Conditional draws

Update the sample space after a reveal and multiply probabilities along a path.

10 minutes · intermediate · Original Poker4Quant lesson

What you will learn

Calculate joint events without replacement using changing conditional denominators.

Condition on what actually happened

The probability of a second event may depend on the first. P(A and B)=P(A)P(B given A). This multiplication rule does not require independence; it explicitly keeps the dependence in the conditional term.

After a heart is drawn from a standard deck and removed, 12 hearts remain among 51 cards. If instead the first card was a spade, 13 hearts remain among 51. Both histories reduce the denominator, but only one reduces the heart count.

A path is not the whole event

For two hearts, one path type is sufficient: heart then heart. For exactly one heart, sum heart-then-nonheart and nonheart-then-heart. Those paths are disjoint.

The experiment matters. Replacement restores the deck composition; no replacement does not. Additional information about how cards were selected can also change the conditional model. Do not assume an unseen opponent card has no bearing merely because its identity is hidden.

Worked example

What is the probability that two sequential uniform draws without replacement are both aces?

The first card is an ace with probability 4/52.
Given that outcome, 3 aces remain among 51 cards.
Multiply: (4/52)(3/51)=12/2652.

The result is 1/221, about 0.4525%. Multiplying 4/52 by itself would describe a different experiment.

Common mistakes

Keeping both numerator and denominator fixed without replacement.
Reducing the favorable count after an unfavorable card.
Forgetting the alternative ordering for exactly-one events.

Keep these ideas

Multiply by a conditional probability along each path.
Update counts according to the actual reveal.
Add disjoint paths when several orderings satisfy the event.