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A287150
Fairest turn sequence for 3 players where the probability of a win for a player on his turn approaches 0.
0
0, 1, 2, 2, 1, 0, 2, 1, 0, 0, 1, 2, 1, 2, 0, 0, 2, 1, 0, 2, 1, 1, 2, 0, 2, 1, 0, 0, 1, 2, 1, 0, 2, 2, 0, 1, 0, 2, 1, 1, 2, 0, 2, 1, 0, 0, 1, 2, 1, 0, 2, 2, 0, 1, 0, 2, 1, 1, 2, 0, 1, 2, 0, 0, 2, 1, 0, 2, 1, 1, 2, 0, 2, 1, 0, 0, 1, 2, 0, 1, 2, 2, 1, 0, 2, 1, 0, 0, 1, 2, 1, 0, 2, 2, 0, 1, 2, 0, 1, 1, 0, 2, 1, 0, 2, 2, 0, 1, 0, 2, 1, 1, 2, 0, 1, 2, 0, 0, 2, 1
OFFSET
0,3
COMMENTS
0, 1, and 2 are three archers of equal skill, but which have infinitesimally low chances of hitting a target. Still, they try: taking turns until one of them hits: 0 goes first, if 0 misses then 1 will try (these are the first two terms). Subsequent terms are found by giving a turn to the player who has the lowest probability of winning so far.
Joshua Cooper and Aaron Dutle showed that for two players this is the Thue-Morse sequence (A010060).
The following observations are made and verified to be true for the first 5000 terms:
- Every 3 terms completes a new round with each player going once per round.
- Every other round is the same as the last but in reverse order.
- The 48th-101st terms form a block of 54 terms which are repeated from then on.
LINKS
Joshua Cooper and Aaron Dutle, Greedy Galois Games, Amer. Math. Monthly, 120 (2013), 441-451, arXiv:1110.1137 [math.CO].
Daniel Hug, Generate fairest turn sequence for n players (web app written in JavaScript)
MATHEMATICA
n=3; a=Range[n]; pw=(1-p)^(Range[n]-1)*p;
Do[
next = First[Select[Range[n], And@@NonNegative/@Limit[Sign[pw-pw[[#]]], p->0]&, 1]];
AppendTo[a, next]; pw[[next]]+=p*(1-p)^k
, {k, n, 50}];
a-1
(* Andrey Zabolotskiy, Jun 05 2017 *)
CROSSREFS
Cf. A010060.
Sequence in context: A328081 A194853 A309866 * A257109 A096830 A319780
KEYWORD
nonn
AUTHOR
Daniel Hug, May 20 2017
STATUS
approved