Who takes how much in these situations:

[yuvallahav]

So we have this situation:

Player A went all in with 1000$.
Player B went all in with 500$.
Player C went all in with 100$.
Player D went all in with 50$.

In our case example 1, player D wins, and we know since he went all in, he can only get back from the pot (which has 1650$ at this point) 200$ (4x the 50$ he went all in with).
Where does the other 1450$ go? back to players? to a second best player?

In case example 2, Players D and A both win with the same hand and kickers, so generally speaking, we are supposed to divide the pot half and half, but player D can still only get 200$ for is all in of 50$, so how much does player A get? half the pot? and if so, where does the rest go? half the pot and the the left overs of player D?

Please help me with these cases as they are a standing argument point where I'm writing this from, and we can't find anything online about these specific cases.

Thanks.

[abstaining]

Situation 1: the rest of the chips would go to the player with the next best hand.

Situation 2: Player A & D would split the $200 down the middle, and Player A would take the rest

[yuvallahav]

Well, there is no 200$ to split, since the pot is 1650$, but from what you are saying I believe you mean that what should happen is that we split the pot in half (825$ each half) of which player D (50$ guy) will get 200$ off his half, and player A will get his while half and also whats left of D's half, since he had 1000$ in and should be getting 4x1000$ (which is anyway impossible since there is no 4000$ in the pot), but the order of action should be that first we split the pot, each player takes from his part, and leftovers will go to who ever can get them buy the amount of money he has put in... am I right?

[abstaining]

No.

Player D ($50 all in) can only win a max of $200, so if he and player A have the same hand, they split that $200 that he is eligible to win. Player A takes the entire remainder of the pot.

[yuvallahav]

I'm not sure I got this... this will be used to program a game online, so I need to get the idea as best I can... (still, a free game, no real money involved, but we're trying to be as realistic as possible).

Why would the 200$ be split? we have 1650$ in the pot, player A (1000$ guy) and player D (50$ guy) both won, so why would the 200$ that player D should take from the pot be split? I mean, I'm looking for a reason and a way to compute this, I can say these things:
pot is 1650$.
player A and D won.
player A can take up to 4000$.
player D can take up to 200$.

I don't see why would the 200$ pot be divided into 2 parts, when it's the win of player D...
if no caps apply, and no limits apply, 2 players winning a pot would divide the pot in half, so thinking of this situation only makes sense that the pot will divide in 2 (825$ each) out of which player D will take his 200$, player A will take the whole half of the pot (again, 825$) and the rest of whats left of player's D pot (which is an extra 625$), or up until he gets the max amount he can take, according to how much he has opted in.
So why is there a division of the 200$ takings of player D?

[Felting]

I'm not sure I got this... this will be used to program a game online, so I need to get the idea as best I can... (still, a free game, no real money involved, but we're trying to be as realistic as possible).

Why would the 200$ be split? we have 1650$ in the pot, player A (1000$ guy) and player D (50$ guy) both won, so why would the 200$ that player D should take from the pot be split? I mean, I'm looking for a reason and a way to compute this, I can say these things:
pot is 1650$.
player A and D won.
player A can take up to 4000$.
player D can take up to 200$.

I don't see why would the 200$ pot be divided into 2 parts, when it's the win of player D...
if no caps apply, and no limits apply, 2 players winning a pot would divide the pot in half, so thinking of this situation only makes sense that the pot will divide in 2 (825$ each) out of which player D will take his 200$, player A will take the whole half of the pot (again, 825$) and the rest of whats left of player's D pot (which is an extra 625$), or up until he gets the max amount he can take, according to how much he has opted in.
So why is there a division of the 200$ takings of player D?

Here's your problem, your not creating separate side pots after players are all in. Once a player is all in he can only compete for the pot that built up until he was all in. If he ties with another player, he's still only intitled to half the pot he's competing for, simple as that.

