Re^3: [OT] Stats problem

What are the odds of a card ending up in its standard position?

That's easy enough - it's 1 in 52.
A card can be in any one of 52 positions, only one of which is its "standard position" - and there's no bias that favours one position over another.

UPDATE What follows doesn't scale correctly ... it aint right IOW - probably the linkage that salva was getting at.

The chance that none of the cards are in their standard position is simply (51/52) ** 52, which is around 0.3643 or approximately 1 in 3.
Therefore the chance that 1 or more cards are in their standard position is 1 - ((51/52) ** 52) which is around 0.6357 or approximately 2 in 3.
(You could turn this into a rather boring game of patience/solitaire.)

As for what you're actually trying to calculate, I'm still trying ot get my head around it.
(Please stick to cards in future ;-)

Cheers,
Rob

Comment on Re^3: [OT] Stats problem

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Re^4: [OT] Stats problem by Anonymous Monk on Feb 26, 2015 at 15:17 UTC
Just for fun, a simulation to confirm: use List::Util qw/shuffle sum/; my $RUNS = 1_000_000; my %counts; for (1..$RUNS) { my @deck = shuffle 0..51; my $in_correct_loc = grep {$_==$deck[$_]} 0..51; $counts{$in_correct_loc}++; } print "Number of cards in the correct position: % of runs\n"; printf "% 4d: %.4f%%\n", $_, 100$counts{$_}/$RUNS for sort {$a<=>$b} keys %counts; my $one_or_more = sum map {$_//0} @counts{1..52}; printf "1-52: %.4f%%\n", 100$one_or_more/$RUNS; __END__ Number of cards in the correct position: % of runs 0: 36.8130% 1: 36.8086% 2: 18.3846% 3: 6.1085% 4: 1.5349% 5: 0.2926% 6: 0.0487% 7: 0.0080% 8: 0.0010% 9: 0.0001% 1-52: 63.1870% [download]	[reply] [d/l]

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Re^4: [OT] Stats problem
by Anonymous Monk on Feb 26, 2015 at 15:17 UTC

Just for fun, a simulation to confirm:

use List::Util qw/shuffle sum/;
my $RUNS = 1_000_000;
my %counts;
for (1..$RUNS) {
    my @deck = shuffle 0..51;
    my $in_correct_loc = grep {$_==$deck[$_]} 0..51;
    $counts{$in_correct_loc}++;
}
print "Number of cards in the correct position: % of runs\n";
printf "% 4d: %.4f%%\n", $_, 100*$counts{$_}/$RUNS
    for sort {$a<=>$b} keys %counts;
my $one_or_more = sum map {$_//0} @counts{1..52};
printf "1-52: %.4f%%\n", 100*$one_or_more/$RUNS;

__END__

Number of cards in the correct position: % of runs
   0: 36.8130%
   1: 36.8086%
   2: 18.3846%
   3: 6.1085%
   4: 1.5349%
   5: 0.2926%
   6: 0.0487%
   7: 0.0080%
   8: 0.0010%
   9: 0.0001%
1-52: 63.1870%
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