Whats the simple way to do "he gets 2/3 of the pot" when player is all in? - Beginning Poker Questions

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Whats the simple way to do "he gets 2/3 of the pot" when player is all in?

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10-01-2018 , 12:47 AM

Damani311

old hand

Join Date: Apr 2006 Posts: 1,560

When it hits that point that two people get all in, run it 3x for whatever reason, and someone gets 2/3

There is an easier way than having to take all the $$ in the middle and make three stacks, right?

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10-01-2018 , 01:06 AM

WereBeer

Carpal \'Tunnel

Join Date: Oct 2013 Posts: 10,700

Count the pot, divide by 3 in your head and cut out that many chips to pay 1/3 and push the rest to the main winner?

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10-01-2018 , 01:29 AM

browni3141

Pooh-Bah

Join Date: Aug 2015 Posts: 5,678

Of course the money taken in already has to be cut into three stacks the way you describe.

For the money left in front, the losing player can cut his effective stack into thirds and give the winning player 1/3 of the effective stack, while the winner takes 2/3 of the collected bets and takes back his all-in bet.

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10-01-2018 , 03:02 AM

Damani311

old hand

Join Date: Apr 2006 Posts: 1,560

Quote:

Originally Posted by browni3141

thanks!

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10-01-2018 , 09:13 AM

answer20

Carpal \'Tunnel

Join Date: Feb 2013 Posts: 10,853

Yep, that's it ... just treat the chips not yet pulled into the middle as 1/2 'pot' each and figure things from there using the nearest common multiple/denominator.

For 3/4 pots ... 1/2 is 2/4, so take 1/2 of the other stack for 2/4 + 1/4 = 3/4

For 2/3 pots ... 1/2 is 3/6, so take 1/3 of the other stack for 3/6 + 1/6 = 4/6 = 2/3

It looks messy on paper and that can cause some issues with Players who can't follow the math, so in those cases just pull it all in and chop it 'correctly' using the whole pot. GL

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10-01-2018 , 02:50 PM

Ray Zee

Carpal \'Tunnel

Join Date: Aug 2002 Posts: 17,920

the way we used to do it in the games with lots of chips in the pots would be to just make 3 pots out of the one without counting the 3 pots. then each pot is played for individually when running out.

otherwise it slows the game down to a crawl.

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10-02-2018 , 09:27 AM

answer20

Carpal \'Tunnel

Join Date: Feb 2013 Posts: 10,853

Players get finicky when Dealers actually can do the math in their head. Have you ever seen the horrifying look on a Player's face when the Dealer creates a side pot exclusively from one Player's stack instead of pulling 'exactly' the right amount from the bigger stacks? GL

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03-21-2020 , 03:18 PM

Damani311

old hand

Join Date: Apr 2006 Posts: 1,560

Quote:

Originally Posted by Ray Zee

wouldn't that actually take the longest of all the options?

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03-21-2020 , 08:39 PM

George Rice

old hand

Join Date: Oct 2002 Posts: 1,873

Quote:

Originally Posted by Damani311

wouldn't that actually take the longest of all the options?

I think the point is that the pots weren't counted, therefore there was no time used counting. And yes, that means they weren't equal values. But so long as they're defined as to which pot goes to which run-out, it's fair in the long run, as anyone can win the largest "third".

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