Probabilistic Thinking Math Example 2

Follow the full solution, then compare it with the other examples linked below.

Example 2

hard
The Monty Hall Problem: 3 doors, prize behind 1. You pick door 1. Host opens door 3 (no prize). Should you switch to door 2? Calculate probabilities for staying vs. switching.

Solution

  1. 1
    Initial: P(prize at door 1)=13P(\text{prize at door 1}) = \frac{1}{3}; P(prize at door 2 or 3)=23P(\text{prize at door 2 or 3}) = \frac{2}{3}
  2. 2
    Host opens door 3 (always reveals no prize): remaining probability from door 3 transfers to door 2
  3. 3
    P(prize at door 1∣door 3 opened)=13P(\text{prize at door 1}|\text{door 3 opened}) = \frac{1}{3} (unchanged — host's action was deterministic given your choice)
  4. 4
    P(prize at door 2∣door 3 opened)=23P(\text{prize at door 2}|\text{door 3 opened}) = \frac{2}{3} (all the original 2/3 probability is now concentrated here)

Answer

Always switch: P(win by switching)=23P(\text{win by switching}) = \frac{2}{3} vs. P(win by staying)=13P(\text{win by staying}) = \frac{1}{3}.
The Monty Hall Problem illustrates how new information (the host's reveal) should update probabilities. Most people's intuition says 50-50, but conditional probability shows switching doubles the win probability. The host's action is not random — it provides information.

About Probabilistic Thinking

Probabilistic thinking is the habit of reasoning about uncertain outcomes in terms of likelihood, expected value, and distributions rather than certainties.

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