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: [email protected] [mailto:[email protected]] 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$n having m > distinct values is at least p? ----- Original Message ----- From: Fraser Jackson <[email protected]> Date: Thursday, May 26, 2011 22:47 Subject: Re: [Jprogramming] roll and deal To: Programming forum <[email protected]> > Hi Roger, > > I am sorry I somehow sent the note to Jprogramming with your name with > lower case. > > I was brought up on the first edition of Feller - and we had > calculators with crank handles to do the sums. If you want to > incorporate any bit of it in your Essay don't hesitate. > > Thanks for all you are doing for J. It is just an amazing tool. > > Kindest regards, > > Fraser > > > ----- Original Message ----- > From: "Roger Hui" <[email protected]> > To: "Programming forum" <[email protected]> > Sent: Friday, May 27, 2011 1:24 PM > Subject: [Jprogramming] roll and deal > > > > What is the least c such that the probability of ?c$n having m > > distinct values is at least p? > > > > ref: http://www.jsoftware.com/jwiki/Essays/Birthday%20Problem ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
