[Statistics 101] (8) Probabilities – Advanced

In this post, we will delve into more advanced features of probabilities.

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Independent Events

Basic Operations of probabilities are like these:

• Addition Rule
  • P (A or B) = P(A) + P(B) – P(A and B)
• Subtraction Rule
  • P(not A) = 1 – P(A)
  • P(A) = 1 – P(not A)
• Multiplication Rule
  • P(A and B) = P(A│B) × P(B) = P(B│A) × P(A)

When events are independent, the rules can be simplified.

• P(A | B) = P(A)
• P(B | A) = P(B)
• Multiplication Rule
  • P(A and B) = P(A│B) × P(B) = P(A) × P(B)
• Addition Rule
  • P (A or B) = P(A) + P(B) – P(A) × P(B)
• Subtraction Rule
  • P(A) = 1 – P(not A)

Another useful rules are:

• P (A or B) = 1 – P(not A and not B) = 1 – P(not A) × P(not B)
• P(A or B or C … or Z) = 1 – P(not A) × P(not B) × P(not C) × …P(not Z)

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[Example 1] Méré’s Paradox

Chevalier de Méré , 17th century gambler, contacted Blaise Pascal to help him solve a gambling problem. He was confused by the probabilities of two games popular in Paris.

• [Game 1] Roll a dice 4 times and win if at least one ace (number 1) appeared
• [Game 2] Roll a pair of dice 24 times and win if at least a double ace (number 1 on both dice) appeared

Chevalier calculated the winning probabilities of 2 games are the same.

• Game 1: (1/6) × 4 = 2/3 (66%)
• Game 2: (1/36) × 24 = 2/3 (66%)

But the right calculation is the like this:

• Game 1: each event (roll a dice) is independent
  • P(A1 or A2 or A3 or A4) = 1 – P(not A1) × P(not A2) × P(not A3) × P(not A4)
  • 1 – (5/6) × (5/6) × (5/6) × (5/6) = 1 – (5/6)⁴ = 0.5177 (52%)
• Game 2: each event (roll a pair of dice) is independent
  • P(A1 or A2 or A3 or … or A24) = 1 – P(not A1) × P(not A2) × P(not A3) × …P(not A24)
  • 1 – (35/36)⁴ = 0.4914 (49%)

Therefore, the winning chances of both games are much smaller than Chevalier thought and there is a bigger chance to win Game 1 than Game 2.

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[Example 2] KIA (Killed In Action)

A WW II pilot has a 2% of chance of being shot down on each mission. So in fifty missions, a pilot was certain to be shot down (2% * 50 = 100%).

This is wrong. What is the right chance of being shot down in 50 missions?

Suppose each mission is independent.

• P(A1 or A2 or … or A50) = 1 – P(not A1) × P(not A2) × … P(not A50)
• 1 – (0.98)⁵⁰ = 0.64 (64%)

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False Positive Paradox