Q:

Probability that any player will receive any similar hand of the previous hand of any player?

I want ask,  how to count this case? I understand that every deal of cards is a separate event, and does not depend on the previous one.

A:

Not sure if I understand the question to 100%, but the chance to receive any specific combination of cards (assumed we are talking about Texas Holdem) is (1/52)*(1/51) => 0.04%.
But in your question you are talking about "similar hand". So if you are talking about another player also getting Aces the chance would be:
(4/52)*(3/51) => 0.45%

