The rules of probability

You roll two dice. You win if at least one of them shows a six. What’s your chance?

The quick answer goes: “Each die has a 1 in 6 chance, so two dice give $\frac{1}{6} + \frac{1}{6} = \frac{1}{3}$.” It sounds right. But follow that logic to six dice and you’d get $\frac{6}{6} = 1$, a guaranteed six, and you know that’s not true: six dice can all miss.

The mistake is double counting. When both dice show a six, the quick answer counts that win twice. In this lesson you’ll see that happen on a grid, then learn the three rules that fix it. They’re the same counting ideas you met in F06 (complements, inclusion–exclusion), now dressed as probabilities.

The real words

• The three basic rules (called the axioms of probability, written down by Andrey Kolmogorov in 1933). Everything else follows from them:

  1. $P(A) \ge 0$ for every event.
  2. $P(S) = 1$: some outcome always happens.
  3. If $A$ and $B$ are disjoint (they can’t both happen), $P(A \text{ or } B) = P(A) + P(B)$.

• Complement rule: “not $A$” (written $A^c$) happens exactly when $A$ doesn’t. $A$ and $A^c$ are disjoint and together make all of $S$, so rules 2 and 3 give

  $$P(A^c) = 1 - P(A)$$
  • “At least one six” is the complement of “no sixes”: $1 - \frac{25}{36} = \frac{11}{36}$. Whenever a question says at least one, try the complement first.

• Union rule (or), the probability version of inclusion–exclusion (F06 lesson 8):

  $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$
  • $\cup$ is “or” (either one, or both), $\cap$ is “and” (both).
  • Red six or blue six: $\frac{6}{36} + \frac{6}{36} - \frac{1}{36} = \frac{11}{36}$. The $-\frac{1}{36}$ removes the double-counted (6, 6).
  • Disjoint events have $P(A \cap B) = 0$, so you just add: sum 2 or sum 12 is $\frac{1}{36} + \frac{1}{36}$.