OFFSET
2,3
COMMENTS
This is a version of the Secret Santa game.
n friends organize a gift exchange. The n names are put into a hat and the first person draws one. If she picks her own name, then she returns it to the bag and draws again, repeating until she has a name that is not her own. Then the second person draws, again returning his own name if it is drawn. This continues down the line. What is the probability p(n) that when the n-th person draws, only her own name will be left in the bag?
I heard about the problem from Gary Thompson at Grove City College in PA.
LINKS
Jon E. Schoenfield, Table of n, a(n) for n = 2..389
Math Forum at Drexel, A variant on the "Secret Santa"
FORMULA
From Jon E. Schoenfield, Sep 30 2006: (Start)
p(n) = Sum_{i=1..n-2} t(n,i)/(n-1)!^2
where
t(n,i) = (n-2)*i^2/(i-1)*t(n-1,i-1) - (n-i-2)*t(n-1,i) for 1 < i < n-1;
t(n,1) = (-1)^(n-1)*(n-1)!/2 for i = 1 and n > 2;
t(n,i) = 0 otherwise.
(End)
Based on the values of p(n) for n <= 1000, it seems plausible that, as n increases, p(n) approaches 1/(n + log(n) + EulerGamma), where EulerGamma = 0.5772156649015... (the Euler-Mascheroni constant). - Jon E. Schoenfield, Dec 11 2021
EXAMPLE
p(2) through p(10) are 0, 1/4, 5/36, 19/144, 203/1800, 4343/43200, 63853/705600, 58129/705600, 160127/2116800.
PROG
(Magma) N:=21; a:=[]; row:=[]; T:=[]; for n in [2..N] do row[n-1]:=0; T[n]:=row; T[n][1]:=(-1)^(n-1)*Factorial(n-1) div 2; for i in [2..n-2] do T[n][i]:=(n-2)*i^2/(i-1)*T[n-1][i-1]-(n-i-2)*T[n-1][i]; end for; p:=0; for i in [1..n-2] do p+:=T[n][i]/Factorial(n-1)^2; end for; a[#a+1]:=Numerator(p); end for; a; // Jon E. Schoenfield, Dec 10 2021
CROSSREFS
KEYWORD
nonn,frac
AUTHOR
Jerrold Grossman, Feb 17 2005
EXTENSIONS
More terms from Jon E. Schoenfield, Sep 30 2006
STATUS
approved