Equally likely outcomes

Choose elementary outcomes carefully before dividing one count by another.

9 minutes · intermediate · Original Poker4Quant lesson

What you will learn

Build an equally likely sample space and distinguish an outcome from an event.

Prerequisites: Reading the deck · Chips, fractions and percentages

An event can contain unequal numbers of outcomes

A uniform deck gives each individual card probability 1/52. Events such as “an ace” and “not an ace” are not equally likely simply because there are two descriptions. One contains 4 cards and the other contains 48.

The ratio favorable/total is valid when the counted elementary outcomes are equally likely. Otherwise, add the probabilities of favorable outcomes rather than counting event names. State the random experiment before calculating.

Order is a modeling choice

Two sequential fair coin flips have four equally likely ordered outcomes: HH, HT, TH and TT. The event “one head” contains two of them. Grouping the possibilities as zero, one or two heads creates three events with different sizes.

Likewise, a shuffled two-card deal may be represented by ordered distinct cards or unordered pairs. Both approaches work when numerator and denominator use the same representation. Mixing them introduces a factor-of-two error.

Worked example

Two independent fair flips are made. What is the probability of exactly one head?

List HH, HT, TH, TT. Independence and fairness give each probability 1/4.
HT and TH are the favorable outcomes.
Add their probabilities: 1/4 + 1/4.

The answer is 1/2, not 1/3. Counting the three possible head counts as equally likely would be wrong.

Common mistakes

Assuming named categories are equally likely.
Using a uniform model without stating it.
Counting ordered wins against an unordered denominator.

Keep these ideas

Define elementary outcomes before events.
Count only when those outcomes have equal probability.
Keep ordered or unordered conventions consistent.