I'd like to know the procedure on chopping an unclosed pot with multiple all ins.
This is a hypothetical situation. Three players all in in a 1/3 game with the two small stacks chopping the pot. It's 1/3 so you can chop down to the dollar.
- Player 1 is all in for $25 and splits the pot
- Player 2 is all in for $51 and splits the pot
- Player 3 is all in for $1,000 and loses to player 1 and 2.
- Pot has $25 in it and no other player has gone all in or is involved in the hand.
- Assume player 1 is closest to the button.
This is what I think would happen but I'm not sure if it's correct.
Dealer takes $25 from each player for $75 total and chops it in half, $37 to player 1 and 2, and then giving the odd chip to player 1. Dealer then splits the main pot, giving $12 to player 1 and 2 and the odd chip to player 1.
Player 1 has won a total of $51 ($37 + $1 + $12 + $1).
The dealer would then take $26 from Player 3 and give it to
player 2 who wins a total of $75 ($37 + $12 + $26).
or would it happen this way:
Dealer takes $25 from each player and combines it with the main pot for a total of $100. Dealer gives
$50 to player 1 and $50 to player 2. Dealer then takes $26 from player and gives it to
player 2 who now has won $76 ($50 + $26).
I know it's
only a difference of a dollar but the first example gives player 1 an extra dollar because he got two odd chips.