RM

MTH331 Lecture Notes 03: Motivation, Conditional Probability, Monty Hall Problem

Motivation

  • Conditional probability studies how the probability of an event A changes when we learn that another event B has occurred.

  • Intuition: More information can change our odds; some information is more informative than others.

  • Key takeaway: Probability updates depend on the relationship between A and B (independence, overlap, etc.).

Conditional probability

  • Definition (general):

    P(A|B)=\frac{P(A\cap B)}{P(B)}

  • Intuition: The probability of A given B is the fraction of outcomes in B for which A also occurs.

  • Generalization: For events A and B with P(B)>0, conditional probability extends to more complex conditioning and to chains of conditioning.

Examples

Die-rolling example (A simple illustration of conditioning)

  • Sample space:

    \Omega = {1,2,3,4,5,6}

  • Each outcome: P({i})=\frac{1}{6}, \quad i=1,2,\dots,6

  • Event A: observing an even number; A = {2,4,6}.

    P(A)=\frac{|A|}{|\Omega|}=\frac{3}{6}=\frac{1}{2}

  • Event B: the number is greater than 3; B = {4,5,6}.

  • If told B occurs, the remaining possibilities are {4,5,6}.

  • Among these, even numbers are {4,6}, so

    P(A|B)=\frac{2}{3}

  • Common insight: Knowing that the number is >3 changes the sample space and can raise P(A|B). For instance, the probability of an even number originally was 1/2, but given the number is greater than 3, the probability becomes 2/3.

  • Another illustration: probability of rain in May is highly information-dependent; if told it’s in California, the probability changes. The pattern: additional information changes probability calculations.

Conditional probability (formal definition, recap)

  • Let A = event of observing even numbers; B = event #dots > 3.

  • Definition:

    P(A|B)=\frac{P(A\cap B)}{P(B)}

  • General form (as used throughout the lecture):

    P(A|B)=\frac{P(A\cap B)}{P(B)}\,.

Examples (continued)

Example 1: Coin-tosses with conditioning

  • Scenario: Toss a fair coin three times.

  • Define A = {more heads than tails}.

  • Define B = {1st toss is a head}.

  • Sample space: \Omega={HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.

  • A = {HHH, HHT, HTH, THH} (4 outcomes).

  • B = {HHH, HHT, HTH, HTT} (4 outcomes).

  • A∩B = {HHH, HHT, HTH} (3 outcomes).

  • The probability of B is P(B) = \frac{|B|}{|\Omega|} = \frac{4}{8} = \frac{1}{2}. The probability of the intersection A∩B is P(A\cap B) = \frac{|A\cap B|}{|\Omega|} = \frac{3}{8}.

  • Therefore, using the formula:

    P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{3/8}{4/8}=\frac{3}{4}.

Next Class Notes:

Example 2: Two rolls of a 4-sided die

  • Setup: A 4-sided fair die is rolled twice. Let X,Y {\in} {1,2,3,4}. Total outcomes = 4 \times 4 = 16.

  • Define A = {max(X,Y) = m} for m {\in} {1,2,3,4}.

  • Define B = {min(X,Y) = 2}.

  • We condition on B and compute P(A|B) for each m.

  • Enumerated outcomes where min = 2 (Event B): {(2,2),(3,2),(4,2),(2,3),(2,4)}. There are 5 such outcomes. So, P(B) = \frac{5}{16}.

  • Now, we find the intersection A∩B for each value of m:

    • For m = 1: A∩B = {(X,Y) | max(X,Y)=1 and min(X,Y)=2}. This is impossible. So A\cap B = \emptyset --> P(A\cap B) = 0.

    • For m = 2: A∩B = {(2,2)}. Only 1 outcome --> P(A\cap B) = \frac{1}{16}.

    • For m = 3: A∩B = {(3,2),(2,3)}. 2 outcomes --> P(A\cap B) = \frac{2}{16}.

    • For m = 4: A∩B = {(4,2),(2,4)}. 2 outcomes --> P(A\cap B) = \frac{2}{16}.

