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
