[lapm] RE: [LA.pm] The odds of finger games

[Robert Spier]

> oblapm: Robert, did you also calculate the probability 
> distribution of your single finger, binary, modulus, algorithm?  
> If I understand it correctly, I think it will give some lower 
> numbered participants an unfair advantage unless the number of 
> participants is a power of 2.

0: 11
1: 11
2: 11
3: 11
4: 10
5: 10

You are correct, sir.

Also, there are some subtle issues in my first program.  It works
correctly because the numbers happened to work out.  (6 states, 6
people.)  It wasn't as general as it should have been, probably
because I wrote it in a post-lunch stupor.  

This version only supports binary digits.  

use strict;
use warnings;
use Set::CrossProduct;
my $states = 2; # 0 fingers, 1 finger
my $hand = [0..$states -1];
my $people = 6;
my $it = Set::CrossProduct->new( [($hand)x $people] );
my %sums = ();
my %modsums = ();

while (my $tuple = $it->get) {
  my $n = join "",@$tuple;
  my $t = unpack("N",pack("B32",substr("0" x 32 . $n,-32)));
  $sums{$t}++;
  $modsums{$t%$people}++;
}

# Output non-mod-ed values (sum of all fingers)
#print "$_: $sums{$_}\n"
#  for (sort {$a <=> $b} keys %sums);

# Output mod-ed values, i.e. "people"
print "$_: $modsums{$_}\n"
  for (sort {$a <=> $b} keys %modsums);


And this version is the general purpose brute force solution:

use strict;
use warnings;
use Set::CrossProduct;
use Math::BaseCalc;
#my $states = 2; # 0 fingers, 1 finger
my $states = 10;
my $hand = [0..$states -1];
my $people = 6;
my $it = Set::CrossProduct->new( [($hand)x $people] );
my $ba = new Math::BaseCalc(digits => $hand);
my %sums = ();
my %modsums = ();

while (my $tuple = $it->get) {
  my $t = $ba->from_base( join "",@$tuple );
  $sums{$t}++;
  $modsums{$t%$people}++;
}

# Output non-mod-ed values (sum of all fingers)
#print "$_: $sums{$_}\n"
#  for (sort {$a <=> $b} keys %sums);

# Output mod-ed values, i.e. "people"
print "$_: $modsums{$_}\n"
  for (sort {$a <=> $b} keys %modsums);