Is this euler 121?

On Wed, Oct 19, 2011 at 6:24 PM, David Vaughan <[email protected]
> wrote:

> 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?
> >>>>>>
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-- 
Of course I can ride in the carpool lane, officer.  Jesus is my constant
companion.
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