The game isOmahahi/lo, structured as a $2-$6 range game with a single $2 blind. Here’s what happened. Dealer puts the flop out; there are 5 players in the pot. The first player goes all-in for $1, the second player says raise and puts out $7. The dealer gives the second player $1 back and informs him that he may make it $6 as the $1 all-in is not a legal bet and that it could not be raised but that the player had the right to make a $6 bet. The floor man agreed with the dealer. Later, when the card room manager was asked, he said that the dealer was wrong and that the second player should have been able to make it $7, raising the $1. Who was right? And Why?
Answer 1:
You’ve probably heard this before Chuck, but it all depends. What is the rule of the house about all-in bets that determines if it is considered a bet or not? At the Shoe inLas Vegasit is considered a bet if the amount is at least half or more. If that’s the rule in the house you were playing, then the $1 all-in bet is half the bet (minimum bet is $2 right?). The next player can 1) fold 2) call 3) call the $1 and raise it 2,3,4,5,or 6 more. So the $7 that the raiser put out there seems correct. He could have put $6 out there and still been correct too.
Answer 2:
The card room manager. The first player is not raising. He’s calling a $2 blind with a $1 bet. Therefore the current bet is $2, and the next player is raising the blind, not the $1 bet. A bet of $7 here is a legal $5 raise of the $2 blind. The exception would be if all bets and raises other than all-ins must be in $2 multiples, which is sometimes the case in these games. In this situation, the raiser could raise to either $6 or $8, but not $7.
Answer 3:
It’s a 2-6 game. The min bet is 2. The 1 is 1/2 that. 1/2 the min bet is a bet and can be raised. The raise stands, make it 7. The error the dealer made was in thinking that the bet size on the flop was 6 (I’m guessing everybody always bets 6 on the flop). The bet size was 2-6.