Advanced Probability Sequences and Binomial Distributions

Probabilistic Modeling of Winning Sequences

• The lecturer utilizes a sports scenario involving the Chicago Bears and the New York Giants to illustrate sequence-based probability and home-field advantage factors.
• Home-Field Advantage Dynamics:   - When playing in New York (at the Meadowlands), the probability of the Bears winning is $30\%$, represented as $0.3$. Conversely, the Giants have a $70\%$ chance or $0.7$ probability of winning.   - When playing in Chicago (at Soldier Field), the Bears have a higher probability of winning, cited as $74\%$ or $0.74$. The Giants' probability of winning at Soldier Field is $26\%$ or $0.26$.   - In any given match, the probabilities of both teams must add up to one ($0.74 + 0.26 = 1.0$).
• Calculating a Specific Sequence:   - Goal: Determine the probability of a specific outcome across three games: The Bears win the first, the Giants win the second, and the Bears win the third.   - The probability of a sequence of independent events occurring together is the product of their individual probabilities (multiplication rule).   - Calculation for (Bears win at home, Giants win in NY, Bears win at home):     - $P(\text{Bears Win 1}) = 0.74$     - $P(\text{Giants Win 2}) = 0.7$     - $P(\text{Bears Win 3}) = 0.74$     - Combined Probability: $0.74 \times 0.7 \times 0.74 = 0.38332$
• Determining the Probability of Winning Two Out of Three Games:   - This requires identifying all possible branches/sequences where the Bears win exactly two games.   - Sequence A: Bears win, Giants win, Bears win.     - Probability: $0.74 \times 0.7 \times 0.74 = 0.38332$   - Sequence B: Bears win, Bears win away, Giants win at home.     - Probability: $P(\text{Bears win home}) \times P(\text{Bears win away}) \times P(\text{Giants win home})$     - Probability: $0.74 \times 0.3 \times 0.26 = 0.05772$   - Sequence C: Giants win at home, Bears win away, Bears win at home.     - Probability: $P(\text{Giants win home}) \times P(\text{Bears win away}) \times P(\text{Bears win home})$     - Probability: $0.26 \times 0.3 \times 0.74 = 0.05772$   - Final Summation: To find the total probability of winning two out of three games, add the probabilities of all successful sequences. The total is approximately $0.5$ or $50\%$     - Total Probability: $0.38332 + 0.05772 + 0.05772 \approx 0.49876$

Conditional Probability in Sports Contexts

• Scenario: What is the probability that the Bears win the third game given they lost the first two games?
• Notation: $P(\text{Bears win game 3} | \text{Bears lost games 1 and 2})$.
• Logical Tree Analysis:   - If the first two games are already lost, we are at a specific node in the decision tree.   - Previous outcomes do not change the probability of the current event if the events are independent.   - If the third game is played in Chicago, the probability of the Bears winning is simply the home-field probability: $0.74$.   - The instructor notes that while one could use a complex formula, looking at the logic tree shows that the condition "lost first two" simply places the team at Soldier Field for the third game, making the answer $74\%$.

Binomial Distribution and Dice Experiments

• Defining the Experiment:   - A single die is rolled six times ($n=6$).   - We want to find the probability of rolling a "1" exactly a certain number of times.
• Probability of Single Outcomes:   - Probability of rolling a one: $P(\text{1}) = \frac{1}{6}$.   - Probability of not rolling a one: $P(\text{not 1}) = \frac{5}{6}$.
• Calculating Exactly Two "1"s:   - Step 1: Specific Sequence Probability: The probability of the first two rolls being 1 and the next four being "not 1" is $(\frac{1}{6}) \times (\frac{1}{6}) \times (\frac{5}{6}) \times (\frac{5}{6}) \times (\frac{5}{6}) \times (\frac{5}{6}) = (\frac{1}{6})^{2} \times (\frac{5}{6})^{4}$.   - Step 2: Accounting for Combinations: There are multiple ways to get two 1s in six rolls (e.g., 1st and 3rd, 2nd and 5th, etc.).   - The number of ways to arrange these outcomes is defined by the combination formula: $\binom{n}{k} = \binom{6}{2}$.   - Final Binomial Probability: $P(X=2) = \binom{6}{2} \times (\frac{1}{6})^{2} \times (\frac{5}{6})^{4}$.   - The result is approximately $0.20$ or $20\%$.
• Binomial Probability for One "1":   - Calculation: $\binom{6}{1} \times (\frac{1}{6})^{1} \times (\frac{5}{6})^{5}$.   - The result is approximately $0.4018$ or $40.18\%$.
• Binomial Probability for Zero "1"s:   - Calculation: $\binom{6}{0} \times (\frac{1}{6})^{0} \times (\frac{5}{6})^{6} = (\frac{5}{6})^{6}$.   - The result is approximately $0.3348$ or $33.48\%$.
• Cumulative Distribution (CDF):   - Probability of getting 0, 1, or 2 "1"s: $P(X=0) + P(X=1) + P(X=2)$.   - Summation: $0.3348 + 0.4018 + 0.2009 \approx 0.9375$.   - This means there is about a $93.75\%$ chance of seeing two or fewer "1"s in six rolls.

TI Calculator Procedures for Binomial Distributions

• To access distributions on a TI-series calculator:   - Press [2ND] followed by [DISTR] (the VARS key).   - Scroll down to the binomial options.
• Binomial PDF (Probability Distribution Function):   - Used for finding the probability of an exact number of successes (e.g., exactly 2).   - Syntax: binompdf(n, p, x).   - Example: binompdf(6, 1/6, 2).
• Binomial CDF (Cumulative Distribution Function):   - Used for finding the probability of a range of successes from 0 up to $x$.   - Syntax: binomcdf(n, p, x).   - Example for 0, 1, or 2 successes: binomcdf(6, 1/6, 2) resulting in $0.9375$.

Real-World Application: Student Selection Probability

• Scenario: A class of 35 students with about 110 to 111 total draws/selections over a semester.
• Variables:   - Number of trials ($n$): $111$   - Probability of being picked ($p$): $\frac{1}{35}$   - Successes ($x$): $1$
• Calculation Analysis:   - The student Solange was picked exactly once.   - Using the Binomial PDF: binompdf(111, 1/35, 1).   - Initial estimate by the professor was $4\%$, which was corrected by a student (Marcella) to $13\%$ ($0.131$).
• Classroom Statistical Validation:   - If $13\%$ of the group should be picked exactly once, in a class of 35, we expect approximately $4.55$ people ($35 \times 0.131$).   - In the actual classroom, 7 people raised their hands as having been picked exactly once, demonstrating relative proximity to the statistical model.

Questions & Discussion

• Question (Marcella): How do we figure that out? (Referring to the probabilities of exactly one selection).
• Answer: The professor explains it is a binomial distribution problem using $n=111$, $p=1/35$, and $x=1$.
• Question (Student): How did you get the distribution? (Referring to the calculator steps).
• Answer: The professor walks through the steps: Second function, DISTR, scroll down to Option A (Binomial PDF) or Option B (Binomial CDF).
• Question (Student): Do you mean the one would only come on the first and second try?
• Answer: No, when we use the combination $\binom{n}{k}$, we count all occurrences regardless of which specific rolls they were. We only care that the count is exactly two out of six.