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1. Why does Player 2 "lose his option" to raise?2. Why is it an autocall of $20? Why can't Player 2 now raise Player 1's bet of $20?
13CARDS, your op was quite clear so I wouldn't worry about that.In response to your question I've always seen the penalty for out-of-turn action as auto-response to player 1's action.therefore, if player 1 checks then player 2 now bets $35 (it's a predetermined action that is binding as stated)if player 1 bets only $20 then it's as if player 2 put in $35 but that would be considered a call since it is less than double the bet.
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13CARDS, your op was quite clear so I wouldn't worry about that.In response to your question I've always seen the penalty for out-of-turn action as auto-response to player 1's action.therefore, if player 1 checks then player 2 now bets $35 (it's a predetermined action that is binding as stated)if player 1 bets only $20 then it's as if player 2 put in $35 but that would be considered a call since it is less than double the bet.
But, then (if you're correct, but I believe that you're not because Player 1 changed the action to Player 2), wouldn't this rule apply?:"If a player puts in a raise of 50% or more of the previous bet but less than the minimum raise, he or she will be required to make a full raise. The raise will be exactly the minimum raise allowed."
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But, then (if you're correct, but I believe that you're not because Player 1 changed the action to Player 2), wouldn't this rule apply?:"If a player puts in a raise of 50% or more of the previous bet but less than the minimum raise, he or she will be required to make a full raise. The raise will be exactly the minimum raise allowed."
If a person declares a raise verbally but does not put in enough then that rule definitely holds true. But in the case where somebody has bet sayy $8 and somebody throws in two nickels then it's assumed as a call unless there's a verbal declaration stating otherwise.
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If a person declares a raise verbally but does not put in enough then that rule definitely holds true. But in the case where somebody has bet sayy $8 and somebody throws in two nickels then it's assumed as a call unless there's a verbal declaration stating otherwise.
$10 is not 50% more than $8. So it is a call.An example of when the 50% rule applies is if someone bets $20, and the next player tosses a quarter and a nickel in, without saying anything, they would be putting in 50% more than the previous bet and they would be required to make the minimum raise to $40.
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But, then (if you're correct, but I believe that you're not because Player 1 changed the action to Player 2), wouldn't this rule apply?:"If a player puts in a raise of 50% or more of the previous bet but less than the minimum raise, he or she will be required to make a full raise. The raise will be exactly the minimum raise allowed."
It is my understanding that this is a limit hold'em rule, unfortunately incorrectly used in some home games and casinos.
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It is my understanding that this is a limit hold'em rule, unfortunately incorrectly used in some home games and casinos.
No. It's for no limit. That's why the rule is phrased "The raise will be exactly the minimum raise allowed." If it were for a limit game, the rule would state that a raise would be required and the word "minimum" would be irrelevant and it would not be used. You're probably thinking of this:"All raises must be equal to or greater than the size of the previous bet or raise on that betting round, except for an all-in wager. A player who has already acted and is not facing a fullsize wager may not subsequently raise an all-in bet that is less than the minimum bet (which is the amount of the minimum bring-in), or less than the full size of the last bet or raise. (The half-the-size rule for reopening the betting is for limit poker only.)"These are two separate rules for ruling on two different occurrences.
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