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
