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 ?

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.