Probability Theory

Sample Space With Equally Likely Outcomes

Samir Orujov, PhD

ADA University, School of Business

Information Communication Technologies Agency, Statistics Unit

2025-10-01

Learning Objectives

🎯 Equally Likely Outcomes

Understand the concept of equiprobable sample spaces

📊 Classical Probability

Apply the classical definition of probability

🎲 Dice & Card Problems

Solve probability problems involving dice and cards

🔄 Combinatorics

Use combinations and permutations in probability

📐 Complex Applications

Master poker hands, birthday problems, and matching problems

🎪 Interactive Demos

Visualize probability concepts through animations

Course Overview

Core Concepts

🎯 Equally Likely Outcomes

Classical Problems

🎲 Classical Examples

Card Games

🃏 Poker Hands

Paradoxes

🎂 Birthday Problem

Advanced Topics

🧩 Complex Applications

Tools

💻 Interactive Demos

What if All Outcomes Are Equally Likely?

Fundamental Question

What if we assume that all outcomes of an experiment are equally likely to occur?

Example 1: Consider an experiment whose sample space \(S\) is a finite set: \[S = \{1, 2, \ldots, N\}\]

If we assume: \[P(\{1\}) = P(\{2\}) = \cdots = P(\{N\})\]

Then by Axioms 2 and 3, we get: \[P(\{i\}) = \frac{1}{N} \quad \text{for } i = 1, 2, \ldots, N\]

Classical Definition of Probability

If all outcomes of an experiment are equally likely to occur, then:

\[P(E) = \frac{\text{number of outcomes in } E}{\text{number of outcomes in } S}\]

where \(E\) is any event and \(S\) is the sample space.

This is the classical or equiprobable definition of probability.

Interactive Demo: Coin Flips

Demonstration: Let’s see how probability stabilizes with equally likely outcomes

Click to flip coins and observe how probability approaches 0.5:

Interactive Demo: Coin Flips with Chart

Demonstration: Let’s visualize how probability converges with a live chart

Manual Mode:

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Think-Pair-Share Activity 1

Dice Rolling Experiment

Question: Two dice are rolled. What is the probability that the sum of the upturned faces will equal 7?

Think Time: 2 minutes
Question: How did you count the favorable outcomes?

Solution:

  • Sample space: \(S = \{(i,j) : i,j = 1,2,3,4,5,6\}\), so \(|S| = 36\)

  • Event \(E = \{(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)\}\), so \(|E| = 6\)

  • Therefore: \(P(E) = \frac{6}{36} = \frac{1}{6}\)

Interactive Dice Simulation

Simulation: Roll two dice and observe sum frequency distribution

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Example: Ball Drawing Problem

Example 3: If 3 balls are “randomly drawn” from a bowl containing 6 white and 5 black balls, what is the probability that one ball is white and the other two are black?

Solution:

  • Total ways to choose 3 balls from 11: \(\binom{11}{3} = \frac{11!}{3! \cdot 8!} = 165\)

  • Ways to choose 1 white from 6 and 2 black from 5: \(\binom{6}{1} \times \binom{5}{2} = 6 \times 10 = 60\)

  • Therefore: \(P(\text{1 white, 2 black}) = \frac{60}{165} = \frac{4}{11}\)

Interactive Ball Drawing Demo

Simulation: Draw 3 balls from 6 white + 5 black balls and observe probability convergence

Manual Mode:

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Unrelated Couples Problem

Example 4: Suppose 5 people are randomly selected from a group of 20 individuals consisting of 10 married couples. What is the probability that all 5 chosen are unrelated (no two are married to each other)?

Solution:

  • Total ways to choose 5 from 20: \(\binom{20}{5}\)

  • Ways to choose 5 unrelated people: Choose 5 couples × Choose 1 person from each couple

  • \(\binom{10}{5} \times 2^5 = 252 \times 32 = 8,064\)

  • Therefore: \(P(\text{all unrelated}) = \frac{8,064}{\binom{20}{5}} = \frac{8,064}{15,504} = \frac{8}{15}\)

Interactive Demo: Married Couples Selection

Demonstration: Visualize how selecting unrelated people affects probability

Manual Selection:

Auto Simulation:

20 People (10 Married Couples)

Key Insights from the Animation

The animation helps build intuition for why the probability is 8/15:

  1. Total Possibilities: There are \(\binom{20}{5} = 15,504\) ways to choose 5 people from 20
  2. Favorable Outcomes: To get all unrelated people, we must:
    • Choose 5 different couples: \(\binom{10}{5} = 252\) ways
    • Pick one person from each chosen couple: \(2^5 = 32\) ways
    • Total favorable: \(252 \times 32 = 8,064\)
  3. Final Probability: \[\frac{8,064}{15,504} = \frac{8}{15} \approx 0.5333\]

Homework Problems

Example 5 (HW): A committee of 5 is to be selected from a group of 6 men and 9 women. If the selection is made randomly, what is the probability that the committee consists of 3 men and 2 women?

