Question: Five integers are selected uniformly at random from the set $\{0, 1, 2, \ldots, 50\}$. Let $ z $ be the sum of the two largest selected integers. What is the probability that $ z \geq 100 $?

Fri, 01 May 2026 03:56:36 +0000

["Title: Probability That the Sum of the Two Largest Random Integers Is at Least 100 (Based on five integers drawn uniformly from {0, 1, ..., 50})", "---", "Introduction When selecting five integers uniformly at random from the set $\{0, 1, 2, \ldots, 50\}$, a fascinating question emerges: What is the probability that the sum of the two largest values among them — denoted $ z $ — is at least 100? This problem combines combinatorics with probabilistic reasoning and offers insight into random sampling over a bounded discrete set. Whether you're analyzing risk events, game theory, or sampling distributions, understanding this probability helps inform decision-making under uncertainty.", "This article explores the problem step by step, explains the key ideas, and delivers a close-to-exact analytical or computational approach to estimate the desired probability.", "---", "### Understanding the Problem", "We select $ n = 5 $ integers independently and uniformly at random from $$ S = \{0, 1, 2, \ldots, 50\}. $$ Let these integers be $ X_{(1)} \leq X_{(2)} \leq X_{(3)} \leq X_{(4)} \leq X_{(5)} $ — the order statistics of the sample. Define $ z = X_{(4)} + X_{(5)} $, the sum of the two largest values.", "We seek: $$ \mathbb{P}(z \geq 100) = \mathbb{P}(X_{(4)} + X_{(5)} \geq 100). $$", "Since the maximum possible value is 50, the sum $ X_{(4)} + X_{(5)} $ can be at most 100 (when both are 50). Thus, $ z \geq 100 $ only if both the fourth and fifth largest values are 50 — impossible unless at least two of the five values are 50, and specifically, both the fourth and fifth are 50, meaning 50 appears at least twice, and the fourth and fifth are 50.", "In fact, for $ z = X_{(4)} + X_{(5)} \geq 100 $, both $ X_{(4)} $ and $ X_{(5)} $ must be at least 50, but since $ X_i \leq 50 $, this forces $$ X_{(4)} = X_{(5)} = 50. $$ Then $ z = 50 + 50 = 100 $, so we are looking for $$ \mathbb{P}(X_{(4)} = 50 \ ext{ and } X_{(5)} = 50) \quad \ ext{and} \quad \ ext{at least two 50s in the sample}. $$ But since $ X_{(5)} = 50 $ means all values $ \leq 50 $, and $ X_{(4)} = 50 $ means at least four values are $ \geq 50 $. Since 50 is the max, $ X_{(4)} = 50 $ and $ X_{(5)} = 50 $ implies both $ X_{(4)} $ and $ X_{(5)} $ are 50, which means all five values must be 50, or at least four of them are 50 and the rest are ≤ 50 — but since 50 is the top, if four values are 50, then $ X_{(4)} = X_{(5)} = 50 $.", "So, correction: $ z \geq 100 $ if and only if $ X_{(4)} = 50 $ and $ X_{(5)} = 50 $, which happens precisely when at least four of the five numbers are 50.", "But wait: if only three values are 50, then the ordered statistics satisfy $ X_{(4)} \leq 50 $, $ X_{(5)} < 50 $ unless the fourth and fifth (i.e., last two) are both 50.", "Thus, $ X_{(4)} = 50 $ means at least four values are $ \geq 50 $ → all are exactly 50, because values ≤ 50. So $ X_{(4)} = 50 $ ⇔ all five values are $ \leq 50 $ and at least four are $ \geq 50 $ → all five are exactly 50.", "Wait — no: if four are 50 and one is less, then $ X_{(4)} = 50 $? No.", "Let’s clarify: - The fifth order statistic $ X_{(5)} $ is the largest. - $ X_{(4)} $ is the second largest.", "So $ X_{(4)} = 50 $ means at least four of the five values are $ \geq 50 $. Since values ≤ 50, this forces $ X_{(4)} = 50 $ ⇔ all five values are $ \geq 50 $ → all values are 50.", "Because maximum is 50, the only way $ X_{(4)} = 50 $ is if all five values are 50 — otherwise, $ X_{(4)} \leq 49 $.", "Similarly, $ X_{(5)} = 50 $ always holds (since max is 50), but $ X_{(4)} + X_{(5)} = z \geq 100 $ requires $ X_{(4)} \geq 50 $, so $ X_{(4)} = 50 $ and $ X_{(5)} = 50 $.", "But $ X_{(5)} = 50 $ always, so $ z = X_{(4)} + 50 $. So $ z \geq 100 $ ⇔ $ X_{(4)} \geq 50 $ ⇔ $ X_{(4)} = 50 $, which again implies all five values are $ \geq 50 $, hence all equal to 50 (since max is 50).", "So $ z \geq 100 $ iff all five selected integers are 50.", "But is that true?", "Wait: suppose two values are 50, and the others are less. Then $ X_{(4)} \leq 