Conditional Probability Example

What this lecture covers

This recitation problem works through several conditional-probability calculations on the classic two-dice sample space, reinforcing both the discrete uniform law and the definition of conditional probability. It also compares two ways of solving the same question: applying the definition directly, versus reasoning about a smaller "conditional universe."

The problem builds up from a simple unconditional probability to conditioning on multiple different events, ending with a clear statement of when the shortcut method works.

Key ideas

• Discrete uniform law: with 36 equally likely outcomes for two dice, any event's probability is its outcome count divided by 36.
• Doubles as a baseline event: the probability that both dice show the same number is 6/36 = 1/6.
• Definition-based conditioning: P(D|sum≤4) = P(D ∩ sum≤4) / P(sum≤4), computed by counting outcomes in numerator and denominator separately.
• Shortcut: conditioning restricts the sample space: since conditioning preserves relative likelihoods, the conditional probability of a uniform model is still uniform, just over the reduced set of outcomes.
• Two independent applications: the same reasoning is used for "sum ≤ 4" and "the two dice land on different numbers," both giving matching answers by either method.

Walkthrough

Setting up the discrete uniform law (0:00)

The video reviews the discrete uniform law — when all outcomes are equally likely, probability reduces to counting — and applies it to find the probability of rolling doubles on two fair dice: 6 favorable outcomes out of 36, or 1/6.

Conditioning on the sum being small (5:28)

The problem asks for the probability of doubles given that the sum of the two rolls is at most 4. It's solved two ways: first by directly applying the definition of conditional probability and counting outcomes in the numerator and denominator, then by treating the six outcomes with sum ≤ 4 as a new, smaller uniform sample space and counting doubles within it. Both methods give 1/3, and the second is shown to be faster because conditioning doesn't change the relative likelihood of remaining outcomes.

Conditioning on rolling different numbers (8:47)

The video finds the unconditional probability of rolling at least one 6 (11/36), then conditions on the event that the two dice show different numbers, which removes the 6 diagonal (doubles) outcomes, leaving 30 equally likely outcomes. Counting how many of those include at least one 6 gives 10/30 = 1/3, again matching what the definition-based method would produce.

Before you watch

• Know the discrete uniform law and the definition of conditional probability, covered in the first two lectures of this course.
• It helps to be comfortable visualizing outcomes for two dice as points on a grid.

Check your understanding

1. Why does conditioning on "sum ≤ 4" let you treat the six matching outcomes as a new uniform sample space?
2. What is the difference between solving a conditional probability using the definition versus using the "conditional universe" shortcut?
3. Why does removing the 6 doubles outcomes leave exactly 30 outcomes when conditioning on "different numbers"?
4. Would the shortcut method still work if the original 36 outcomes were not equally likely?