accumulates. Suppose that from a standard deck you are shown 3 of the 4 aces, and the rest of the cards are unseen. The remaining 49 cards include exactly 1 ace, so the chance the next card drawn from them is an ace is 1/49, about 2.0%, well below the original 1/13 (about 7.7%).
Conversely, if you have seen many low cards leave a deck, the remaining cards are richer in high ones. This is the arithmetic idea behind card counting in blackjack, which is discussed separately; the point here is only that the mathematics of conditional probability is straightforward, while any practical use is limited by casino rules, shuffling and the fact that casinos may refuse service. Counting never removes the built-in house edge entirely or guarantees an outcome.
A simple method for dependent draws
For a card question, try this routine:
- Write down the current number of cards and how many favourable ones remain.
- Compute the first probability as favourable ÷ total.
- Update both numbers according to what was drawn, assuming the first event happened.
- Compute the second probability from the updated counts.
- Multiply along the path.
If several paths lead to the same outcome, add their results. Drawing this as a diagram with branches makes it harder to lose track, which is the idea behind probability trees. The fact that totals across all branches must come to 1 gives a handy check.
Reversing a conditional
Be careful: P(A | B) is generally not equal to P(B | A). The chance that a card is an ace given that it is a heart is 1/13, while the chance that it is a heart given that it is an ace is 1/4. Confusing the two is a recurring error in reasoning about risk, and the fix involves Bayes theorem, which relates the two directions properly.
Key points
- Conditional probability narrows the sample space to what you know.
- Independent events leave conditional probabilities unchanged (dice, coin flips).
- Draws without replacement are dependent; each draw alters the next.
- With no information about earlier draws, each position in a shuffled deck has the same chance of holding a given card.
These ideas describe fair random processes. They do not offer a route to winning at games where the odds favour the operator, and gambling can lead to real harm. Support services exist locally for anyone who needs them.