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Probability Calculator — Combos & Perms

Calculate probability, combinations, and permutations

At Least Once

99.90%

P(A)

50.00%

Expected

5.00

Results

Probability P(A)
50.00%
Complement P(A')
50.00%
At Least Once in 10 Trials
99.90%
Never in 10 Trials
0.10%
Expected Occurrences
5.00
out of 10 trials

Visual Breakdown

Success Probability50%
Failure Probability50%
At Least Once in 10 Trials100%

Formulas Used

Combinations

C(n,r) = n! / (r! × (n-r)!)

Count the number of ways to choose r items from n items where order does not matter.

Where:

n= Total number of items
r= Number of items chosen

Permutations

P(n,r) = n! / (n-r)!

Count the number of ordered arrangements of r items chosen from n items.

Where:

n= Total number of items
r= Number of items chosen

Complement Rule

P(at least one) = 1 - (1 - p)^n

Calculate the probability of an event occurring at least once over multiple independent trials.

Where:

p= Probability of the event in a single trial
n= Number of independent trials

Example Calculations

1Coin Flip - At Least One Heads in 5 Flips

Inputs

Probability0.5
Trials5

Result

At Least Once96.88%
Complement50.00%
Never in 5 Trials3.13%
Expected Occurrences2.50

P(at least one heads) = 1 - (1 - 0.5)^5 = 1 - 0.03125 = 0.96875 = 96.88%.

2Lottery: Choose 6 from 49

Inputs

n (total)49
r (chosen)6

Result

Combinations C(49,6)13,983,816
Permutations P(49,6)10,068,347,520

C(49,6) = 49! / (6! × 43!) = 13,983,816. This means there are nearly 14 million possible lottery combinations.

Frequently Asked Questions

Q

What is the difference between combinations and permutations?

Combinations count the number of ways to choose items where order does not matter (e.g., picking 3 team members from 10 people). Permutations count arrangements where order matters (e.g., assigning 1st, 2nd, 3rd place). C(10,3) = 120 but P(10,3) = 720, because each combination of 3 can be arranged in 6 different orders.

  • C(10,3) = 120 — choosing 3 people from 10 for a committee (order irrelevant)
  • P(10,3) = 720 — assigning president, VP, secretary from 10 people (order matters)
  • P(n,r) = C(n,r) × r! — each combination has r! arrangements
  • Lottery "pick 6 from 49": C(49,6) = 13,983,816 possible tickets
  • A 4-digit PIN (digits 0–9): P(10,4) = 5,040 if no repeats; 10⁴ = 10,000 with repeats
ScenarioTypeFormulaResult
3 from 10 (committee)CombinationC(10,3)120
3 from 10 (ranked)PermutationP(10,3)720
5 from 52 (poker hand)CombinationC(52,5)2,598,960
6 from 49 (lottery)CombinationC(49,6)13,983,816
Q

How do I calculate the probability of at least one success?

Use the complement rule: P(at least one) = 1 - P(none). For independent events with probability p over n trials: P(at least once) = 1 - (1-p)^n. For example, the chance of rolling at least one 6 in 4 rolls is 1 - (5/6)^4 = 51.77%.

  • Rolling at least one 6 in 4 dice rolls: 1 – (5/6)⁴ = 51.77%
  • Flipping at least one heads in 5 coin flips: 1 – (0.5)⁵ = 96.88%
  • At least one defective in 20 items (2% defect rate): 1 – (0.98)²⁰ = 33.24%
  • At least one birthday match in a group of 23 people: ~50.73% (birthday paradox)
  • This shortcut avoids summing P(1) + P(2) + P(3) + ... individually
Q

What is a complement in probability?

The complement of an event A, written P(A'), is the probability that event A does NOT occur. It equals 1 - P(A). If the probability of rain is 0.30 (30%), the complement (no rain) is 0.70 (70%). The sum of an event and its complement always equals 1.

  • P(rain) = 0.30 means P(no rain) = 0.70 — they always sum to 1.0
  • P(passing an exam) = 0.85 means P(failing) = 0.15
  • Often easier to calculate what you do NOT want, then subtract from 1
  • Example: probability of NOT rolling a 6 on a die = 5/6 ≈ 83.33%
Q

What does C(n,r) mean?

C(n,r) means 'n choose r' — the number of ways to choose r items from n items without regard to order. The formula is C(n,r) = n! / (r! × (n-r)!). For example, C(5,2) = 5! / (2! × 3!) = 120 / (2 × 6) = 10.

  • C(5,2) = 10 — there are 10 ways to pick 2 items from 5
  • C(n,0) = 1 and C(n,n) = 1 — there is always exactly 1 way to choose nothing or everything
  • C(n,r) = C(n, n–r) — choosing 3 from 10 is the same count as choosing 7 from 10
  • Excel/Sheets: use =COMBIN(n, r) to calculate directly
Q

When would I use permutations vs combinations?

Use permutations when the order of selection matters: arranging books on a shelf, assigning ranked positions, creating passwords. Use combinations when order does not matter: choosing team members, selecting lottery numbers, picking items from a menu.

  • Permutation examples: race finishing order, locker combination codes, seating arrangements
  • Combination examples: pizza toppings from a menu, raffle winners, card hands
  • Passwords are permutations — "ABC" and "CBA" are different passwords
  • Team selection is a combination — {Alice, Bob} = {Bob, Alice} is the same team
  • When in doubt: ask "Does rearranging the selection create a different outcome?"

