1
Count with the same tools
Permutations, combinations, complements and inclusion–exclusion. The denominator is a count too and is often the easier one.
Lesson · Discrete Probability · The model
Computing Probabilities
With equally likely outcomes, every probability is a ratio of two counts, so Combinatorics and Counting is the whole technique. Complementary counting is as useful here as it was there.
P(E) = |E| / |S|
P(Ē) = 1 − P(E) complement
"at least one" ⟹ 1 − P(none)
The hard part is the counting, not the division.
Two counts and a division
Core ideas
What this lesson establishes
Nothing new is needed beyond Combinatorics and Counting. What is new is that the answer must lie between 0 and 1, which is a free check.
2
The complement rule saves work
P(at least one) = 1 − P(none), exactly as in “Counting with Restrictions” in Combinatorics and Counting and for the same reason.
3
The answer is bounded
Any probability outside [0, 1] is a counting error, and that check costs nothing.
The material
Probability as a ratio
The rule, the complement, and three classic calculations.
Theorem · Equally likely outcomes
If S is finite and its outcomes are equally likely, then for any event E, P(E) = |E|/|S|.
Theorem · Complement rule
P(Ē) = 1 − P(E), because |Ē| = |S| − |E|.
Example · Poker: a flush
Five cards from 52: C(52,5) = 2 598 960 hands. A flush is five cards of one suit: choose the suit (4 ways) and five of its thirteen cards (C(13,5) = 1287), giving 5148. So P = 5148/2598960 ≈ 0.00198. (This includes straight flushes; excluding them subtracts 40.)
Example · At least one six
Roll four dice. P(at least one 6) = 1 − P(no 6) = 1 − (5/6)⁴ = 1 − 625/1296 = 671/1296 ≈ 0.518. Counting the ways to get at least one six directly requires inclusion and exclusion over the four dice; the complement is one line.
The answer most people give first is 4 × 1/6 = 2/3. That adds the four events "die i shows a 6", which overlap — the roll (6,6,1,2) is counted twice — and the same reasoning with six dice would give a probability of 1. Adding gives an upper bound, which “The Probability Axioms” names the union bound, and it gives the answer only when the events are disjoint.
The birthday problem
Theorem · Shared birthdays
With n people and 365 equally likely birthdays, P(some pair shares) = 1 − (365 · 364 ⋯ (365−n+1)) / 365ⁿ.
n P(a shared birthday)
10 0.117
20 0.411
23 0.507 ← passes one half
30 0.706
50 0.970
70 0.999
Twenty-three people suffice for a better-than-even chance, which surprises nearly everyone. The reason is that the question is about pairs: 23 people form C(23,2) = 253 pairs, and each has about a 1/365 chance of matching. The intuition that fails is comparing 23 with 365 rather than 253 with 365.
This is a different question from the pigeonhole statement in “The Pigeonhole Principle” in Sets, Relations, and Functions. That one said 367 people guarantee a match; this one says 23 make it likely. Certainty and probability are different claims and neither implies the other.
When outcomes are not equally likely
Then probabilities are assigned to outcomes directly, subject to being non-negative and summing to 1, and P(E) = Σ_{s ∈ E} P(s). A loaded die, a biased coin and a weighted lottery all work this way, and “The Probability Axioms” states the axioms that govern them.
Interactive
Sample space, event, condition
Exact fractions, every outcome listed
Method
Computing a probability
Denominator first; it is often easier and it anchors the model.
1
Count the sample space
This confirms what an outcome is. If |S| is hard to write down, the model is probably wrong.
2
Count the event
Using Combinatorics and Counting. Classify by order and repetition exactly as there.
3
Consider the complement
"At least one" and "not all" almost always count more easily the other way round.
4
Check the answer lies in [0,1]
And that it is plausible. A probability of 0.99 for a rare event is a counting error announcing itself.
Computing probabilities
Score 0 / 0Choose the answer you can justify.
Two aces
Five cards from a standard deck. Probability of exactly two aces.
Sample space: C(52,5) = 2 598 960 hands
Event: exactly two aces
choose 2 of the 4 aces: C(4,2) = 6
choose 3 of the 48 non-aces: C(48,3) = 17 296
product: 103 776
P = 103 776 / 2 598 960 ≈ 0.0399
CHECK by summing over the number of aces:
C(4,0)C(48,5) + C(4,1)C(48,4) + C(4,2)C(48,3)
+ C(4,3)C(48,2) + C(4,4)C(48,1)
= 1712304 + 778320 + 103776 + 4512 + 48 = 2 598 960 ✓
The check is Vandermonde's identity from “Binomial Coefficients and Pascal's Triangle” in Combinatorics and Counting, appearing as a probability check: the five disjoint cases must exhaust the hands. It would catch a wrong C(48,3) immediately.
Common mistakes
1
Counting the event with a different model from the spaceIf the denominator counts ordered outcomes, the numerator must too. Mixing ordered and unordered counts is the commonest arithmetic-looking error.
2
Counting "at least one" directlyIt produces overlapping cases. 1 − P(none) cannot double-count.
3
Confusing certainty with high probability367 people guarantee a shared birthday; 23 make it likely. Sets, Relations, and Functions's pigeonhole and this lesson answer different questions.
Check your understanding
✓
Finish when you check your answer against a complementary count.
Compute the probability that a five-card hand contains at least one ace, both directly (four cases) and by complement. The complement is one line and the agreement is the check.
Note: The birthday calculation assumes birthdays are uniform and independent, and neither is exactly true. Real birth data makes coincidences slightly MORE likely, so 23 remains a safe answer — but the assumptions should be stated rather than assumed away.