Puzzle Archive

October 2020 solution

The probability that every child would have strictly more of a candy than each other trick-or-treater was precisely 318281087/8016470462, which comes to just under 4%.  The most straightforward way to compute this was to assign a particular candy to each child and compute the probability of that (writing some code certainly helped here) and then multiplying the result by 120 to account for the disjoint other assignments of candy to trick-or-treater.

Congrats to the following solvers who managed to submit the precise probability correctly!

Guillermo Wildschut
Rahul Saxena
Harrison W
Kilian B.
Mrs Nesbitt
Glauber Guarinello
Dimas Ramos
Keith Schneider
Alex Lang
Carl Placid
Conor Mac Amhlaoibh
Sébastien Geeraert
Konstantin Vladimirov
Jonah Goldstein
Walter Sebastian Gisler
Senthil Rajasekaran
Will C
Ben Reiniger
Neil Thistlethwaite
Konstantin Gukov
Alen Abdrakhmanov
Sergey Serebryakov
Rishabh Vaid
Yongxing Wang
Karl Mahlburg
Leif Metcalf
Maria Kormacheva
Edward Wall
Ian Sleightholme
Calvin Pozderac
Alex Kalbach
Will Liao
David Molay
Nikolay Zakharov
Tomer Tzadok
Michael Ore
Kelvin Davis
Andrew Argatkiny
Henry Heffan
Sergey Tychinin
Stefan Glock
Martin Hesselborn
Lazar Ilic
Aleksandar Bukva
Janko Sustersic
Sean Egan
Iman Hosseini
Jordi G Rodríguez
Valerii Gusman
Jiba Da
Anton Puhach