The Timpson Lyceum · Discrete Math (2025)

Probabilities and Sample Spaces - Lecture 14 (Discrete Math)

Overview

This lecture opens the unit on probability, the mathematics of how likely an outcome is. The central lesson is that probabilities are deeply unintuitive: you cannot trust gut instinct and must instead reason carefully through every possible outcome. We build the formal object that makes this reasoning precise — the sample space, an ordered pair of a set of outcomes together with a probability function. We practice modeling coins, dice, spinners, and poker hands as sample spaces, then apply the idea to a real medical-testing scenario to show why a "90% accurate" test does not mean a positive result is right 90% of the time. The recurring theme: think it out, do not assume you already know the answer.

Key Concepts

Counting outcomes. To find a probability, list every equally likely outcome, then count how many are the ones you care about. The probability is (favorable outcomes) / (total outcomes).

• Flipping a fair coin $n$ times gives $2^n$ equally likely sequences (2 flips → 4 outcomes, 3 flips → 8, 4 flips → 16).
• $P(\text{heads, heads}) = \tfrac14$, $P(\text{HHH}) = \tfrac18$, $P(\text{HHHH}) = \tfrac{1}{16}$.

Order matters in a sequence. A single specific sequence (e.g. HTH) has probability $\tfrac18$, but an event described by a count (e.g. "two heads and one tail") collects several sequences. There are three sequences with exactly two heads (HHT, HTH, THH), so $P(\text{exactly two heads in three flips}) = \tfrac38$, while $P(\text{HHH}) = \tfrac18$. You are three times more likely to get two-heads-one-tail than three heads.

The two-children puzzle. A couple has two children and at least one is a boy; the chance the other is also a boy is $\tfrac13$, not $\tfrac12$. The ordered outcomes are BB, BG, GB, GG. Knowing one child is a boy rules out GG, leaving three equally likely cases, only one of which (BB) is two boys.

Sample space (the key definition). A sample space is an ordered pair $(S, P)$ where:

• $S$ is a non-empty set — typically the set of all outcomes.
• $P$ is a function $P : S \to \mathbb{R}$ assigning each outcome a real number (its probability), subject to two conditions:
  • Non-negativity: $P(s) \ge 0$ for every $s \in S$ (a probability is never negative).
  • Normalization: $\displaystyle\sum_{s \in S} P(s) = 1$ (all the probabilities add up to one, i.e. 100%).

$P$ is called the probability function and $S$ the set of outcomes.

There is no "right" or "wrong" sample space — only more or less useful ones. The same scenario can be modeled many ways. A fair die can be modeled as $S = \{1,2,3,4,5,6\}$ with $P(s) = \tfrac16$, or with lopsided probabilities (as long as they are non-negative and sum to 1, it is still a valid sample space — just not a faithful model of a fair die). The "useful" model is the one that fits your purpose.

Shorthand notation. When every outcome shares the same probability, write $P(s) = \tfrac16 \text{ for all } s \in S$ instead of listing each outcome separately.

Standard sample spaces:

• One fair die: $S = \{1,2,3,4,5,6\}$, $P(s) = \tfrac16$.
• Two dice: $S = \{1,\dots,6\} \times \{1,\dots,6\}$, $|S| = 36$, $P(s) = \tfrac{1}{36}$. If instead you take $S$ to be the possible sums $\{2,3,\dots,12\}$, the probabilities are no longer uniform — e.g. the sum 7 occurs 6 ways so $P(7) = \tfrac{6}{36} = \tfrac16$, while the sum 2 occurs once so $P(2) = \tfrac{1}{36}$. (This is why a 7 is the most common roll, and Settlers of Catan prints more pips on the more probable numbers.)
• A poker hand: the 52-card deck is $\{1,\dots,13\} \times \{H,D,S,C\}$. But for hands, $S$ is the set of all 5-card hands, of size $\binom{52}{5}$, each with probability $\tfrac{1}{\binom{52}{5}}$. You may write $S$ in words ("the set of all possible poker hands") rather than enumerate it.

Independent trials and the gambler's fallacy. After spinning red five times in a row, the next spin is still equally likely red or black — past independent results do not change future ones. The confusion comes from mixing two different questions: before spinning four times, "three reds and one black" is more likely than "four blacks"; but given you have already seen reds, the next single spin is 50/50. The same logic explains why, among many mothers each guessing the sexes of three children, a majority guess "more than half" correct purely by chance — and the lucky ones wrongly credit a method.

Reading a probability function as a table. Given a list like $P(1)=0.1, P(2)=0.4, P(3)=0.5, P(4)=0, P(5)=0, P(6)=0$, check it is a sample space: each value is $\ge 0$ (zero is allowed — an outcome can be impossible) and they sum to 1.

Conditional reasoning: accuracy is not predictive value. "90% accurate" means: if you have the condition, the test reads positive 90% of the time, and if you don't, it reads negative 90% of the time. It does not mean a positive result is correct 90% of the time. To find the chance you actually have the condition given a positive test, you must account for how rare the condition is in the whole population — base rates dominate when the condition is uncommon.

All scientific knowledge is probabilistic. Only in mathematics is a proven result certain. Empirical fields carry confidence levels — roughly physics 99.99%, chemistry 99.7–99.99%, biology 90–95%, social sciences 60–80%. Combined with the fact that only a small fraction of all proposed theories are correct, a "proven" social-science result can have a surprisingly low chance of actually being true (worked below).

Worked Example

Positive test, low prevalence. 10% of a population has a condition, and a test is 90% accurate. If your test comes back positive, what is the probability you actually have the condition? Use a population of 100 people and split by true status, then apply the test's accuracy to each group; the answer is (true positives) / (all positives).

Even with a "90% accurate" test, a positive result means only a 50% chance of actually having the condition — because most positives come from the much larger healthy group. Lowering the prevalence below 10% makes this even worse.

A social-science theory "proven true." Assume (generously) 10% of all proposed theories are correct, and the field's tests run at 60% accuracy. Among 100 theories:

So a "proven" result in such a field has only about a 14% chance of being correct.

Summary

• To compute a probability, enumerate all equally likely outcomes and count the favorable ones; do not rely on intuition.
• A sample space is an ordered pair $(S, P)$ with $S$ a non-empty set of outcomes and $P : S \to \mathbb{R}$ a probability function where $P(s) \ge 0$ for all $s$ and $\sum_{s \in S} P(s) = 1$.
• The same scenario has many valid sample spaces; there is no "correct" one, only models that are more or less useful.
• A specific ordered sequence and an event defined by a count are different events: $P(\text{HHH}) = \tfrac18$ but $P(\text{exactly two heads in three flips}) = \tfrac38$.
• Independent trials have no memory — the gambler's fallacy comes from confusing "before any trials" probabilities with "given past results" probabilities.
• Test accuracy is not the same as the chance a positive result is correct; rare conditions make positives misleading, and outside mathematics all knowledge is probabilistic.