Same chipcount, who bubbles and who gets paid

[DutchTD076]

Hi all,

There is a 3-way all-in during the final stages of a tournament, 21 players left and 20 places paid, the biggest stack wins the pot, eliminating the two other players.
After a careful count it turns out that both of them have exactly the same chipcount, who will be bubble and who finish in 20th place (and get paid)?

My “pokerinstincts” tell me to give the prizemoney to the player with the best hand but since he didn’t actually beat anything in that hand, I’m doubting that decision. Some say you should split the prizemoney between the two players but that doesn't feel right as well. 

Can you help me out here? Thnx in advance for your thoughts!

[Brian Vickers]

If they both had exact same chips I would split 20th place prize evenly between them both.  I have written it in my rules that way as well.

[Steff0111]

Splitting the money is the fairest way!

[Dave Miller]

Your first instinct is correct. This sort of thing happens frequently in the poker league I deal for.

There should be a house rule covering this, and there probably is, based upon your description.

This is kinda like a side-pot resolution, but to determine a winner among the payers that are only interested in the main pot.

It sounds like the winner not only had the biggest stack, it was obviously bigger. Therefore, why are you bothering to count the stacks of the losers? Because the rule states that the cash would go to whomever had the bigger stack - even though they both lost.

Here's the reasoning. Had the big stack only had as many chips as the shortest stack, the loser with the bigger stack would still be in the game. Had the big stack actually have less than the two equal size stacks, then the one of those with the better hand would still be in the game.

The only time the rules would state that the prize needs to be shared would be if the two losers had the same stack and the same hand.

[Brian Vickers]

This will probably be one of those "split down the middle" opinions where you're just going to have to decide on what your house rule is going to be and go with it.  If it's your room, run it your way when there isn't a precedent set or very common way of doing it.

[Max D]

I have seen it before where the money gets split evenly.

[BillM16]

Welcome Dutch, good question!

The current TDA rules do not cover this ... perhaps the next should?  So, house rules should apply.  The 2015 WSOP rules offer this:

119. Hand for Hand:
  Upon nearing the “Money” (the first level of the prize pool payout), a “Hand-for-Hand” method of determining participant
placement within the prize pool and the actual amount of prize pool disbursement within that event will be utilized. This will begin by completing the current hand in progress at all tables. Once all hands are complete, the dealer at each table will deal one hand only, then – once the hand is complete – suspend play. This process will continue until enough participants have been eliminated to reach the money.

  During the Hand-for-Hand process, more than one participant may be eliminated during the same hand. If two or more participants are eliminated during the same hand at different tables, those participants will “tie” for that place finish. If two or more participants are eliminated during the same hand at the same table, the participant(s) who began the hand with the highest chip count will receive the higher place finish.

My ruling would be a tie and split the $$.

Regards,
B~

[Dave Miller]

Wow. It REALLY surprises me to see a rule that treats two players differently based solely on whether they're at the same table or different tables.

[BillM16]

Dave,

I can rationalize the single table situation like this:  When three players at the same table choose to play a hand, they do so realizing their own stack size and how that compares to their opponents. In other words, they are knowingly assuming the risk.  If two players lose all of their chips to the third player, it seems correct to me that the losing player who began the hand with the most chips should finish in a position ahead of the losing player who began with the least chips.  The losing chip leader assumed a larger risk and in one sense finished ahead of the losing small stack by paying off the larger percentage of the amount due to the winner.

In the multi-table situation the same cannot be said:  When the two losers are at different tables their stack sizes relative to each other are irrelevant as they were not playing against each other.  In other words, they did not knowingly assume the risk.  In fact, it is mathematically possible that the losing chip leader paid only 10% percentage of the chip amount due to his winner, while the losing small stack might have paid 99% of the total chip amount due to his winner.  In both cases, the losers paid 100% of there chips but in one sense the smaller stack finished ahead of the larger stack by paying a larger percentage of the amount due to his winner.  I'm sure there are other factors that could be distinguished when players are at different tables.

Put simply, without contrived mathematical scenarios, according to WSOP, you can only compare losing stack sizes when both players are at the same table and playing in the same hand.  You cannot compare two losing stack sizes at different tables playing in different hands.

Regards,
B~

[Uniden32]

Bill's reply covers it pretty well, and it's the method that we use here as well.

I agree, maybe we can add this to the TDA rule set.

[Dave Miller]

Bill -

When I wrote my initial response, I hadn't considered that the players were at different tables. (And, re-reading the initial post, it remains unclear.)

Since I deal in a poker league, my thinking was all about a single table situation. Or at least where only final table players get paid.

Upon reading your response, your solution, at least for tournaments where at least everyone at the final table gets paid, makes sense.

I said everyone because of the scenario where two players are eliminated on the bubble.


On a semi-related note, I recall something similar several years ago, during the final table of a televised event, with three players remaining, and one player all-in. The player with the second biggest chip stack was thinking about going all-in and asked for a rule: If both short stacks are eliminated, would he get 2nd place money? The answer was yes.

[DutchTD076]

It sounds like the winner not only had the biggest stack, it was obviously bigger. Therefore, why are you bothering to count the stacks of the losers? Because the rule states that the cash would go to whomever had the bigger stack - even though they both lost.

Here's the reasoning. Had the big stack only had as many chips as the shortest stack, the loser with the bigger stack would still be in the game. Had the big stack actually have less than the two equal size stacks, then the one of those with the better hand would still be in the game.

The only time the rules would state that the prize needs to be shared would be if the two losers had the same stack and the same hand.

Dave,

I counted the stacks of the losers to determine who came in 21nd place and who in 20th. If they hadn't had the same stacks, the one with the bigger stack would obviously be 20th and the smaller stack 21nd, but since they have the same chipcount it is impossible to determine it that way. My question was if the handranking of the two losers should be leading then.

By the way, I forgot to mention in my initial post that all three players were at the same table, you filled that in correctly.

Put simply, without contrived mathematical scenarios, according to WSOP, you can only compare losing stack sizes when both players are at the same table and playing in the same hand.  You cannot compare two losing stack sizes at different tables playing in different hands.

Bill,

I think this would be a fair addition to the TDA rule set.


That being said, I would like to express my gratitude to to all responders, it strengthens me to see that there isn't an obvious solution for this situation.   

The bigger/smaller stack arguments have set me on the right track I guess. When there is a bigger and a smaller stack, the bigger one gets eliminated "later" than the smaller stack. When the stacks are the same, they are eliminated at the same time. Therefor is the handranking irrelevant in this situation. My ruling in the future will be that it is a tie and split the money.
Ofcourse I'm going to try to get this added to my company's house rules

Regards,

T.

P.s. English is my second language, if it is somewhat crooked, please don't be offended.