## mathematical expectation of how many times to get all cards

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

You need to clarify the process.
mathman
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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!
I'm just an imagination of your figment...
Denis
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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)
newme
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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
newme
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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 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
Last edited by BookInquiry on Fri Jan 13, 2012 5:04 am, edited 1 time in total.
BookInquiry
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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.
mathman
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### Re: mathematical expectation of how many times to get all ca

mathman wrote: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.
So this is how liberty dies... with thunderous applause.

Erimess

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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.
BookInquiry
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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?
So this is how liberty dies... with thunderous applause.

Erimess

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### 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
newme
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### 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.
mathman
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### Re: mathematical expectation of how many times to get all ca

mathman wrote: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
newme
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### Re: mathematical expectation of how many times to get all ca

newme wrote:
mathman wrote: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...mathman 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.
Living in the pools, They soon forget about the sea...— Rush, "Natural Science" (1980)

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

MarkFL wrote: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?
I'm just an imagination of your figment...
Denis
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