[HalfSugar]

Let's simplify the example to try and add clarity:

Let's say there are three players with chip stacks as follows:

Player A - 3,000 chips
Player B - 500 chips
Player C - 500 chips

Player A goes all-in for 3,000 chips. Players B and C also go all-in for their 500 chips each.

Here's the important part - Player A immediately gets 2,500 chips back because no other player can compete with him for those chips.

Your example is the equivalent of this with regard to Player A - as soon as it is determined that he and Player D tie with the best hand, whatever Player D is eligible to win (eg $200) is removed from the pot and the entire rest of it goes to Player A. The $200 is then split between Players A and D.

In reality, the pots would be split as the chips enter the pot but for clarity, I've explained it this way and we'll assume there is just a big pile of $1,650 chips in the middle of the table.

The other important bit is this - no player can EVER win more chips from a single player than he has to wager so Player D could only ever win a maximum of $200 chips in your example.

[yuvallahav]

You see, the "Player A immediately gets 2,500 chips back because no other" is the part I was missing, I didn't see that anywhere I was looking for the right rules... if that is the case then it's more understandable how the division goes...

but again, I don't understand why the 200$ should be split...

lets make this super simple, 2 players, A and B, so lets say player A goes all in with 1000$, and player B goes all in with 100$.
So the next step is to give player A 900$ back (since 100$ is the max amount any other player, in this case, B, can win off him), and the we are actually left with a pot of 200$ total, which is split half way... this makes sense... in a way, so returning to our first example of 4 players, what do we do once all 4 players went all in with the sums 1000,500,100 and 50, and in that order, since in that order, our main pot is capped to 1000, but since all other players have less then that, they can all go all in without creating any side pot, so in the case, what happens when players A and D both win? if we go by your example, since A and D won, and A can win a max of 50 off player D, he gets back 950$ right away (1000 minus the 50 to stay in with player D), so now what remains in the pot is 1650-950, which is 700$, and he also get 500 he won off player B, but what about the winnings from player C?? as I calculate this, we are coming short... this is the reason I gave that specific example in my need to understand the rules, can you please break it down to me step by step for my first example of 4 players playing 1000, 500, 100, and 50??

[HalfSugar]

Sure. Here goes:

Player A is all-in for 1,000 chips. All 1,000 chips get pushed forward.

Player B goes all-in for 500 chips. If the action were to stop there, Player A would take back 500 chips and the pot of 1,000 chips (500 from each player) would be contested.

When Player C goes all-in for 100 chips, the following happens:

A 'main pot' is created in the value of 300 chips. This is 3 x 100 chips from each player. The reason it is 3 x 100 is because Player C only has 100 chips and therefore can only win 100 from each player.

A 'side pot' is created in the value of 800 chips. This is made up of 400 chips from Player B (the remainder of his stack) and 400 chips from Player A to match it. Again, Player A gets 500 chips back because no-one can contest his additional chips.

In this case, if there were no player D, the three players would contest the pots. All three players are able to win the main pot based on who has the best hand but only Players A and B can win the side pot depending on who out of the two of them has the better hand, even if Player C's is better than both of them. Player C cannot win the side pot because he wagered no money into it.

Moving on, IGNORING WHAT HAS GONE BEFORE, when Player D goes all-in for 50 chips, the 'main pot' is created to the value of 200 chips. This is 4 x 50 chips being the value of Player D's stack taken from every player.

'Side pot 1' is created to the value of 150 chips which is 3 x 50 chips. The reason it is this value is because Player C only has 50 chips left after contributing to the main pot and Players A and B can match it. Players A, B and C can contest side pot 1, Player D cannot.

'Side pot 2' is created to the value of 800 chips which consists of the remainder of Player B's stack (400 chips) and a matching amount from Player A. As always, Player A gets 500 chips back.

