Real Numbers – Episode 15, The Half Court Shot

Episode 1: Basketball Expected Value Problem Setup

• Host introduces a new problem involving a basketball player, Danielle, who takes half-court shots until making one.
• Probability of success per shot is 20% ($p = 0.2$).
• Primary question asks for the expected number of attempts to make a single shot.
• Bonus question asks for the expected attempts required to make two consecutive shots.

Episode 1: Solution to Previous Week's Problem (Two Final Exams)

• Scenario: Jayden has two exams scheduled uniformly at random on one of 5 weekdays (Monday–Friday).
• Method 1: Direct Enumeration
  • Total possibilities for two exams: $5 \times 5 = 25$ equally probable outcomes.
  • Days between exams distribution:
    • 0 days (same day): 5 cases.
    • 1 day (consecutive): 8 cases.
    • 2 days: 6 cases.
    • 3 days: 4 cases.
    • 4 days (Monday/Friday): 2 cases.
  • Weighted sum calculation: $(5\times0 + 8\times1 + 6\times2 + 4\times3 + 2\times4) / 25 = 40 / 25$.
  • Result: Expected days between two exams = $1.6$ (or $8/5$).
• Method 2: Random Variable Decomposition
  • Defined $L$ as the day of the later exam and $K$ as the day of the earlier exam.
  • Target calculation: $E[L - K] = E[L] - E[K]$ via linearity of expectation.
  • Calculation of $E[L]$: Sum of entries in table where entries represent "max" of two days; sum found to be 95.
  • $E[L] = 95 / 25 = 3.8$ (or $19/5$).
  • Calculation of $E[K]$: Derived via symmetry ($E[K] = 6 - E[L]$), resulting in $2.2$ (or $11/5$).
  • Final result: $3.8 - 2.2 = 1.6$, confirming Method 1.
• General Formula Derived:
  • For non-negative integer random variable $X$, $E[X] = \sum_{i=1}^{\infty} P(X \ge i)$.

Episode 1: Extension to Three Final Exams

• Scenario: Jayden now has three exams (Math, Statistics, Advanced Math) on 5 weekdays.
• Method 1: Cubic Table (Geometric approach)
  • Total possibilities: $5^3 = 125$.
  • $E[\text{Latest}]$: Calculated as $5 - \sum_{i=1}^{4} (i/5)^3 = 5 - 100/125 = 4.2$.
  • $E[\text{Earliest}]$: Derived via symmetry ($6 - 4.2$) = $1.8$.
  • Result: Expected days between earliest and latest = $4.2 - 1.8 = 2.4$.
• Method 2: Linearity of Expectation via Pairs
  • Logic: Sum of days between all pairs (Early-Middle, Middle-Latest, Early-Latest) equals 2 times the span (Early-Latest).
  • Equation: $2 \times E[\text{Span}] = \sum E[\text{Pair Distances}]$.
  • There are 3 pairs total; $E[\text{Pair Distance}]$ is known from the 2-exam case ($1.6$).
  • Calculation: $2 \times E[\text{Span}] = 3 \times 1.6 \rightarrow E[\text{Span}] = 2.4$.

Episode 2: Current Problem Solution (Pending)

• The transcript concludes by restating the basketball problem without providing the solution.
• Danielle makes 20% of shots; expected attempts to make one shot is requested.
• Bonus question: Expected attempts to make two consecutive shots.
• Listener is instructed to submit answers via the webpage.