My Math Forum mathematical expectation of how many times to get all cards

 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: 3,832 Thanks: 33 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: 3,089 Thanks: 36 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: 3,832 Thanks: 33 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:
 Originally Posted by mathman 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.
Don't think so. By saying the card that appeared in trial 1 is not allowed is somewhat implying that it could be picked (just not "allowed"), and therefore was indeed put back. So one of the other 5 would have to be picked in trial 2. Etc.

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?
 January 12th, 2012, 02:52 AM #11 Joined: Jan 2012 Posts: 8 Thanks: 0 Re: mathematical expectation of how many times to get all ca it takes at least 6 times to get all cards i think the expected values is 6*K6 + 7*K7 + ... + n*Kn Kn is possibility you get all cards at nth time. n->infinite
 January 12th, 2012, 04:20 PM #12 Global Moderator   Joined: May 2007 Posts: 3,832 Thanks: 33 Re: mathematical expectation of how many times to get all ca The problem statement is unclear. There are six cards. A card is picked at random. IS THE CARD PUT BACK??????????? If not, then the solution is the trivial one that I suggested. If it is put back, then it is an interesting problem.
January 12th, 2012, 05:38 PM   #13

Joined: Jan 2012

Posts: 8
Thanks: 0

Re: mathematical expectation of how many times to get all ca

Quote:
 Originally Posted by mathman The problem statement is unclear. There are six cards. A card is picked at random. IS THE CARD PUT BACK??????????? If not, then the solution is the trivial one that I suggested. If it is put back, then it is an interesting problem.
there are 6 kinds of cards, and number of each kind of card is unlimited

January 13th, 2012, 12:44 AM   #14
Global Moderator

Joined: Jul 2010
From: St. Augustine, FL., U.S.A.'s oldest city

Posts: 11,566
Thanks: 108

Math Focus: The calculus
Re: mathematical expectation of how many times to get all ca

Quote:
Originally Posted by newme
Quote:
 Originally Posted by mathman The problem statement is unclear. There are six cards. A card is picked at random. IS THE CARD PUT BACK??????????? If not, then the solution is the trivial one that I suggested. If it is put back, then it is an interesting problem.
there are 6 kinds of cards, and number of each kind of card is unlimited
Say, what?

So...is each card is marked uniquely and chosen with replacement? (i.e, when a card is chosen, is it put back in the group, possibly chosen again?) These are the kinds of things we need to know to solve the problem.

A well-formed question gets a well-formed response...not being harsh, that's just the way it is...[color=#00BF00]mathman[/color] raises valid concerns that must be addressed before the question makes sense...otherwise people are really just guessing at what you mean. Help us to help you.

January 13th, 2012, 08:50 AM   #15
Math Team

Joined: Oct 2011

Posts: 3,089
Thanks: 36

Re: mathematical expectation of how many times to get all ca

Quote:
 Originally Posted by MarkFL So...is each card is marked uniquely and chosen with replacement? (i.e, when a card is chosen, is it put back in the group, possibly chosen again?) These are the kinds of things we need to know to solve the problem.
I'll answer on his behalf: yes, oui, si, ...

1- take 6 cards at radom from a regular 52-card deck
2- throw the other 46 in the garbage
3- pick a card (from the li'l pack of 6, of course; anyway the others are in the garbage...)
4- LOOK at the card, write down what it is
5- put it back in the li'l pack of 5 (so you now have 6), and shuffle 'em well
6- repeat steps 3 to 5 five more times (note that 5th step need not be done at end!)

What is the probability that you wrote down 6 different cards?

 Tags cards, expectation, mathematical, times

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post rubis Advanced Statistics 5 January 5th, 2014 05:08 PM helloprajna Math Books 0 November 18th, 2012 10:50 PM Sara so Algebra 3 January 4th, 2011 12:27 PM wkmrachmistrz Math Books 0 September 18th, 2010 04:46 AM wkmrachmistrz Math Books 0 August 22nd, 2009 01:32 AM

 Contact - Home - Top