running the board more than once

T

TT Swindlehurst

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OK I am a noob so if this is a repeated discussion or an overused topic forgive me. i did do a search for the answer to my question but could not find a satisfactory answer.

I am a tournament director (a very experienced one) at a poker room in india. I was told when i arrived that they allow to run the board more than once here, which is ok i suppose but as i was not familiar with the concept i had it explained to me.

OK

so after a few weeks a situation occurs whereby two players all in decide to run the turn and river. one player has aces another queens. Running the first board brings up a straight (split pot) and the second run nothing spectacular and the aces wins.

Ok here is where it gets tricky. when the running the board more than once was explained to me no mention of a split pot situation was ever made so when the dealer proceeded to split the pot into two and award half of this pot to each player and then give the other split to the aces i questioned the decision and the logic of this.

what the dealer was effectively doing was awarding 75% of the pot to the aces and 25% to the player with queens.

However my logic after nearly 30 years in the business was this....

A player has the right to claim the 'entire' pot for a winning hand unless his hand is either beaten or the same as his opponents in which case it is generally split in equal amounts. every player has a claim on the ENTIRE pot until his cards are beaten. The reason the pot is generally split into two when the board is run more than once is (my thinking) that probability dictates both players winning a share (right?)

in the situation i was faced with we had effectively three winning hands in play (two claims on the split pot and one on the second run) as such logic dictates (and i believe the rules of the game whereby every player has a claim on the entire pot in which he is involved) that as there were three claims on the entire pot that the pot should be divided into three equal shares with aces taking two thirds and queens taking a third as effectively had they not run more than once then the pot would have been divided ' in its entirety 50-50! My supervisors said that i was wrong and that the pot should have been split 75% 25% but that would not have been mathematically correct as 75% is 3/4's of the pot not 1/3rd.

my logic is that the pot is a pot in its entirety and both players have equal claims until their hand is defeated. the way it was being done is that the pot was split effectively making it two pots and not one. only three winning hands were shown NOT 4 so how could the pot be divided into four parts?

i maintain that i was correct and that as the players were happy it was the correct decision however i would appreciate some clarification by a knowledgeable member of the forum.
 
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Dank Hugh

Dank Hugh

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your supervisors are correct. the 75-25 split is correct.

your "logic" is so fuzzy I cannot follow it.

"The reason the pot is generally split into two when the board is run more than once is (my thinking) that probability dictates both players winning a share (right?) " WRONG !!

when you run it twice you run it for 50% of the pot each time end of story
 
dd_decker

dd_decker

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Seems to me that the split was done correctly and that you are overthinking it.
 
Dorkus Malorkus

Dorkus Malorkus

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i've read your 'logic' about 6 times now and i still have no idea what your point is or what you're trying to say.
 
WVHillbilly

WVHillbilly

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Honestly I think the AA hand should have one the entire pot. In a situation without the tie the pot would go to either hand that won twice or it would be split if each hand won once. Since the 1st pot was a tie to my mind everything should have been riding on the 2nd run of the cards.
 
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Sori

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100% agree with WV. I do believe that the split was technically done correctly, but wouldn't it be so much better if the 2nd run of the cards determined the full pot amount? Way more exciting.
 
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baudib1

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AA gets 75%, QQ 25%. You are dividing the pot into 2 and each run is for half of it.

Player A has $100
Player B has $100
Pot is $200

First run = board is straight, chop. Player A gets $50, Player B gets $50
Second run = AA wins. Player A wins $100

I can't see how this can be any more straightforward or interpreted any other way.

If you want to do it the way WV suggests, what happens if you have a freeroll situation, i.e. both players flop the nut straight but one has a flush draw.
 
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