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.


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%

  • Does it mean that if I have (e.g. Js 10h), then any player can get Js 10t in next hand with probability of 0.45% ? Aug 3 '20 at 12:27
  • If it's any combination of JT, then the chance is (4/52)*(4/51)=>0.6%, for a specific combination of suits it's (1/52)*(1/51)=0.04%.
    – Feroc
    Aug 3 '20 at 13:24
  • Ok, thanks a lot! Aug 3 '20 at 13:34

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