in the C=3, N=2 case:
f(3) = f(1) + 1;//when you are miss 3 card, you will get 2 new cards when
you buy a package;
f(1) = 1/3*f(1) + 2/3*f(0) + 1;//when you are missing one card, the
propobility that you will not get a new card is 1/3.
So, f(0)=0; f(1)=1.5; f(3) = 2.5;


On Sat, Sep 12, 2009 at 6:33 PM, Roshan Amadoru <[email protected]>wrote:

>
> can you show how to get the answer when C=3, N=2.. i'm still a little
> confused to be honest.
>
> bas wrote:
> > Let f(k) be the expected number of packs we need to buy if we still
> > don't have k cards (or we have C-k cards).
> > To solve the problem we need to find f(C). Note: f(0)=0.
> >
> > In the first example C=2;
> > When we buy the first pack, we will get one of the cards we need with
> > the probability 1.
> > f(2)=1*f(1)+1
> > means that we expect to buy that many packs to get all the cards we
> > need. But we still don't know f(1).
> > For k=1 we have one card we need, and we need to find the other one.
> > When we buy this next pack, we will get the card we already have with
> > a probability 1/2 and the card we need with prob 1/2.
> > f(1)=1/2*f(1)+1/2*f(0)+1
> > Then: f(1)=2; f(2)=3
> >
> >
> >
> >
> >
> > On Sep 11, 9:54 pm, qasim zeeshan <[email protected]> wrote:
> >
> >> Hi All,
> >> First of all congrats to all who made it to next round.
> >>
> >> I have seen some codes for Problem C. but didn't understand. Can anyone
> >> please explain both example test cases for Problem C.
> >>
> >> Thanks
> >>
> >> --
> >> Qasim Zeeshan
> >>
> > >
> >
> >
>
> >
>


-- 
有D

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