Hint: Use combinations to count favorable outcomes.

Example 6 (HW): An urn contains n balls, one of which is special. If k of these balls are withdrawn one at a time, with each selection being equally likely to be any of the balls that remain at the time, what is the probability that the special ball is chosen?

Hint: Think about the symmetry of the problem.

Poker Hands: Straight

Example 8: A poker hand consists of 5 cards. If the cards have distinct consecutive values and are not all of the same suit, we say that the hand is a straight. What is the probability that one is dealt a straight?

Solution: - Total possible hands: \(\binom{52}{5} = 2,598,960\) - Number of straights: 10 possible sequences (A-2-3-4-5 through 10-J-Q-K-A) - Each sequence: \(4^5\) ways to choose suits minus 4 straight flushes = \(1024 - 4 = 1020\) - Total straights: \(10 \times 1020 = 10,200\)

  • Therefore: \[P(\text{straight}) = \frac{10,200}{2,598,960} \approx 0.0039\]

Homework: Full House

Example 9 (HW): A 5-card poker hand is said to be a full house if it consists of 3 cards of the same denomination and 2 other cards of the same denomination (different from the first). What is the probability that one is dealt a full house?

Hint: - Choose the denomination for the triple: 13 ways - Choose 3 cards from that denomination: \(\binom{4}{3}\) ways
- Choose the denomination for the pair: 12 ways - Choose 2 cards from that denomination: \(\binom{4}{2}\) ways

Bridge Problem

Example 10 (HW): In the game of bridge, the entire deck of 52 cards is dealt out to 4 players. What is the probability that:

  1. One of the players receives all 13 spades?
  2. Each player receives 1 ace?

Think-Pair-Share: Discuss the difference in complexity between parts (a) and (b).

Think-Pair-Share Activity

Prompt: Which part seems harder to calculate and why?

Time: 3 minutes to think, 5 minutes to discuss

The Birthday Problem

Example 11: If n people are present in a room, what is the probability that no two of them celebrate their birthday on the same day of the year? How large need n be so that this probability is less than 1/2?

Solution: - Probability of no shared birthdays: \(P(n) = \frac{365!}{(365-n)! \cdot 365^n}\) - Equivalently: \(P(n) = \frac{365}{365} \times \frac{364}{365} \times \frac{363}{365} \times \cdots \times \frac{365-n+1}{365}\) - For \(n = 23\): \(P(23) \approx 0.493 < 0.5\)

Interactive Birthday Simulation

Simulation: Test the birthday paradox with random groups

Football Team Roommates

Example 13: A football team consists of 20 offensive and 20 defensive players. The players are to be paired in groups of 2 for roommates. If the pairing is done at random, what is the probability that there are no offensive-defensive roommate pairs?

Solution: This is equivalent to a perfect matching problem. - Total ways to pair 40 players: \(\frac{40!}{2^{20} \times 20!}\) - Ways with no offensive-defensive pairs: \(\frac{20!}{2^{10} \times 10!} \times \frac{20!}{2^{10} \times 10!}\)

Note

Stirling Approximation: For large n: \(n! \approx \sqrt{2\pi n} \left(\frac{n}{e}\right)^n\)

The Matching Problem

Example 15 (The matching problem): Suppose that each of N men at a party throws his hat into the center of the room. The hats are first mixed up, and then each man randomly selects a hat. What is the probability that none of the men selects his own hat?

Solution: This is the famous derangement problem. - Total permutations: \(N!\) - Number of derangements: \(D_N = N! \sum_{k=0}^{N} \frac{(-1)^k}{k!}\) - For large N: \(P(\text{no matches}) \approx \frac{1}{e} \approx 0.368\)

Interactive Matching Demo

Simulation: Test the matching problem with different group sizes

Sequences and Limits

Definition 2: A sequence of events \(\{E_n, n \geq 1\}\) is an increasing sequence if: \[E_1 \subset E_2 \subset E_3 \subset \ldots \subset E_n \subset E_{n+1} \subset \ldots\]

Definition 3: A sequence of events \(\{E_n, n \geq 1\}\) is a decreasing sequence if: \[E_1 \supset E_2 \supset E_3 \supset \ldots \supset E_n \supset E_{n+1} \supset \ldots\]

Example 18: Consider \(E_n = \{z \in \mathbb{Z} | 2^n \text{ divides } z\}\). Then \(\{E_n, n \geq 1\}\) is a decreasing sequence.

Limits of Sequences

Definition 4: If \(\{E_n, n \geq 1\}\) is an increasing sequence, then: \[\lim_{n \to \infty} E_n := \bigcup_{i=1}^{\infty} E_i\]

Definition 5: If \(\{E_n, n \geq 1\}\) is a decreasing sequence, then: \[\lim_{n \to \infty} E_n := \bigcap_{i=1}^{\infty} E_i\]

Continuity of Probability

Proposition 2 (Continuity of Probability)

If \(\{E_n, n \geq 1\}\) is either an increasing or decreasing sequence of events, then:

\[\lim_{n \to \infty} P(E_n) = P\left(\lim_{n \to \infty} E_n\right)\]

Remark 1: If \(\{E_n, n \geq 1\}\) is decreasing, then \(\{E_n^c, n \geq 1\}\) is increasing.

