0

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.

1

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%

3
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.