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March 11th, 2012, 05:56 AM  #21 
Newbie Joined: Mar 2012 Posts: 11 Thanks: 0  Re: mathematical expectation of how many times to get all ca
Recently,I also meet this problem ,I'm is the problem is difficult! I don't think the solution "6/6+6/5+6/4+6/3+6/2+6/1" isn't right! In traditional understanding of expected value,we can easily to write :E=p(6)*6+p(7)*7+......+p(n)*n; but for each f(n) is not easy to Calculate,so it is difficult! 
March 11th, 2012, 02:15 PM  #22  
Global Moderator Joined: May 2007 Posts: 6,807 Thanks: 717  Re: mathematical expectation of how many times to get all ca Quote:
 
March 11th, 2012, 03:52 PM  #23 
Newbie Joined: Mar 2012 Posts: 11 Thanks: 0  Re: mathematical expectation of how many times to get all ca
here may be the question he really want to ask: we often some companies promote in this way:if you can collect all kinds of "card"(refer to a kind of mark,not cards) in the package of good,you will be give a prize!Our question is how many goods we need to buy in average to collect all kinds of cards? In this model,that is to solve the the expected value of how many cards we need until a set cards number 1~n appeared,there are no limit of the cards(you can buy as many as you want if you have enough money, ). So, you clear ? 
June 2nd, 2012, 05:46 AM  #24 
Newbie Joined: Mar 2012 Posts: 11 Thanks: 0  Re: mathematical expectation of how many times to get all ca
I have solved this problem.If you think in this way,it's not too difficult. First it's a Geometric distribution http://en.wikipedia.org/wiki/Geometric_distribution model,we want to gets all cards,so we first need to get a kind may marked 1,2,3.....,6,then put it in a set named "Getted",then we need get a kind different from the "Getted" set. This means if we firstly get the card marked 2,we next need gets a card one of 1,3,4,5,6.Then we repeat this,finally we will gets all kinds cards! In this process,every time when we get a new kinds cards to put it in the "Getted" set,this is a Geometric distribution model.First we get any kind cards,it a new kind,so its mathematical expecation is 1,next time the probability we get a card different from the first is,so its expecation is,repeat this ,finally we can work out the answer is . 
June 2nd, 2012, 08:15 AM  #25 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038  Re: mathematical expectation of how many times to get all ca
Sigh....

June 11th, 2012, 06:15 AM  #26 
Newbie Joined: Oct 2011 Posts: 11 Thanks: 0  Re: mathematical expectation of how many times to get all ca
Ok, I am a novice so someone please correct me if my thinking is off. This is just how my brain puts the logic together. 6(n/(n1)) 6 representing the desired outcome with picking all six cards n representing the 1st card picked n1 representing the allowed card after the 1st. (n/(n1) representing the 1st card picked but not allowed thus negating the possible outcome of cards by minus 1 ie 6/5, 6/4 ....etc. If anyone can help me better formulate this as well as explain their reasoning I would greatly appreciate it. A lot of the times people post these long equations but I'm not always sure I can follow them even though I know the logic behind the problem. Maybe that's where I get tripped up..... 
June 11th, 2012, 09:48 AM  #27  
Newbie Joined: Oct 2011 Posts: 11 Thanks: 0  Re: mathematical expectation of how many times to get all ca Quote:
 
June 11th, 2012, 10:28 AM  #28  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038  Re: mathematical expectation of how many times to get all ca Quote:
However, I think you would get a better chance at a reply if you started your own thread; tacking on to an existing inactive thread means usually that no one realises a question is being asked...  

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