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

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

Denis wrote: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?

sorry for my poor english.
i think this is what i mean.
newme
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### Re: mathematical expectation of how many times to get all ca

Well then, probability is: 6/6 * 5/6 * 4/6 * 3/6 * 2/6 * 1/6 = 5 / 324
I'm just an imagination of your figment...
Denis
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### Re: mathematical expectation of how many times to get all ca

The problem statement originally was unclear. There are two different problems.

1) (Which has been answered) what is the probability that you will get six different cards when selecting six cards at random?

2) (Difficult) What is the average number of selections required to get six different cards?
mathman
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### Re: mathematical expectation of how many times to get all ca

Denis wrote: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?

sorry for my poor english.
i think this is what i mean.
newme
Newcomer

Posts: 8
Joined: Tue Jan 10, 2012 12:32 pm

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

mathman wrote:The problem statement originally was unclear. There are two different problems.

1) (Which has been answered) what is the probability that you will get six different cards when selecting six cards at random?

2) (Difficult) What is the average number of selections required to get six different cards?

second one.
newme
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### 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!
rihkddd
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### Re: mathematical expectation of how many times to get all ca

Denis wrote:
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?

your question is easy to answer (5!/6^5), but it is not the same question as originally asked.
mathman
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### 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 ?
rihkddd
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### 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$\frac{5}{6}$,so its expecation is$\frac{1}{\frac{5}{6}}=\frac{6}{5}$,repeat this ,finally we can work out the answer is $\frac{6}{6}+\frac{6}{5}+\frac{6}{4}+\frac{6}{3}+\frac{6}{2}+\frac{6}{1}$.
rihkddd
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### Re: mathematical expectation of how many times to get all ca

Sigh....
I'm just an imagination of your figment...
Denis
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Joined: Mon Oct 31, 2011 7:29 pm

### 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/(n-1))

6 representing the desired outcome with picking all six cards
n representing the 1st card picked
n-1 representing the allowed card after the 1st.
(n/(n-1) 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.....
Jabby J
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### Re: mathematical expectation of how many times to get all ca

Denis wrote:Sigh....

I believe you had it right the first time with the orginal question. Of course here I come in with my garbage... sorry to complicate the matter.
Jabby J
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### Re: mathematical expectation of how many times to get all ca

Jabby J wrote:I believe you had it right the first time with the orginal question. Of course here I come in with my garbage... sorry to complicate the matter.

Hey, no problems...
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...
I'm just an imagination of your figment...
Denis
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