Two Players with Same Chip Count Bust Out in the Money
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Re: Two Players with Same Chip Count Busts Out in the Money
you are right i misread the OP..sorry guys..
Quote:
Originally Posted by hachkc
There are still 2 players playing after these 2 went out so you can't touch 2nd place money. I'm gathering that they were on the bubble and they are only paying 3 places when 3rd and 4th go out; not stated but its a safe assumption.
As to the original question, split 3rd between them unless they agree to a coinflip or something.
Re: Two Players with Same Chip Count Busts Out in the Money
Thanks for all the responses.
Some additional details which I left out of the OP...
- 3 places pay. 4th gets nothing.
- 2 players left after the 2 bust out so I can't really touch the 2nd place money.
- questionable bubble play not up for scrutiny/criticism (even though it should be).
Re: Two Players with Same Chip Count Busts Out in the Money
Quote:
Originally Posted by MeridianFC
Split is correct....
Correct, at least according to Robert’s Rules. Under section 15. TOURNAMENTS:
#35.... Players eliminated on the same deal who start their final hand with an equal amount of chips receive equal prize money, with the best hand on that deal receiving any non-divisible award.
Not sure about TDA, but I'll bet it says (pretty much) the same thing
Re: Two Players with Same Chip Count Busts Out in the Money
Quote:
Originally Posted by DaiTauHa
Thanks for all the responses.
Some additional details which I left out of the OP...
- 3 places pay. 4th gets nothing.
- 2 players left after the 2 bust out so I can't really touch the 2nd place money.
- questionable bubble play not up for scrutiny/criticism (even though it should be).
read: I was one of the all-ins, so go easy on me...I thought a 7-2 offsuit was good in position! (j/k)