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
