 gekko13
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| Joined: 02 Dec 2011 |
| Total Posts: 3 |
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7 Card Stud betting structure question...
Posted: 12-02-2011 05:13 AM |
Hello folks. I have a question about 7 Stud.
In a $4-$8 game, I know that the forced bring-in would be $2. My question is, can the second player raise it to $6 or can he only complete it to $4? If he can raise it to $6, would that make the cap at $14?
Which now brings me to another question...
If it is a $5-$10 game with a $2 forced bring-in, can the next player raise it to $7? I would have to say NO because the $2 is not half of the full bet.
Thanks for the help.
J.C. |
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Re: 7 Card Stud betting structure question...
Posted: 01-09-2012 05:03 PM |
"In a $4-$8 game, I know that the forced bring-in would be $2. My question is, can the second player raise it to $6 or can he only complete it to $4? If he can raise it to $6, would that make the cap at $14? "
2nd player can only call $2 or complete to $4. Third player can make it $8, etc. Cap on 3rd street is always going to be $4 x 4, with the usual betting structure. So bring in starts with $2. Next player completes to $4. Everyone else loving their hands make it $8, $12 and then $16. That's the cap for 3rd street, and we move to 4th...
"If it is a $5-$10 game with a $2 forced bring-in, can the next player raise it to $7? I would have to say NO because the $2 is not half of the full bet. "
When the first player brings in the minimum forced bet ($2 in this case), the next player has 3 options. Fold, call $2 or complete to $5. Only after someone "completes" the bring in bet to $5 can someone else bump it to $10, $15 and then $20....
Both of these questions are related. If you just remember that someone has to complete the bring in before the usual betting limits kick in, that'll answer your questions I think... |
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