🔄 Symmetry Probability

Using symmetry principles to find probabilities without explicit counting, especially for random choices.

🎯 Recognition Cues

  • Keywords: “random”, “symmetry”, “equally likely”, “uniform”, “fair”
  • Setup: Random selection or arrangement with symmetry
  • Constraints: All outcomes equally likely, symmetric situations

đź“‹ Solution Template

  1. Identify the symmetry:

    • Equal probability: All outcomes equally likely
    • Symmetric positions: Equivalent positions in arrangement
    • Symmetric choices: Equivalent choices in selection

  2. Apply symmetry principles:

    • Equal probability: $P(\text{event}) = \frac{\text{favorable outcomes}}{\text{total outcomes}}$
    • Symmetric positions: Use relative positions
    • Symmetric choices: Use relative probabilities

  3. Check for common patterns:

    • Random permutations
    • Random selections
    • Random arrangements

đź’ˇ Micro-Examples

Equal Probability

  • Problem: What’s the probability that a random 3-digit number has its digits in increasing order?
  • Solution: $\frac{1}{3!} = \frac{1}{6}$ (only 1 out of 6 permutations is increasing)

Symmetric Positions

  • Problem: What’s the probability that a random person in a line of 10 people is in the middle position?
  • Solution: $\frac{1}{10}$ (all positions equally likely)

Random Permutations

  • Problem: What’s the probability that a random permutation of ${1,2,3,4,5}$ has 1 in the first position?
  • Solution: $\frac{1}{5}$ (all positions equally likely for 1)

⚠️ Common Pitfalls & Variants

Pitfall: Assuming symmetry when it doesn’t exist

  • Wrong: “Random 3-digit number” = All numbers equally likely
  • Right: Check if the selection process is truly uniform.

Pitfall: Forgetting to account for all outcomes

  • Wrong: “Probability of increasing digits” = $\frac{1}{2}$ (only considering increasing vs. decreasing)
  • Right: $\frac{1}{3!} = \frac{1}{6}$ (considering all permutations)

Pitfall: Misapplying symmetry in non-symmetric situations

  • Wrong: “Probability of heads” = $\frac{1}{2}$ (always true for fair coins)
  • Right: This is actually correct! But make sure the symmetry assumption is valid.

🏆 AMC-Style Worked Example

Problem: What’s the probability that a random permutation of ${1,2,3,4,5}$ has exactly 2 fixed points?

Solution:

  • Step 1: Total permutations = $5! = 120$
  • Step 2: Choose 2 positions to be fixed: $\binom{5}{2} = 10$ ways
  • Step 3: The remaining 3 elements must form a derangement
  • Step 4: Number of derangements of 3 elements = $!3 = 2$
  • Step 5: Total ways = $10 \cdot 2 = 20$
  • Step 6: Probability = $\frac{20}{120} = \frac{1}{6}$

Key insight: Use symmetry to avoid casework on which specific elements are fixed!

đź”— Related Topics

Next: Hypergeo vs Binomial → Digit Counting