  • Now we can compute P(max(X,Y)=m\;|\;min(X,Y)=2) for each m using the conditional probability formula P(A|B)=\frac{P(A\cap B)}{P(B)}:

    • If m = 1: P(max(X,Y)=1\;|\;min(X,Y)=2)=\frac{0/16}{5/16}=0

    • If m = 2: P(max(X,Y)=2\;|\;min(X,Y)=2)=\frac{1/16}{5/16}=\frac{1}{5}

    • If m = 3: P(max(X,Y)=3\;|\;min(X,Y)=2)=\frac{2/16}{5/16}=\frac{2}{5}

    • If m = 4: P(max(X,Y)=4\;|\;min(X,Y)=2)=\frac{2/16}{5/16}=\frac{2}{5}

  • Summary of probabilities given B (min(X,Y)=2):

    P(max(X,Y)=m\;|\;min(X,Y)=2)=\begin{cases}0,&m=1\frac{1}{5},&m=2\frac{2}{5},&m=3\frac{2}{5},&m=4\end{cases}

Example 3: Two teams A and B with competing probabilities

  • Setup: A succeeds with probability P(A)=\frac{2}{3}; B succeeds with P(B)=\frac{1}{2}.

  • Given: probability that at least one team succeeds is P(A\cup B)=\frac{3}{4}.

  • Compute intersections and related quantities:

    • Probability that both A and B succeed (intersection) using the Inclusion-Exclusion Principle:
      P(A\cap B)=P(A)+P(B)-P(A\cup B)=\frac{2}{3}+\frac{1}{2}-\frac{3}{4}=\frac{8}{12}+\frac{6}{12}-\frac{9}{12}=\frac{5}{12}

    • Probability that exactly one team succeeds (exclusive-or):
      P(A\oplus B)=P(A\cup B)-P(A\cap B)=\frac{3}{4}-\frac{5}{12}=\frac{9}{12}-\frac{5}{12}=\frac{4}{12}=\frac{1}{3}

    • Probability that only A succeeds: P(A\cap B^c) = P(A) - P(A\cap B) = \frac{2}{3} - \frac{5}{12} = \frac{8}{12} - \frac{5}{12} = \frac{3}{12} = \frac{1}{4}

    • Probability that only B succeeds: P(B\cap A^c) = P(B) - P(A\cap B) = \frac{1}{2} - \frac{5}{12} = \frac{6}{12} - \frac{5}{12} = \frac{1}{12}

  • Now, let's calculate conditional probabilities based on these values. If we condition on the event A\oplus B (exactly one succeeds):

    • Probability that A succeeded exclusively, given that exactly one team succeeded:
      P(A\cap B^c\;|\;A\oplus B)=\frac{P((A\cap B^c)\cap(A\oplus B))}{P(A\oplus B)} = \frac{P(A\cap B^c)}{P(A\oplus B)} = \frac{1/4}{1/3} = \frac{3}{4}
      (Note: The event (A\cap B^c) is a subset of (A\oplus B), so (A\cap B^c)\cap(A\oplus B) = A\cap B^c)

    • Probability that B succeeded exclusively, given that exactly one team succeeded:
      P(B\cap A^c\;|\;A\oplus B)=\frac{P((B\cap A^c)\cap(A\oplus B))}{P(A\oplus B)} = \frac{P(B\cap A^c)}{P(A\oplus B)} = \frac{1/12}{1/3} = \frac{1}{4}
      (Note: The event (B\cap A^c) is a subset of (A\oplus B), so (B\cap A^c)\cap(A\oplus B) = B\cap A^c)

The Monty Hall Problem

Description

  • Original setup: Three doors; prize behind one door uniformly at random; other two have goats.

  • You pick one door (e.g., door 1).

  • Host Monty Hall then opens one of the other two doors revealing a goat.

  • You are asked whether to switch to the remaining unopened door.

  • Goal: Determine the best strategy based on conditional probability given the host’s action.

Understanding the information flow

  • The host’s action depends on where the prize is; their choice is not random with respect to the prize location.

  • Conditioning on Monty’s move provides information about where the prize is likely to be.

  • Classic result: Switching doors improves odds, because the initial choice had a 1/3 chance of being correct, and the unopened remaining door then accumulates the remaining 2/3 probability when Monty always reveals a goat.

Strategy 1: Stick with the initial choice

  • If you pick door 1, the probability the prize is behind door 1 is 1/3. If you stick, you win if your initial choice was correct.

  • Probability of winning by sticking: P(\text{win }|\text{stick})=\frac{1}{3}.

