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### Re^3: [OT] Stats problem

by syphilis (Bishop)
 on Feb 26, 2015 at 14:04 UTC ( #1117954=note: print w/replies, xml ) Need Help??

in reply to Re^2: [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

Replies are listed 'Best First'.
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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