Coins and dice

Coins and dice are the canonical probability experiments. Their sample spaces are clean, the outcomes are equally likely (for fair coins and dice), and the enumeration is mechanical. Master these and a wide variety of probability problems become routine.

One coin

Sample space: $S = \{H, T\}$. Size $2$.

For a fair coin, $P(H) = P(T) = 1/2$.

Two coins (or one coin tossed twice)

Sample space: $S = \{HH, HT, TH, TT\}$. Size $4$.

Three coins (or one coin tossed three times)

Sample space: $S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$. Size $8$.

A useful observation: with $n$ tosses, the count of "exactly $k$ heads" is $\binom{n}{k}$ (read "$n$ choose $k$"). For $n = 3$: $\binom{3}{0}, \binom{3}{1}, \binom{3}{2}, \binom{3}{3} = 1, 3, 3, 1$.

One die

Sample space: $S = \{1, 2, 3, 4, 5, 6\}$. Size $6$.

Two dice

Sample space: ordered pairs $(d_1, d_2)$ where each $d_i$ is one of $1, \ldots, 6$. Size $36$.

The most important distribution to know: the sum of the two dice.

Notice the symmetric triangular distribution: counts go $1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1$, summing to $36$. The most likely sum is $7$.

Other useful events:

Worked examples

Example 1. A coin is tossed twice. Find $P($head appears at least once$)$.

$P = 3/4$.

Example 2. Three coins are tossed. Find $P($exactly two heads$)$.

$P = 3/8$.

Example 3. Two dice are rolled. Find $P($sum is 8$$).

$5$ combinations. $P = 5/36$.

Example 4. Two dice are rolled. Find $P($product is a multiple of 6$$).

Outcomes where $d_1 \cdot d_2$ is a multiple of $6$: enumerate. Multiples of $6$ in $\{1, \ldots, 36\}$ are $6, 12, 18, 24, 30, 36$. Pairs: $(1,6)(6,1)(2,3)(3,2) \Rightarrow$ for $6$, $4$ pairs. For $12$: $(2,6)(6,2)(3,4)(4,3) \Rightarrow 4$. For $18$: $(3,6)(6,3) \Rightarrow 2$. For $24$: $(4,6)(6,4) \Rightarrow 2$. For $30$: $(5,6)(6,5) \Rightarrow 2$. For $36$: $(6,6) \Rightarrow 1$. Total $15$. $P = 15/36 = 5/12$.

Example 5. Two dice are rolled. Find $P($same number on both$)$.

Doubles: $(1,1), (2,2), \ldots, (6,6) \Rightarrow 6$. $P = 6/36 = 1/6$.

Example 6. A die is rolled. Find $P($neither prime nor multiple of 3$$).

Primes: $\{2, 3, 5\}$. Multiples of $3$: $\{3, 6\}$. Union: $\{2, 3, 5, 6\}$. Complement: $\{1, 4\}$, $2$ outcomes. $P = 2/6 = 1/3$.

Example 7. Two dice are rolled. Find $P($sum is less than 6$$).

Sums $2, 3, 4, 5$: $1 + 2 + 3 + 4 = 10$ outcomes. $P = 10/36 = 5/18$.

Try it yourself

1. A coin is tossed three times. $P($all three heads$) = ?$
2. A coin is tossed three times. $P($exactly two heads$) = ?$
3. A coin is tossed three times. $P($at least one head$) = ?$
4. A die is rolled. $P($prime number$) = ?$
5. A die is rolled. $P($even or multiple of 3$$) = ?
6. Two dice are rolled. $P($sum = 7$$) = ?
7. Two dice are rolled. $P($sum is even$) = ?$
8. Two dice are rolled. $P($sum is \geq 9$$) = ?
9. Two dice are rolled. $P($at least one shows 6$$) = ?
10. Two dice are rolled. $P($product is odd$) = ?$
11. Three coins are tossed. $P($no tails$) = ?$
12. A coin is tossed $n$ times. $P($all heads$) = ?$ (express in terms of $n$).

Pitfalls / Insight

(1) For two coins or two dice, treat them as ordered (first and second). The sample space is $4$ or $36$, not $3$ or $21$.

(2) For "at least" events, often easier to use complement: $P($at least one$) = 1 - P($none$)$.

(3) For dice sums, memorise the triangular distribution. The sum $7$ has the most ways ($6$); the further from $7$, the fewer ways.

(4) Always reduce fractions to lowest terms. The board examiner expects $1/12$, not $3/36$.