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January 10th, 2012, 09:37 AM   #1
 
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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.
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January 10th, 2012, 01:20 PM   #2
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Re: mathematical expectation of how many times to get all ca

You need to clarify the process.
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January 10th, 2012, 01:23 PM   #3
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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!
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January 10th, 2012, 05:53 PM   #4
 
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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)
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January 10th, 2012, 09:24 PM   #5
 
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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
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January 11th, 2012, 09:48 AM   #6
 
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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
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January 11th, 2012, 04:13 PM   #7
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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.
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January 11th, 2012, 06:48 PM   #8
 
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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.
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January 11th, 2012, 09:42 PM   #9
 
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This problem is more difficult than it seems. Don't look down on it.
I just remember the solution, and don't understand before.
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January 11th, 2012, 11:34 PM   #10
 
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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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