Strategy 2: Switch to the other unopened door

  • Case analysis:

    • If prize behind door 1 (probability 1/3): Monty opens door 2 or 3; switching loses as you move from the correct door.

    • If prize behind door 2 (probability 1/3): Monty must open door 3 (since you picked door 1, and the prize is at door 2); switching to door 2 wins.

    • If prize behind door 3 (probability 1/3): Monty must open door 2 (since you picked door 1, and the prize is at door 3); switching to door 3 wins.

  • Total probability of winning by switching: P(\text{win by switch})=\frac{1}{3} + \frac{1}{3} = \frac{2}{3}.

Strategy 3: Choose door 1 and switch if door 3 opens

  • Assumptions: Host randomly opens door 2 or 3 when prize is behind door 1.

  • Case analysis:

    • Prize behind door 1 (probability 1/3): Host opens door 2 or 3 with equal probability (1/2 for each). If door 3 opens, you switch, which means you switch from door 1 to door 2, losing. The probability of this specific win scenario is \frac{1}{3} \cdot 0 = 0.
      Correction for Strategy 3 based on original intention: If prize behind door 1 and door 3 opens, you switch. By switching, you move from door 1 to door 2, losing. Win = 0. So the original calculation implies a win if door 3 opens and you switch. Let's re-evaluate the original line: 'you switch only if door 3 opens'. This means if door 2 opens, you stick with door 1. If prize is behind door 1, P(win if door 3 opens and switch) = 0. The original calculation was strange: \frac{1}{3} \cdot \frac{1}{2} gives the probability of Host opening door 3 and prize being behind door 1. But switching loses in this case. The original text had a winning probability of 1/6, which would happen if switching when door 3 opens led to a win. This is only if the prize was at door 2. Let me follow the structure as given but clarify the win/loss condition:

    • Prize behind door 1 (probability 1/3): Host randomly opens door 2 or 3 (each with 1/2 probability). If door 3 opens, you switch (from door 1 to door 2) and lose. Probability of this specific scenario leading to a win is 0.

    • Prize behind door 2 (probability 1/3): Host must open door 3. You switch (from door 1 to door 2) and win. Probability of this scenario leading to a win is \frac{1}{3}.

    • Prize behind door 3 (probability 1/3): Host must open door 2. You do not switch if door 3 does not open (as per strategy). So you stick with door 1 and lose. Probability of this scenario leading to a win is 0.

  • Therefore, the Total probability of winning with Strategy 3 (switching only if door 3 opens) is P(\text{win})=0+\frac{1}{3}+0=\frac{1}{3}.. (The original text had 1/2, which likely implies a different interpretation of the strategy or a mistake in the original calculation. I will correct this to what makes sense with the interpretation.)

Summary of Monty Hall results

  • Best strategy: Switch to the other unopened door (Strategy 2) with probability of winning \frac{2}{3}.

  • Alternative strategy (Strategy 1, sticking with initial choice) yields \frac{1}{3}. The revised Strategy 3 (switching only if door 3 opens) also yields \frac{1}{3}.

  • Key takeaway: The host’s action provides information that changes the conditional probabilities of where the prize is located.

Connections and implications

  • The Monty Hall problem is a canonical example of conditional probability and information update.

  • Demonstrates that intuitive reasoning can be misleading when conditioning on nontrivial information (host behavior matters).

  • Real-world relevance: updating beliefs as new information arrives (e.g., diagnostic tests, decision under uncertainty).

  • Ethical/philosophical note: In probabilistic reasoning, how information is revealed (the mechanism) matters as much as the information itself.

Formulas and key relations (summary)

  • Conditional probability: P(A|B)=\frac{P(A\cap B)}{P(B)}

  • For disjoint events: if A and B are disjoint (A\cap B = \emptyset), then P(A|B)=0. An example is in Example 2 for m=1.

  • Inclusion-exclusion (used in Example 3): P(A\cup B)=P(A)+P(B)-P(A\cap B)

  • Exclusive OR: P(A\oplus B)=P(A\cup B)-P(A\cap B)

  • Monty Hall probabilities:

    • Strategy 1 (Stick): 1/3

    • Strategy 2 (Switch): 2/3

    • Strategy 3 (Switch if door 3 opens): 1/3 (based on corrected logical flow)