Understanding Probability, Combinations, and Permutations

1

Combinations vs. Permutations: When Order Matters

C(52,5) = 2,598,960 possible poker hands but P(52,5) = 311,875,200 possible ordered draws — 120× more, because each hand of 5 cards can be arranged in 5! = 120 different orders. This 120× factor is the core difference: combinations ignore arrangement, permutations count it.

The formula C(n,r) = n! / (r! × (n−r)!) divides out the duplicate orderings. For a 6/49 lottery, C(49,6) = 13,983,816 possible tickets. At $2 per ticket, buying every combination costs $27,967,632 — which actually happened in the 1992 Virginia lottery when an Australian syndicate attempted to purchase all combinations for a $27 million jackpot.

Permutations P(n,r) = n! / (n−r)! count ordered arrangements. A 4-digit PIN using digits 0–9 without repeats: P(10,4) = 5,040 possibilities. With repeats allowed: 10⁴ = 10,000 possibilities. The Standard Deviation Calculator uses combinatorial formulas internally when computing binomial distribution parameters.

Combinations vs. PermutationsC(5,3) = 10Order does NOT matter{A,B,C} = {C,B,A}= {B,A,C} (same group)P(5,3) = 60Order DOES matterABC ≠ CBA ≠ BAC(3! = 6 arrangements each)P(n,r) = C(n,r) × r!60 = 10 × 6Each combination of 3 items can bearranged in 3! = 6 different ordersCombinations (groups)Permutations (arrangements)
Permutations always produce larger counts than combinations for the same n and r
ScenarioFormulaOrder Matters?Result
Poker hand (5 from 52)C(52,5)No2,598,960
Lottery (6 from 49)C(49,6)No13,983,816
Race podium (3 from 10)P(10,3)Yes720
4-digit PIN (no repeats)P(10,4)Yes5,040
2

The Complement Rule and "At Least One" Problems

P(at least one) = 1 − P(none) is the most powerful shortcut in applied probability. Rolling at least one 6 in 4 dice rolls: 1 − (5/6)⁴ = 1 − 0.4823 = 51.77%. Calculating this directly would require summing P(exactly 1) + P(exactly 2) + P(exactly 3) + P(exactly 4) — four separate binomial terms instead of one subtraction.

The birthday paradox illustrates why the complement rule is counterintuitive. In a group of 23 people, P(at least one shared birthday) = 1 − P(all different) = 1 − (365/365 × 364/365 × ... × 343/365) ≈ 50.73%. Most people guess you need about 183 people for a 50% chance, but the quadratic growth of pairwise comparisons (23 people = 253 pairs) makes collisions likely much sooner.

In quality control, the complement rule determines defect probability. If a manufacturing line has a 2% defect rate, the chance of at least one defective item in a batch of 50 is 1 − (0.98)⁵⁰ = 63.58%. This is why 100% inspection is standard for safety-critical components. The Z-Score Calculator extends these probability concepts to continuous normal distributions.

P(at least one) = 1 − P(none)Dice: At Least One 64 rolls, p = 1/61 − (5/6)⁴ = 51.77%Coins: At Least 1 Head5 flips, p = 0.51 − (0.5)⁵ = 96.88%Birthday Paradox23 people, 365 daysP(match) ≈ 50.73%Quality Control50 items, 2% defect1 − (0.98)⁵⁰ = 63.58%Complement = P(none) subtracted from 1
3

Practical Applications of Counting Formulas

Password security depends directly on permutation counts. An 8-character password using lowercase letters (26) has 26⁸ = 208.8 billion combinations. Adding uppercase doubles the alphabet to 52, giving 52⁸ = 53.5 trillion — a 256× increase. Adding digits (62 chars) reaches 218.3 trillion, and special characters (95 printable ASCII) reaches 6.63 quadrillion. At 10 billion guesses per second, brute-forcing the 95-character version takes 7.7 days versus 0.003 seconds for 26-character lowercase.

Poker probability uses combinations heavily. The probability of a royal flush is C(4,1)/C(52,5) = 4/2,598,960 = 1 in 649,740. A full house: C(13,1)×C(4,3)×C(12,1)×C(4,2) / C(52,5) = 3,744/2,598,960 = 1 in 694. Knowing these probabilities is the mathematical foundation of optimal poker strategy.

In genetics, combination formulas calculate genotype probabilities. If both parents carry one copy of a recessive allele (Aa), the probability of the child being aa (affected) is C(2,2)×(0.5)² = 0.25 = 25%. For 4 children, P(at least one affected) = 1 − (0.75)⁴ = 68.36%. Genetic counselors use these exact calculations when advising families about inherited conditions.

  • Lottery: C(49,6) = 13,983,816 tickets × $2 = $28M to buy all combinations
  • Poker: Royal flush probability = 0.000154% (1 in 649,740 hands)
  • Passwords: Each additional character multiplies possibilities by alphabet size
  • Genetics: P(recessive trait) = 0.25 per child from two carriers
  • Survey sampling: C(1000,50) ways to select a sample from a population

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Last Updated: Mar 26, 2026

This calculator is provided for informational and educational purposes only. Results are estimates and should not be considered professional financial, medical, legal, or other advice. Always consult a qualified professional before making important decisions. UseCalcPro is not responsible for any actions taken based on calculator results.

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