Neat. That solved my problem. Thanks bas & 有D .

On Sat, Sep 12, 2009 at 5:39 PM, 有D <[email protected]> wrote:

> 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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