What is the fundamental difference between choosing a ball and rolling a die type of problems in probability? Suppose, I have a box where I have $n$ balls out of which $b$ are blue. Hence, the probability of picking up a blue ball at random is $p=\frac{b}{n}$.
Now suppose, I know the total number of balls, and the probability of getting a blue ball, I could easily calculate the number of blue balls originally in the bag.
However consider that I have thrown $n$ dies on the floor. I pick up a die randomly from the floor. I know the probability of getting a six is $1/6$. However, unlike the ball scenario, I can't reverse engineer the situation and find out the exact number of dies on the floor that show $6$ on their face.
In the first scenario, by knowing the total number of balls, and the probability of picking up a blue ball, I could very well calculate the total number of blue balls.
In the second scenario, I know the probability of getting a six, and the number of dies on the floor. However, I still can't say how many of them rolled a six.
These two situations seem exactly analogous to me, and yet there is a fundamental difference between them that I can't seem to grasp for some reason. In the second case, the probability of getting a six doesn't seem to depend on how many dies on the floor actually show a six against the total number of dies. In a sense, it is impossible to know the actual number of dies on the floor that have a six. This takes a form of a distribution.
But then, if the ball case is alike, as it sounds like, why can we be so sure of the number of blue balls.
Is it something like, in the ball scenario, we know exactly what the probability of getting a blue ball is. However, in the die case, we are first tossing $n$ dies and then picking one up randomly to check if it is a six. However, the true probability of getting a six would actually depend on the actual number of sixes on the floor, and since during every roll of $n$ dies, we can expect to get a different number of sixes, the true probability of picking a single six at random would change every single time we do the experiment. The value that we take as the probability i.e. $\frac{1}{6}$ is not the true probability of picking up a six from the floor. Rather, it is our best guess of what the true probability is.
Hence we can't reverse engineer this situation to get the actual total number of sixes on the floor at any time. Rather, we only get an estimate. The real number of sixes on the floor keep on changing., and follow some distribution.
Is this the fundamental difference between the two situations ?
 A: 
consider that I have thrown $n$ dies on the floor. I pick up a die randomly from the floor. I know the probability of getting a six is $1/6$.
the probability of getting a six doesn't seem to depend on how many dies on the floor actually show a six against the total number of dies.

Of course the $\displaystyle\frac16$ probability of an arbitrary die showing six doesn't depend on how many dies are on the floor.

However, in the die case, we are first tossing $n$ dies and then picking one up randomly to check if it is a six. However, the true probability of getting a six would actually depend on the actual number of sixes on the floor, and since during every roll of $n$ dies, we can expect to get a different number of sixes, the true probability of picking a single six at random would change every single time we do the experiment. The value that we take as the probability i.e. $\frac{1}{6}$ is not the true probability of picking up a six from the floor. Rather, it is our best guess of what the true probability is.

The probability of the $n$ dies on the floor showing exactly one six equals $$\frac n5\left(\frac56\right)^n.$$
Your error is in conflating picking up one die and picking up multiple dies.

However, I still can't say how many of them rolled a six.

If the number $n$ of dies on the floor is large enough, then you can reasonably expect $\displaystyle\frac n6$ (rounded to the nearest integer) of them to have rolled a six. Slightly worse guesses are the integers near $\displaystyle\frac n6.$

But then, if the ball case is alike, as it sounds like, why can we be so sure of the number of blue balls. Is it something like, in the ball scenario, we know exactly what the probability of getting a blue ball is.

You had obtained that number of blue balls from definition.
A: It seems that my confusion arises from a misunderstanding of how the probability of an event is defined. To resolve this, let us go back, and talk about two events - rolling a fair die, and picking a numbered marble at random from a bag of $6$ numbered marbles.
Before we can talk about probability, let us talk about the sample space. It is the set of possible outcomes. As you can expect, in both the above cases, the sample space is the same for both our experiments. We can write it as : $$S=\{1,2,3,4,5,6\} $$
To make this slightly less trivial, consider that in the die, instead of the second face having a $2$, it is also labeled $1$. Similarly in case of our marbles, two of the marbles have been labelled $1$ and none of them have been labelled $2$. The new sample space for both would now be : $$S=\{1,1,3,4,5,6\}$$
This denotes the all the possible outcomes of the two experiments. Now suppose, we want to know that what is the probability of rolling the die and getting a $1$. Or maybe you want to pick a marble at random, and find the likelihood of it having $1$ written on it.
This would just be the number of favourable outcomes in the sample space, divided by the total number of outcomes. Since, $2$ elements in the sample space have that number $1$ in it, we can say there are $2$ out of $6$ outcomes where the die rolls a $1$ or the marble we pick up has a $1$ written on it.
Hence the probability is defined to be :
$$P(1)=\frac{2}{6}$$
Now comes the part where we failed to interpret this properly.
Note that, in case of the die, the sample space listed the possible outcomes of rolling a 'single die'. Each outcome here represents, which face of the single die comes out on top. Each face of the die might have an unique number associated with it, but as a whole, the die doesn't have a unique number. Any of the faces have some chance of coming on top.
In case of the marbles, the situation is different. Each marble has an unique number associated, and so, by definition, we are working with $6$ different marbles. The outcome here represents which one of the $6$ marbles comes out. Each of them have some chance, and we are calculating that.
So, these marbles are like the faces of the die. But here is the thing, we don't treat the faces of the die as separate objects, they are part of a single object. In the marble case, each of the numbered marbles, is a separate physical object. So, it is silly to associated a single number to a die as a whole. However, in case of the marbles, there is a single number or colour associated with them.
In a sense, rolling a die is the same thing as choosing one of the faces to come out on top. In that sense, rolling a die and choosing a marble problems are similar. However, it is important to remember that a single die is not equivalent to a single marble. It is actually, one of the faces of the die, that you can associate with a marble.
Now you can understand the difference between rolling a die and picking a card type of problem in probability. From the mathematical perspective, there is no difference. However from the philosophical/physical perspective, be careful to understand what you are dealing with, or you'd end up in a rabbit hole of misunderstanding and misinterpretations as I did.
Now you can begin to understand why the probability of picking up a blue marbe depends on the number of blue marbles, but the probability of getting a $6$ on a die, is independent of any other die that you roll alongside it. This is because it would be wrong to compare a die to a marble as a whole. You can compare the die to the bag that contains all the marbles. Then you can compare the different marbles to the different faces of the die. So, two blue marbles for example, are part of the same system. However, two die are completely different systems and are thus, independent of each other.
