Capture the last row of c
]l=:{:c
106 249 7 326 308 274 81 347 276 113 108 20...
And use grade up to arrange the numbers in sequence
(/:l){l
0 1 3 3 4 4 5 5 6 6 7 7 7 7 8 8 8 13 13...
This should be i.365
I have an idea on how to get what you want but I'll wait to see what you do
next.
I have to use a deal without a fixed seed for the deal.
I happened to remember 24 and haven't gone looking for the number and
haven't the foggiest notion of how I proved it!
Linda
-----Original Message-----
From: Linda Alvord [mailto:lindaalv...@verizon.net]
Sent: Tuesday, January 17, 2012 11:25 PM
To: 'Programming forum'
Subject: RE: [Jprogramming] Challenge 4 Bountiful Birthdays
I think we need sortleaf.
Linda
-----Original Message-----
From: programming-boun...@jsoftware.com
[mailto:programming-boun...@jsoftware.com] On Behalf Of Brian Schott
Sent: Tuesday, January 17, 2012 10:14 PM
To: Programming forum
Subject: Re: [Jprogramming] Challenge 4 Bountiful Birthdays
$c=.500 365?.@$365
$D=. i.&0"1(=&#~.)\"1 c
mean D+1 NB. 25.086
pretty 5 SL sortleaf 1+D
$c=. 10 500 365?.@$365
$D=. i.&0"1(=&#~.)\"1 c
mean, D+1. NB. 24.6856
pretty"2] 5 SL"0 1 sortleaf"1] 1+D
I am suspicious of these results because my memory is that the mean is
closer to 18.5 people. Then again, other measures of the center look closer
to 19, just looking at the stemplots.
---
(B=)
On Jan 15, 2012, at 5:40 AM, "Linda Alvord" <lindaalv...@verizon.net> wrote:
>
> For this challenge, as usual, do not use @ but you may use whatever
style
> and strategy you like to accomplish the task.
>
>
>
> The problem is to simulate the classic birthday problem.
>
>
>
> A single trial works this way. People enter a room one by one and declare
> their birth date. Suppose the 29th person is the first person to match a
> birthday of someone in the room. The result of the first trial is 29.
>
>
>
> Repeat for 500 trials.
>
>
>
> If you use Kip's frequency distribution of the results you have lots of
> information.
>
>
>
> fd=:[: /:~ ~. ,. [: +/"1 ~. =/ ]
>
>
>
> But the mean gives a more concise summary of the data. So, the final part
> of the problem is to obtain a list of 10 means of 500 trials and the
> mean of the means.
>
>
>
> Linda
>
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