Thanks, this was the method I had envisaged. I wasn't thinking about it in the J way originally - this is something I'm trying to work on.
On 19 Oct 2011, at 20:21, Marshall Lochbaum wrote: > Sorry--a few off-by-one errors. Here's code that gives you the correct > result: > p =. 4 :'(+/ */"1 >: y comb x) % (!>:x)' > 4x p"0 i.>.-:4 > 1r120 1r12 > +/ 4x p"0 i.>.-:4 > 11r120 > > Marshall > > On Wed, Oct 19, 2011 at 12:38 PM, David Vaughan < > [email protected]> wrote: > >> Ignore the domain error thing that was just one of my usual careless >> mistakes. >> >> But even so, the result is not right. >> >> (i.<.-:<:4) p 4 >> 1r120 >> (>:i.<.-:<:4) p 4 >> 1r20 >> >> On 19 Oct 2011, at 17:03, Marshall Lochbaum wrote: >> >>> The number of ways to choose k red balls is the sum over all combinations >> of >>> bags of the probability you draw a red ball from each of those bags, >>> multiplied by the number of ways to choose the red balls, which is 1. So, >> I >>> suggest you find one of the various comb verbs running around, and then >>> something like >>> (+/ */"1 k comb n) % (!>:n) >>> will give you that probability. Then make this into a verb dependent on >> k, >>> apply it with rank zero to (i.<.-:<:n), and add those results up. >>> >>> Marshall >>> >>> On Wed, Oct 19, 2011 at 11:09 AM, David Vaughan < >>> [email protected]> wrote: >>> >>>> That was untested - hence errors. >>>> >>>> If r and b are two choices, and you start with a bag containing r,b and >>>> after each turn where you take one item from the bag you add another r. >> The >>>> goal is to find the probability of choosing more b's than r's. As I >> probably >>>> haven't explained well, here's an example: >>>> >>>> b,r - take b, return it and add another r >>>> b,r,r - take r, return it and add another r >>>> b,r,r,r - take b, return it. >>>> >>>> In this case there were 3 turns, and 2 b's were chosen vs 1 r, so that >> is a >>>> success. The probability of taking more b's than r's is the probability >> of >>>> taking 3 b's = %24 plus the probability of taking 2 b's and 1 r = >> 24%~1+2+3 >>>> which gives 7r24. Though I'm not actually sure my idea for the general >> case >>>> is totally correct. >>>> >>>> On 19 Oct 2011, at 15:48, Marshall Lochbaum wrote: >>>> >>>>> The correct way is to compute all the terms in the expression and then >>>>> multiply them together. Something like >>>>> n* > *&.>/ >:@i.&.> n->:i.k >>>>> However, the expressions you gave will return length errors, since >>>> (>:i.n-2) >>>>> is patently not the same length as (>:i.n-1). What exactly are you >> trying >>>> to >>>>> compute? >>>>> >>>>> Marshall >>>>> >>>>> On Wed, Oct 19, 2011 at 10:39 AM, David Vaughan < >>>>> [email protected]> wrote: >>>>> >>>>>> I couldn't really think of an appropriate title for this. My issue is >>>> that >>>>>> I want to compute an expression that has a different number of terms >>>>>> depending on y. >>>>>> >>>>>> +/(>:i.n-1)*n NB. for all n, y >: n > 1 >>>>>> +/(>:i.n-2)*(>:i.n-1)*n NB. for all n, y >: n > 2 >>>>>> >>>>>> and so on, so that: >>>>>> >>>>>> lim =. <:<.y%2 >>>>>> >>>>>> and we carry on the style of expression above until we are doing it >> for >>>> all >>>>>> n, y >: n > lim. >>>>>> >>>>>> So in the case for y=.5, lim=.1 and we only do the first of the lines >>>>>> above. For y=.7, we would do the second one as well. For y=.9 we would >>>> do >>>>>> the same as the second but with a (>:i.n-3) multiplied with it all as >>>> well. >>>>>> I guess/hope there is some way of achieving this with power? >>>>>> ---------------------------------------------------------------------- >>>>>> For information about J forums see >> http://www.jsoftware.com/forums.htm >>>>>> >>>>> ---------------------------------------------------------------------- >>>>> For information about J forums see http://www.jsoftware.com/forums.htm >>>> >>>> ---------------------------------------------------------------------- >>>> For information about J forums see http://www.jsoftware.com/forums.htm >>>> >>> ---------------------------------------------------------------------- >>> For information about J forums see http://www.jsoftware.com/forums.htm >> >> ---------------------------------------------------------------------- >> For information about J forums see http://www.jsoftware.com/forums.htm >> > ---------------------------------------------------------------------- > For information about J forums see http://www.jsoftware.com/forums.htm ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
