Calculate the probability of getting items from a chest?

Question

Probability theory

I am weak in probability theory, but such a task has arrived.
There is a game in which players have the opportunity to open chests. From each chest, with a certain probability, you can get an item of different quality and, therefore, prices.
The price of chest X is a known value.
Condition - the probabilities of dropping items cheaper or equal in value X are the same. The probabilities of dropping objects more expensive than X obey some law - the more expensive, the less chance.
Win - the player got an item more expensive than the cost of the chest, so win = the cost of the item - X. Loss - the player got the item cheaper than the cost of the chest, then loss = X - the cost of the item.
Question: how to calculate the probability of dropping each item in such a way that the player's total gain is equal to his total loss with a large number of openings?
How I did:
1. Probability for items cheaper than X - P_low;
2. Let's say the probability distribution law for items more expensive than X is P(cost) = k/(cost + c) where k and c are unknown
3. P(X) <= P_low
4. The sum of the probabilities for all items (the number of which is known ) = 1, which is logical.
And then the plug.

Answer 1

P ₁ + P ₂ + ... + P _n = 1
P ₁ *C ₁ + P ₂ *C ₂ + ... + P _n *C _n = X

Answer 2

It is necessary to build an equation from the parameters you specified and solve it (most likely by a numerical method).
Since you want 2 frequency distribution parameters, you need to know the values ​​at 2 points and that is the system. If I were you, I would reduce it to one parameter and then I would solve it by any numerical method.
Be that as it may, the task is not yet fully formalized by you, or you describe something not quite clearly in the formulation.