[Jprogramming] roll and deal

[Marshall Lochbaum]

Provided you mean at least m distinct values, the number of ways to have
exactly m distinct values is
(m!n)*(m^c)
And the number of ways total is
n^c
Which gives us a probability of
(m!n)* (n-m)^c

Then the answer is
c=. >.(n-m)^. p% m!n

Marshall

-----Original Message-----
From: programming-bounces at jsoftware.com
[mailto:programming-bounces at jsoftware.com] On Behalf Of Roger Hui
Sent: Friday, May 27, 2011 2:00 AM
To: Programming forum
Subject: Re: [Jprogramming] roll and deal

But what about the problem I posed?

> What is the least c such that the probability of ?c$n having m 
> distinct values is at least p?



----- Original Message -----
From: Fraser Jackson <fraser.jackson at xtra.co.nz>
Date: Thursday, May 26, 2011 22:47
Subject: Re: [Jprogramming] roll and deal
To: Programming forum <programming at jsoftware.com>

> Hi Roger,
> 
> I am sorry I somehow sent the note to Jprogramming with your name with 
> lower case.
> 
> I was brought up on the first edition of Feller - and we had 
> calculators with crank handles to do the sums.  If you want to 
> incorporate any bit of it in your Essay don't hesitate.
> 
> Thanks for all you are doing for J.  It is just an amazing tool.
> 
> Kindest regards,
> 
> Fraser
> 
> 
> ----- Original Message -----
> From: "Roger Hui" <rhui000 at shaw.ca>
> To: "Programming forum" <programming at jsoftware.com>
> Sent: Friday, May 27, 2011 1:24 PM
> Subject: [Jprogramming] roll and deal
> 
> 
> > What is the least c such that the probability of ?c$n having m 
> > distinct values is at least p?
> >
> > ref: http://www.jsoftware.com/jwiki/Essays/Birthday%20Problem

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