The Infinite Ball Paradox

Example 19: We have an infinitely large box and infinite balls numbered 1, 2, 3, …

Process: - Put balls 1-10 in box, remove ball 10 - Put balls 11-20 in box, remove ball 20
- Continue indefinitely…

Question: How many balls are in the box at the end?

Think-Pair-Share Activity

Prompt: What’s your intuition? Will there be infinitely many balls, or some finite number, or zero balls?

Time: 5 minutes to discuss with a partner

More Infinite Ball Problems

Example 20 (HW): Same setup, but now remove ball 1 in the first step, ball 2 in the second step, etc. How many balls remain?

Example 21 (HW): Same setup, but whenever a ball is withdrawn, it’s randomly selected from those present. How many balls remain on average?

These problems illustrate the subtleties of infinite processes and limiting behavior!

Group Problem-Solving Activity

Sports Club Problem

Example 14: A total of 36 members play tennis, 28 play squash, and 18 play badminton. Furthermore, 22 play both tennis and squash, 12 play both tennis and badminton, 9 play both squash and badminton, and 4 play all three sports.

Tasks (15 minutes): 1. Draw a Venn diagram 2. How many play at least one sport? 3. How many play exactly two sports? 4. What’s the probability a randomly selected member plays tennis given they play squash?

Groups: 4-5 students each

Interactive Venn Diagram Tool

Sports Club Analysis: Tennis (T), Squash (S), Badminton (B)

T
S
B
6
1
1
18
8
5
4
T only:6 S only:1 B only:1 All:4
T∩S:18 T∩B:8 S∩B:5

Goal: Understand |T ∪ S ∪ B| = 43 members play at least one sport

Key Formulas Summary

Concept Formula Application
Classical Probability \(P(E) = \frac{|E|}{|S|}\) Equally likely outcomes
Combinations \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\) Selecting k from n
Permutations \(P(n,k) = \frac{n!}{(n-k)!}\) Arranging k from n
Birthday Problem \(P(n) = \prod_{i=0}^{n-1} \frac{365-i}{365}\) No shared birthdays
Derangements \(D_n = n! \sum_{k=0}^{n} \frac{(-1)^k}{k!}\) No matches
Stirling’s Approximation \(n! \approx \sqrt{2\pi n} \left(\frac{n}{e}\right)^n\) Large factorials

Interactive Quiz Challenge

Quick Fire Questions

Round 1: What’s the probability of getting exactly one head in three coin flips?

A. 1/8 B. 3/8 C. 1/2 D. 5/8

Round 2: In a class of 30 students, what’s the approximate probability that at least two share a birthday?

A. 0.3 B. 0.5 C. 0.7 D. 0.9

Answers: 1. B. 3/8 (outcomes: HTT, THT, TTH out of 8 total) 2. C. 0.7 (birthday paradox - quite high!)

Real-World Applications

Quality Control

Sampling products to estimate defect rates using classical probability principles.

Medical Testing

Calculating false positive rates when test results are equiprobable.

Cryptography

Birthday attacks exploit the birthday paradox to find hash collisions.

Sports Analytics

Calculating probabilities of various game outcomes and player statistics.

Risk Assessment

Evaluating unlikely but catastrophic events using extreme probability calculations.

Network Security

Analyzing the probability of random password matches or system failures.

Take-Home Messages

Key Insights

  1. Equiprobable outcomes make probability calculations straightforward using counting principles

  2. The birthday paradox shows our intuition about probability can be misleading

  3. Combinatorics (combinations and permutations) are essential tools for complex probability problems

  4. Interactive simulations help verify theoretical calculations and build intuition

  5. Real-world applications are everywhere - from card games to cryptography

Mastery Check

Self-Assessment Checklist

Rate your confidence level (1-5) for each skill:

  • Apply classical probability formula (Essential)
  • Use combinations and permutations correctly (Essential)
  • Solve dice and card problems (Essential)
  • Understand the birthday paradox (Proficient)
  • Analyze poker hand probabilities (Proficient)
  • Work with infinite processes and limits (Advanced)
  • Apply probability to real-world scenarios (Advanced)

Thank You & Questions

Summary

Today we explored equally likely outcomes and their applications through: - 🎲 Interactive simulations and visualizations - 🃏 Classic probability problems (poker, birthdays, matching) - 📊 Combinatorial analysis and counting principles - 🧮 Hands-on problem-solving activities

Next Steps

  • Practice with the interactive tools at home
  • Work through the homework problems
  • Explore probability simulations in R or Python
  • Prepare for next lecture on conditional probability

Questions?

Let’s discuss these fascinating probability phenomena!

Next Lecture: Conditional Probability and Independence

Mathematical Statistics - ADA