IOAI Training: Learn AI by solving problems

The problem in brief

You roll an ordinary six-sided die three times. What is the probability that at least one of the three rolls shows a six? The answer must be written as a decimal with four digits after the point. It is a warm-up question of the Fizmat AI Olympiad 2025 (Kazakhstan) online selection, and its real lesson is a reflex you will use constantly: when "at least one" appears, look at the opposite event.

How you are scored

One exact answer, worth 10 points. Ours, checked four independent ways in solution.py:

\boxed{0.4213}

The key idea

"At least one six" is the complement of "no six at all", and "no six" is easy because the rolls are independent. One roll avoids the six with probability $5/6$ ; three independent rolls all avoid it with probability $(5/6)^3$ . Therefore

P(\text{at least one six}) = 1 - P(\text{no six}) = 1 - \left(\frac{5}{6}\right)^3 = 1 - \frac{125}{216} = \frac{91}{216} \approx 0.421296 .

Rounded to four decimals: $0.4213$ .

Why is the direct route harder? "At least one six" splits into many cases: exactly one six, exactly two, exactly three, and each case in several positions. The opposite event is a single, clean case. The complement turns an "or" of many overlapping events into an "and" of independent ones, and independent "and"s multiply.

Checking it three other ways

Counting outcomes. The $6^3 = 216$ ordered results are equally likely (the classical definition $P(A) = m/n$ of the statement). The outcomes without any six are $5^3 = 125$ , so $m = 216 - 125 = 91$ . Splitting by the number $k$ of sixes, $\binom{3}{k} 5^{3-k}$ outcomes have exactly $k$ sixes:

sixes $k$	0	1	2	3	total
outcomes	125	75	15	1	216

and $75 + 15 + 1 = 91$ , as expected.

Inclusion-exclusion. Let $A_i$ be "roll $i$ is a six". Adding $P(A_1) + P(A_2) + P(A_3) = 3/6$ counts the outcomes with two sixes twice and the one with three sixes three times, so we correct:

P(A_1 \cup A_2 \cup A_3) = 3 \cdot \frac{1}{6} - 3 \cdot \frac{1}{36} + \frac{1}{216} = \frac{108 - 18 + 1}{216} = \frac{91}{216}.

Simulation. One million simulated triples of rolls give $0.4211 \pm 0.0010$ (95 % interval), which contains $0.4213$ .

The classic trap

The tempting answer is $3 \times \frac{1}{6} = 0.5$ . It adds the three events as if they could not happen together, but rolling two sixes is counted twice. The figure shows how badly this goes as the number of rolls grows: with 6 rolls the naive rule promises certainty, with 12 rolls a "probability" of 2.

Probability of at least one six in n rolls: the exact value 1 - (5/6)^n rises towards 1 but never reaches it, while the naive n/6 grows without limit and passes 1 at n = 6. At n = 3 the exact value is 91/216 = 0.4213.

This exact confusion puzzled the Chevalier de Méré in the 17th century and led Pascal and Fermat to lay the foundations of probability theory.

Results

approach	value	verdict
complement, $1 - (5/6)^3$	0.4213	correct
counting 91 of 216 outcomes	0.4213	correct, agrees
inclusion-exclusion	0.4213	correct, agrees
Monte Carlo, $10^6$ triples	$0.4211 \pm 0.0010$	agrees
adding probabilities, $3/6$	0.5000	wrong: double counting
exactly one six only, $75/216$	0.3472	wrong: "at least one" includes two and three sixes
$(1/6)^3$	0.0046	wrong: that is "three sixes"

What to remember

"At least one" means "use the complement": $P(\text{at least one}) = 1 - P(\text{none})$ .
Independent events multiply: $P(\text{no six in } n \text{ rolls}) = (5/6)^n$ .
Probabilities of overlapping events do not add; inclusion-exclusion corrects the double counting.
The same formula is everywhere in ML: if each of $n$ independent models (or test cases, or attempts) succeeds with probability $p$ , at least one succeeds with probability $1 - (1-p)^n$ . It is why "best of $n$ samples" from a language model (pass@n) climbs so fast, and why testing many hypotheses at once produces false discoveries.

"""FAIO 2025 (Kazakhstan) qualification, Problem 2 "Probability Theory 101".

Three rolls of a fair die: probability of at least one six, 4 decimals.
Four independent computations (complement, enumeration, inclusion-exclusion,
Monte Carlo) and the figure of the correction.  Run:  python solution.py
"""
from fractions import Fraction as F
from itertools import product
import math
import random
import matplotlib
matplotlib.use("Agg")
import matplotlib.pyplot as plt

# 1. complement: "at least one six" is the opposite of "no six at all"
p_complement = 1 - F(5, 6) ** 3

# 2. brute force over the 6^3 = 216 equally likely outcomes
outcomes = list(product(range(1, 7), repeat=3))
p_enum = F(sum(6 in o for o in outcomes), len(outcomes))

# 3. inclusion-exclusion over the events A_i = "roll i is a six"
p_ie = 3 * F(1, 6) - 3 * F(1, 36) + F(1, 216)

# exactly k sixes (binomial), to see where the 91 outcomes come from
exact = {k: math.comb(3, k) * 5 ** (3 - k) for k in range(4)}     # counts out of 216
assert sum(exact[k] for k in (1, 2, 3)) == 91

assert p_complement == p_enum == p_ie == F(91, 216)
print("exact:", p_enum, "=", float(p_enum), "; counts of k sixes:", exact)

# 4. Monte Carlo
rng = random.Random(0); N = 1_000_000
hits = sum(any(rng.randint(1, 6) == 6 for _ in range(3)) for _ in range(N))
p_mc = hits / N; half = 1.96 * math.sqrt(p_mc * (1 - p_mc) / N)
print(f"Monte Carlo: {p_mc:.4f} +- {half:.4f}")
print("ANSWER:", f"{float(p_enum):.4f}")              # 0.4213

# figure: the complement rule against the tempting n/6
plt.rcParams.update({"font.size": 12, "axes.spines.top": False, "axes.spines.right": False})
ns = list(range(1, 13))
fig, ax = plt.subplots(figsize=(9, 4.4), dpi=150)
ax.plot(ns, [1 - (5 / 6) ** n for n in ns], "o-", color="#2e86ab", lw=2.2,
        label="exact: 1 - (5/6)^n")
ax.plot(ns, [n / 6 for n in ns], "s--", color="#d1495b", lw=1.8,
        label="wrong: n/6 (adds overlapping cases)")
ax.axhline(1, color="#999", lw=1)
ax.scatter([3], [91 / 216], s=140, facecolor="none", edgecolor="black", lw=2, zorder=5)
ax.annotate("n = 3: 91/216 = 0.4213", (3, 91 / 216), xytext=(4.5, 0.2),
            arrowprops=dict(arrowstyle="->"))
ax.set_xlabel("number of rolls n"); ax.set_ylabel("P(at least one six)")
ax.set_xticks(ns); ax.set_ylim(0, 2.05)
ax.set_title("At least one six in n rolls")
ax.legend(frameon=False, loc="upper left")
fig.tight_layout(); fig.savefig("figures/at_least_one.png", facecolor="white")
print("figure written")

Problem 2: Probability Theory 101