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3 votes
Accepted

Facing a 4-bet with QQ

3bet is good behind a loose player. If you think you have UTG beat then you want to isolate. You want to fold out suited connectors. An early raise and 3bet is pretty strong. Good chance one of ...
paparazzo's user avatar
  • 6,921
2 votes

Facing a 4-bet with QQ

Think for a moment that you were in this guy's with KK position and you see an early loose bet and a 3-bet. With what ranges are you going all in with? An average player would say something like: ...
koita_pisw_sou's user avatar
2 votes

Facing a 4-bet with QQ

On the one hand you had the third best starting Hand in Poker No Limit Hold'em. That makes me feel that you did nothing wrong here. But on the other hand whenever I got 4 bettet and I hold Queens and ...
RayofCommand's user avatar
2 votes

Three players, three pocket pairs?

That number seems high to me so I ran it. The chance of a straight flush is 72,192:1 and three pocket pairs seems harder to make than a straight flush. The chance of 4 of a kind is 4,164:1 and that ...
paparazzo's user avatar
  • 6,921
1 vote
Accepted

Confusion about two pair rule in poker, can someone help me who will win

A poker hand is built with 5 cards (and no more than 5). There is no limit to how many of those cards come from your hole cards, and how many of those come from the table (community cards) For ...
David's user avatar
  • 923
1 vote

Is it correct to go all-in when you have KK?

You are only behind KK+. You have 50% equity on QQ+. Mathematically you should call if their range is QQ+. If they raise you a lot pre flop no way their range is KK+. If their range is KK+ then they ...
paparazzo's user avatar
  • 6,921
1 vote

Playing AA against multiple opponents

In my case it would depend on the opponent. I always keep track of my play style and how it's perceived in the table and keep as much information about my opponents as possible. Don't enable reads on ...
Marcio's user avatar
  • 639
1 vote

Three players, three pocket pairs?

For three pocket pairs: One player has a pair: 13*4c2 / 52c2. Another player has a pair: (12*4c2 + 1) / 50c2. For the case of having the same value of first player, I add 1 (there is only one ...
Luis Masuelli's user avatar

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