In wine tasting, you count Each item (+/,) hit between two sets.
It is different from Any-item (+/+./) or All-items (+/*./) matching.
Let's look at All-items case, similar could applied to other cases.
We have permutations (?~). And when considering All-items match
permutation can be replaced with its index.
So what is the universe and match events?
For n=2 items we have:
0 : 0 1 indices
1 : 1 0
0 0 0 1 toss pairs
1 0 1 1
There are 2 matches out of 4 tosses, a 0.5 chance.
For number of n=3 items.
([EMAIL PROTECTED] A. i.)3
0 1 2
0 2 1
1 0 2
1 2 0
2 0 1
2 1 0
we have 6 = !3 permutations
<"2,"0/~i.6
+---+---+---+---+---+---+
|0 0|1 0|2 0|3 0|4 0|5 0|
|0 1|1 1|2 1|3 1|4 1|5 1|
|0 2|1 2|2 2|3 2|4 2|5 2|
|0 3|1 3|2 3|3 3|4 3|5 3|
|0 4|1 4|2 4|3 4|4 4|5 4|
|0 5|1 5|2 5|3 5|4 5|5 5|
+---+---+---+---+---+---+
we notice that total number of pairs is
(!n)^2 and total matches is !n. So probability
is 1%!n. Here it's 1%6.
$ , /,"0/~i.!3
36 2
+/=/|:,/,"0/~i.!3
6
6%36
0.166667
%!3
0.166667
For n=5,
$ , /,"0/~i.!5
14400 2
+/=/|:,/,"0/~i.!5
120
(, *:) !5
120 14400
x:(% *:) !5
1r120
Experimentally, with All-items compare, matches reduce
proportional to ! of set size.
(!2)* +/-:"1/?~2 1000$2
1032
(!2)* +/-:"1/?~2 1000$2
982
(!3)* +/-:"1/?~2 1000$3
1044
(!3)* +/-:"1/?~2 1000$3
906
(!5)* +/-:"1/?~2 1000$5
1440
(!5)* +/-:"1/?~2 1000$5
600
----- Original Message ----
> From: Devon McCormick <[EMAIL PROTECTED]>
>
> Members of the Forum -
>
> I had a blind-tasting last night where we sampled five different wines.
> After two rounds of tasting the wines in the same order, we had a round
> where we sampled them in a random, unknown order. Afterwards, we tried to
> guess which wines from the random order corresponded to their original
> numbering. On average, we were able to do this only once each.
>
> Not knowing how this shakes out by chance, I did the following to simulate
> comparing pairs of random permutations of five items to see how our results
> compared to random selection.
>
> ?~5 NB. 5-permutation
> 0 4 3 1 2
> $rs=. ?~2 1000$5 NB. 1000 pairs of 5-permutations
> 2 1000 5
> 0{=/rs NB. Look at the first comparison
> 1 0 0 0 0
> 0{"2 rs NB. and permutation pair.
> 1 0 4 3 2
> 1 3 2 0 4
> +/,=/rs NB. How many total matches?
> 992 NB. Average of 1.
>
> So, our ability to recognize the wines we'd just tasted twice is no better
> than random. It occurred to us that perhaps five is too many to distinguish
> and maybe we should taste only two at a time if we do this again.
>
> So, what would be the random match if we did this with only two wines?
>
> +/,=/?~2 1000$2
> 986
>
> Huh? It's the same: about one on average. Try this for other permutation
> lengths....
>
> +/,=/?~2 1000$3
> 992
> +/,=/?~2 1000$10
> 982
>
> Try larger sample size too.
>
> +/,=/?~2 10000$5
> 9886
> +/,=/?~2 10000$10
> 10185
> +/,=/?~2 10000$20
> 10058
> +/,=/?~2 10000$100
> 10049
>
> No matter what size the permutation, the average chance of a match is about
> one. This seems counter-intuitive to me. Am I doing something wrong in my J
> code or is this one of those well-known theorems of statistics about which
> I'm completely ignorant?
>
> You decide.
>
> Regards,
>
> Devon McCormick
> ^me^ at acm.
> org is my
> preferred e-mail
> ----------------------------------------------------------------------
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