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#1
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If a player bets 40$, and I raise by 40$ (to 80$), is the player after me forced to raise by another 40$ (for a total of 120$) or by 80$ (for a total of 160$)?
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#2
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I mean, ofcourse, if he chooses to re-raise. Sorry for not clarifying.
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#3
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He has to raise at least the same as the last raise, which was $40. He can still move all-in with less than that $40 though [img]/images/graemlins/smile.gif[/img]
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#4
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Perfect! Thank you Nightlight [img]/images/graemlins/smile.gif[/img]
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#5
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Each bet must be 2x the bet before it. So, he must raise by $80.
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#6
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Umm... ok. That's 2 different answers. Who's right?
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#7
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[ QUOTE ]
Each bet must be 2x the bet before it. So, he must raise by $80. [/ QUOTE ]Not in any game I have ever seen or played in. |
#8
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[ QUOTE ]
Each bet must be 2x the bet before it. So, he must raise by $80. [/ QUOTE ] that is a complete wrong answer. no idea what weird houserules you play with. |
#9
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Ok I think I get it. The minimum raise in any round is always either the BB or the last raise amount that round. In the next round, it resets back to the BB, right?
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#10
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Yeh that's right. Just think of the BB as the initial bet, so you either have to call, or raise it by at least as much, or fold.
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