A394762 Minimum number of draws without replacement from an urn with n white and n black balls to ensure the probability of drawing at least one ball of each color is >= n/(n+1).

2, 2, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8

Offset

1,1

Comments

Equivalently, the smallest number of draws k such that the probability of drawing balls of all the same color is <= 1/(n+1).

The sequence grows logarithmically; asymptotically, a(n) ~ log_2(n).

Links

Table of n, a(n) for n=1..87.

Formula

a(n) = min { k >= 1 : 2 * binomial(n, k) / binomial(2*n, k) <= 1/(n+1) }.

Mathematica

a[n_]:=Module[{k=1}, While[2Binomial[n, k]/Binomial[2n, k]>1/(n+1), k++]; k]; Array[a, 75] (* Stefano Spezia, Apr 02 2026 *)

Prog

(PARI) a(n) = for(k=1, oo, if(2*binomial(n, k)*(n+1)<=binomial(2*n, k), return(k))) \\ Jason Yuen, Mar 31 2026
(R)
a <- numeric(100)
for(n in 1:100) {
  ### The threshold for drawing all the same color
  threshold <- 1 / (n + 1)
  for(k in 1:(2*n)) {
    ### P(all same color) = 2 * choose(n, k) / choose(2*n, k)
    if(k <= n) p_same <- 2 * choose(n, k) / choose(2 * n, k)
    else p_same <- 0
    ### Check if the probability meets the threshold condition
    if(p_same <= threshold) {a[n] <- k; break}
  }
}
cat(paste(a, collapse = ", "), "\n")

Crossrefs

Sequence in context: A137300 A201052 A278044 * A372475 A255121 A095791

Adjacent sequences: A394759 A394760 A394761 * A394763 A394764 A394765

Keyword

nonn

Author

Michael Shmoish, Mar 31 2026

Status

approved