What path should I choose? Imagine that I am participating in a contest.
It consists of:


*

*There are 3 types of objects in a box, and each type, is the same number of times in the box.

*To win, I must grab the same amount of each object.

*Initially, I can only grab 3 objects, therefore to win, I must grab each item once (once type 1, once type 2 and once type 3).


If I do not win the first time, they offer me the following:
1) You can grab 3 more objects, to try to match the quantities (knowing that the quantities are different, but without knowing the quantities)
2) You can grab 6 more objects, to try to match the quantities (knowing that the quantities are different, but without knowing the quantities)
Quickly, I might think that grabbing 6 more objects will give me more opportunities to match, but it will also give me more opportunities to make the difference between the amounts, even bigger.
So, what is better? Pick 6 more objects to grab or 3?
Also assuming they are equiprobable.
My thought was that, it is better to choose 6 objects following the logic of the law of big numbers, however I realized that this law applies when there are many attempts. So, what is the correct one?
 A: Let's get one thing cleared up! The "law of large numbers" tells you -- roughly speaking -- that with more events you are more likely to get approximately what you expect, rather than exactly what you expect. For example, it's far harder to get exactly $1000$ tails and $1000$ heads out of $2000$ tosses than $1$ tail and $1$ head out of $2$ tosses.
As a very, very rough approximation, for problems such as this, with $n$ independent events, you can think that about half of the "probability mass" is concentrated within the average plus or minus a quantity that is proportional to $\sqrt{n}$; and within this "standard deviation" all results are more or less as likely as each other (within a constant factor).
Then, getting results within an interval proportional to the number of events (say, average $\pm 10\%$) becomes easier with large numbers. Getting results within an interval proportional to the square root of the number of events (say, average $\pm 20\%$ for a given $n$, or average $\pm 5\%$ when $n$ is $16$ times larger $(20/5=\sqrt{16})$ stays roughly the same. Getting results within an interval that is independent of the number of events (say, average $\pm 10$ - note, $10$, not $10\%$) becomes harder with large numbers.
But note that for subtle situations like this one, with conditional probabilities etc. such rough reasoning is dangerous and easily leads to mistakes!
A: Remark: This answer was given assuming you can see what you wound up with on the first grab, but user Anonymous has pointed out that the OP only wants you to know whether you did or did not win, so that you have to base your choice of grabbing another three or another six without knowing whether you are currently holding three of one type or two of one type and one of another. I'm leaving the answer as is for now, in case it's of use for the OP or anyone else. End remark.
As saulspatz observed in comments, if on the first round you wind up with three of one type, you cannot win if you only grab another three objects, so you must opt to grab six more objects to have any hope of winding up with three of each type. So we need only concern ourselves with what to do if we wind up with two of one type, say type A, and one of another, say type B.  
If we imagine the box started with an unknown number N of each type, we are left with $N-2$ objects of type A, $N-1$ objects of type B, and $N$ objects of type C.  If we grab three more items, hoping to get two type C's and one more type B, the probability of winning is
$${N-2\choose0}{N-1\choose1}{N\choose2}\over{3N-3\choose3}$$
On the other hand, if we grab six more items, hoping to get three of type C, two of type B, and one of type A, the probability of winning is
$${N-2\choose1}{N-1\choose2}{N\choose3}\over{3N-3\choose6}$$
(Note: we must assume $N\ge3$ in order for this option to make sense.) The ratio of the first to the second probability works out to be
$${(3N-6)(3N-7)(3N-8)\over20(N-2)^3}={3(3N-7)(3N-8)\over20(N-2)^2}$$
Some experimentation and/or algebra shows that this ratio is less than $1$ for $3\le N\le5$ and greater than $1$ for $N\ge6$.  So in general you are better off grabbing just another three items if your first grab gave you two of one type and one of another, unless you have reason to believe the box started out with only $9$, $12$, or $15$ items altogether.