50 $, $ X_{(5)} \leq 50 $, but $ X_{(4)} $ could be 49. So maximum possible $ X_{(4)} + X_{(5)} $ when two values are 50 is: Say $ X_{(5)} = 50 $, $ X_{(4)} = 50 $ only if at least four values are 50.", "But suppose values are: 50, 50, 50, 40, 30 → then $ X_{(4)} = 50 $, $ X_{(5)} = 50 $, so $ z = 100 $. Yes.", "But if values are 50,50,40,40,30 → still $ X_{(4)} = 40 $, $ X_{(5)} = 50 $, $ z = 90 < 100 $.", "So $ z \geq 100 $ requires $ X_{(4)} + X_{(5)} \geq 100 $. Since $ X_i \leq 50 $, we must have $ X_{(4)} \geq 50 $ or $ X_{(5)} \geq 50 $, but both ≤ 50 → sum ≥ 100 only if $ X_{(4)} \geq 50 $ and $ X_{(5)} \geq 50 $? No — sum of two numbers ≤ 100, each ≤ 50, sum ≤ 100.", "Maximum sum of two values is 100 (50+50). So $ z \leq 100 $. Thus $ z \geq 100 $ iff $ z = 100 $.", "And $ z = 100 $ iff $ X_{(4)} + X_{(5)} = 100 $. Since both ≤ 50, this requires $$ X_{(4)} = 50 \quad \ ext{and} \quad X_{(5)} = 50. $$", "- $ X_{(5)} = 50 $ ⇔ at least four values are $ \geq 50 $ ⇔ at least four values are 50 (since values ≤ 50). - $ X_{(4)} = 50 $ ⇒ at least four values are $ \geq 50 $ → same condition.", "So both conditions collapse to: at least four of the five numbers are 50.", "Let’s verify:", "- If exactly four values are 50: ordered: $ a \leq 50 $, $ b = 50 $, $ c = 50 $, $ d = 50 $, $ e = 50 $ → but values ≤ 50, so $ e = 50 $. Then $ X_{(4)} = 50 $, $ X_{(5)} = 50 $, $ z = 100 $. - If all five are 50: same, $ z = 100 $. - If fewer than four 50s: at most three 50s → then at most three values ≥ 50 → $ X_{(5)} \leq 49 $ → $ z \leq 49 + 49 = 98 < 100 $.", "Therefore, $$ \mathbb{P}(z \geq 100) = \mathbb{P}(\ ext{at least four of the five numbers are 50}). $$", "---", "### Computation of Probability", "Each integer is selected independently and uniformly from $\{0,1,2,\ldots,50\}$ → 51 possible values.", "Let $ p = \frac{1}{51} $ be the probability that a single draw is 50.", "Let $ q = 1 - p = \frac{50}{51} $.", "We want the probability that at least four of the five draws are 50.", "This is a binomial probability: $$ \mathbb{P}(\ ext{2 or more 50s}) = \mathbb{P}(X = 4) + \mathbb{P}(X = 5), $$ where $ X \sim \ ext{Binomial}(5, \frac{1}{51}) $.", "Compute:", "1. $ \mathbb{P}(X = 5) = \binom{5}{5} p^5 q^0 = \left(\frac{1}{51}\right)^5 \approx \frac{1}{357911} $", "2. $ \mathbb{P}(X = 4) = \binom{5}{4} p^4 q^1 = 5 \cdot \left(\frac{1}{51}\right)^4 \cdot \frac{50}{51} = 5 \cdot \frac{50}{51^5} $", "So total: $$ \mathbb{P}(z \geq 100) = \frac{5 \cdot 50}{51^5} + \frac{1}{51^5} = \frac{250 + 1}{51^5} = \frac{251}{51^5} $$", "Now compute $ 51^5 $:", "- $ 51^2 = 2601 $ - $ 51^3 = 51 \cdot 2601 = 132651 $ - $ 51^4 = 51 \cdot 132651 = 680 reap1000 $ Better: $ 132651 \cdot 51 $", "Break: $ 132651 \cdot 50 = 6,632,550 $ $ 132651 \cdot 1 = 132,651 $ Sum: $ 6,765,201 $", "Then $ 51^4 = 6,765,201 $ Now $ 51^5 = 6,765,201 \cdot 51 $", "Compute: $ 6,765,201 \cdot 50 = 338,260,050 $ $ 6,765,201 \cdot 1 = 6,765,201 $ Sum: $ 345,025,251 $", "Thus: $$ \mathbb{P}(z \geq 100) = \frac{251}{345,025,251} $$", "This fraction is already in simplest form: 251 is prime (check: not divisible by primes ≤ √251 ≈ 15.8 → not divisible by 2,3,5,7,11,13), and 345,025,251 not divisible by 251 (do not divide), so fraction is reduced.", "---", "### Conclusion", "The probability that the sum of the two largest among five randomly selected integers from $\{0, 1, \ldots, 50\}$ is at least 100 is exactly $$ \frac{251}{345,025,251} $$", "This result explains why such high-sum events are exceedingly rare — requiring at least four 50s, a near-ceiling event in this bounded distribution. Understanding event probabilities helps in risk modeling, simulation design, and statistical inference.", "---", "Keywords: probability, binomial distribution, order statistics, five random integers, sum of two largest, combinatorics, discrete probability, $X_{(4)} + X_{(5)} \geq 100$, $\{0,1,\ldots,50\}$, $ \frac{251}{51^5} $", "---", "Further Reading: - Order statistics in uniform discrete distributions - Extreme value analysis for bounded samples - Probability of rare events in sampling with replacement", "---", "Note: For numerical intuition, $ \frac{251}{345,025,251} \approx 7.27 \ imes 10^{-7} $, an extremely small chance."]