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January 10th, 2012, 09:37 AM  #1 
Joined: Jan 2012 Posts: 8 Thanks: 0  mathematical expectation of how many times to get all cards
there are 6 cards. a user picks a card from the 6 cards randomly each time. chance of getting each card is equal. what is the expected value of how many times a user get all of 6 cards? thanks in advance. 
January 10th, 2012, 01:20 PM  #2 
Global Moderator Joined: May 2007 Posts: 4,923 Thanks: 285  Re: mathematical expectation of how many times to get all ca
You need to clarify the process.

January 10th, 2012, 01:23 PM  #3 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 4,284 Thanks: 254  Re: mathematical expectation of how many times to get all ca
6/6 * 5/6 * 4/6 * 3/6 * 2/6 * 1/6 = 5 / 324 : with replacement, which is what I think you mean!

January 10th, 2012, 05:53 PM  #4 
Joined: Jan 2012 Posts: 8 Thanks: 0  Re: mathematical expectation of how many times to get all ca
sorry for my poor english. i wanted to know the expected value of how many times it takes for a user to get all cards. i think i just figured out the formula. 6 * (1/6)6 + 7 * C(6, 1) * (1/6)7 + 8 * (C(6,1) + C(6,2)) * (1/6)8 + ... + 11 * (C(6,1) + C(6,2) + C(6,3) + C(6,4) + C(,6,5)) * (1/6)11 + 12 * (2^6  1) * (1/6)12 + ... + n * (2^6  1) * (1/6)n C = combination (1/6)n = pow((1/6), n) 
January 10th, 2012, 09:24 PM  #5 
Joined: Jan 2012 Posts: 8 Thanks: 0  Re: mathematical expectation of how many times to get all ca
~= 63 * (1*(1/6) + 2*(1/6)^2 + ... + n*(1/6)^n) = 63 * 26 / 25 ~= 65 
January 11th, 2012, 09:48 AM  #6 
Joined: Jan 2012 Posts: 21 Thanks: 0 
I am asking for any books/ materials discussing the background of these formula (especially the mechanics, statistics, numerical method parts): http://www.mathsnetalevel.com/index.php ... I_formulae They look terrific. I don't find any nice book talking about this. I just learnt the solution in an educational TV. cards numbered 1, 2, 3, 4, 5, 6 In order to collect all cards: 1st trial: Any card could be chosen. P=6/6 The expectation of no. of trials=1 2nd trial: the card appeared in trial 1 is not allowed P=5/6 The expectation of no. of trials to obtain the other cards=6/5 3rd trial: the cards appeared in trial 1, trial 2 are not allowed P=4/6 The expectation of no. of trials to obtain the other cards =6/4 4nd trial: the cards appeared in trial 1, trial 2, trial 3 are not allowed P=3/6 The expectation of no. of trials to obtain the other cards =6/3 ... 6nd (last) trial: the 5 cards appeared in previous trials are not allowed. The probability to pick this card =1/6. We are expected to draw 6/1 times to get this card. trial expectation= 6/6+6/5+6/4+6/3+6/2+6/1 
January 11th, 2012, 04:13 PM  #7 
Global Moderator Joined: May 2007 Posts: 4,923 Thanks: 285  Re: mathematical expectation of how many times to get all ca
The process you described looks like picking cards at random without putting them back. All I see then is you pick 6 cards and get them all  there is no probability involved.

January 11th, 2012, 06:48 PM  #8  
Joined: Apr 2011 From: USA Posts: 782 Thanks: 1  Re: mathematical expectation of how many times to get all ca Quote:
EDIT: Never mind. I just realized what I was quoting was not by the OP. Although I think that probably is a correct interpretation, like what Denis did.  
January 11th, 2012, 09:42 PM  #9 
Joined: Jan 2012 Posts: 21 Thanks: 0 
This problem is more difficult than it seems. Don't look down on it. I just remember the solution, and don't understand before. 
January 11th, 2012, 11:34 PM  #10 
Joined: Apr 2011 From: USA Posts: 782 Thanks: 1  Re: mathematical expectation of how many times to get all ca
Who was looking down on it? Where'd that come from?


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