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 A136300 Numerator of ratio (the denominator being (n-1)!^2 = A001044(n-1)) giving the probability that the last of n persons drawing names randomly from a set of names draws their own, given that each person previously has drawn in succession and did not keep their own name. (Probability of derangements when allocated / rejected sequentially.) 4
 1, 0, 1, 5, 76, 1624, 52116, 2298708, 133929216, 9961180416, 921248743680, 103715841415680, 13967643016085760, 2217449301162071040, 409861043056032503040, 87262626872384643052800, 21202798521768886355558400, 5831660090586059239329792000, 1802587564536011525042697830400 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,4 COMMENTS In "Secret Santa", if a person picks their own name, they pick another name and they throw their own name back in. If the last person draws their own name, there's a problem. What is that probability as a function of the number of people participating? LINKS Brian Parsonnet, Table of n, a(n) for n = 1..30 Brian Parsonnet, Probability of Derangements FORMULA Sum of H(i, N-2) * X(i, N-2) for i=0..2^(N-3), N is the number of people and H(r,c) = sum of H(T(r),L(r)+j) * M(c-T(r)-1,j) for j = 0..c-L(r)-1 and X(r,c) = product of (3 + k - b(r,k)) for k = 0..(c-2) and M(y,z) = binomial distribution (y,z) when y - 1 > z and (y,z)-1 when (y-1)<=z and b(r,k) = bit k of r in base 2 and T(r) = A053645 and L(r) = A000523. a(n) = (n-1)!*A102262(n)/A102263(n) for n > 1. EXAMPLE If there is one person, the chance of the last person getting their own name is 100%, or 1 over 0!^2. For 2 people, it is 0 / 1!^2. For 3 people, it is 1 / 2!^2, creating a more interesting case. The possible drawings are {2,1,3}, {2,3,1} and {3,1,2}. All other drawings can't happen because the name is rejected and redrawn. But these 3 outcomes don't have equal probability, rather, they are 25%, 25% and 50% respectively. The first outcome is the only one in which the last person draws their own name. The first person has a 50% chance of drawing a 2 or 3. If 2, the second person has a 50% chance of drawing 1 or 3, for a total outcome probability of 1/4. Similarly with 4 people, the chance is 5/36, followed by 76/576 for 5 people, etc. For the case of 5 people, the above equations boil down to this end calculation: {1,5,2,1} * {12,8,9,6} summed, or 12 + 40 + 18 + 6 = 76. MATHEMATICA maxP = 22; rows = Range[1, 2^(nP = maxP - 3)]; pasc = Table[ Binomial[p + 1, i] - If[i >= p, 1, 0], {p, nP}, {i, 0, p}]; sFreq = Table[0, {maxP - 1}, {2^nP}]; sFreq[[2 ;; maxP - 1, 1]] = 1; For[p = 1, p <= nP, p++, For[s = 1, s <= p, s++, rS = Range[2^(s - 1) + 1, 2^s]; sFreq[[p + 2, rS]] = pasc[[p + 1 - s, 1 ;; p + 2 - s]] . sFreq[[s ;; p + 1, 1 ;; 2^(s - 1)]]]]; sProb = Table[p + 2 - BitGet[rows - 1, p - 1], {p, nP}]; sProb = Table[Product[sProb[[i]], {i, p}], {p, nP}]* Table[If[r <= 2^p, 1, 0], {p, nP}, {r, rows}]; rslt = Flatten[ Prepend[Table[sProb[[p]] . sFreq[[p + 2]], {p, nP}], {1, 0, 1}]] prob = N[rslt/Array[(#1 - 1)!^2 & , maxP]] (* Brian Parsonnet, Feb 22 2011 *) CROSSREFS The sequence frequency table (sFreq) is A136301. Cf. A000523, A053645, A102262, A102263. Sequence in context: A364323 A011918 A209095 * A196689 A196725 A197063 Adjacent sequences: A136297 A136298 A136299 * A136301 A136302 A136303 KEYWORD nonn AUTHOR Brian Parsonnet, Mar 22 2008 EXTENSIONS Corrected and extended to a(19) by Brian Parsonnet, Feb 22 2011 STATUS approved

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Last modified April 14 14:09 EDT 2024. Contains 371665 sequences. (Running on oeis4.)