All four players then contest the pots with the following possible winners:

Main pot - all four players
Side pot 1 - Players A, B and C only
Side pot 2 - Players A and B only

To show how each of them got their chips in:

Player A - 50 in the main pot, 50 in side pot 1, 400 in side pot 2 = 500 total
Player B - 50 in the main pot, 50 in side pot 1, 400 in side pot 2 = 500 total
Player C - 50 in the main pot, 50 in side pot 1 = 100 total
Player D - 50 in the main pot = 50 total

In your example, Players A and D tie for the winning hand. The first pot that is awarded is always the main pot so in this example, the 200 chips are split with 100 going to Player A and 100 going to Player D.

Side pot 1 is awarded next and because Player A has a better hand than Players B and C, he takes it all.

Side pot 2 is awarded next and because Player A has a better hand than Player B, he takes it all.

Chips won/lost are as follows:

Player A - won 100 from the main pot, 150 from side pot 1 and 800 from side pot 2. Having got 500 chips back at the start due to no-one having enough chips to put them at risk, he has a total of 1,550 chips after the hand is over.

Player B - lost all three pots he was eligible for and is eliminated.

Player C - lost both pots he was eligible for and is eliminated.

Player D - won 100 from the main pot and has a total of 100 chips after the hand is over.

The total chips after the hand is still 1,650 (1,550 + 100) and Players B and C are eliminated.

As a side note, regardless of what cards they had, Player B would finish ahead of Player C by virtue of starting the hand with more chips.

[yuvallahav]

Wow... I'm going to have to process all this before I get back to you on it, but on a side note, I was under the impression that a side pot is created only if someone had gone into a bet while there is a pot with a cap, in whats more, a pot has a cap only if some one went all in on that pot while betting the highest sum, was all this a wrong assumption?

I mean, lets say player D plays first, 50$, all in, so we have right away a "main" pot with 50 in it, then player C plays his 100, or less, but more then 50, so 50 goes in the main pot, and the rest will go to a side pot, if the he went all in, then the side pot will be capped to what ever that left over sum is, in our case, 50$ again, now player B plays his all in with 500, so 50 to main pot, 50 to second pot, and 400 to a new side pot which will be capped to 400, in this order of playing, player A now can only bet 500 anyway (he can't all in or raise, just call or fold, since he's the only player with remaining chips), so he bets 500, 50+50+400) and the hand is dealt to completion.
I didn't know that when betting the other way around will create in the end the same situation of 3 pots...
I was always sure that if the betting took place in a reverse order, then 1000 will make the main pot where all the rest of the players going either all in or fold, and all the bets go into the single main pot... that was why I had the problem of trying to figure out who get what (and more important, why...).

Thanks again, I'll go over what you wrote and I'll come back if I have a comment or a question.

Yuval.

[yuvallahav]

OK, I've went through what you explained, and now it makes sense... now I just need to write the algorithm to do it for me, so here I go... again..
Thank you so much for the explanation, and the time you took to write it, and I'll be sure to let you know when the work is done and the game goes on our site!

Yuval Lahav.

[HalfSugar]

Cool, we look forward to seeing the result.

The algorithm should be fairly easy, it's just a matter of ranking players by smallest stack first and extracting matching sums of chips from players with larger stacks until someone is left with a surplus to be returned.

[yuvallahav]

Yep, already done, since I had the algorithm already written, but it worked only one way, bottom to top (as I explained, if some one went all in with 50, then second player went all in with 100, then it would have created 2 pots, one with 100, and one with 50, and so on), the only thing I actually needed to change was to reset all my pots after each bet, reset the total bet value for the current hand with the new bet sum, and sort it numerically in ascending order, simple

Thanks again, and I'll keep you updated!

Yuval Lahav.

[yuvallahav]

Hell there...
So I'm almost done with our game, last stages, and I've come to this situation I don't know what to do with...

Player A goes all in with 50$.
Player B plays normally and have 1000$ on the table.
Player C plays normally and have 1000$ on the table.

So far we have 2 pots, one with 150$ and one with 1900$.

In this point, for unknown reasons we don't care about, player B and C both fold....

My logic tells me that player A, anyway, can win just the 150$ pot... what happens to the 1900$ pot?? or when a player is left alone at the table he takes all the pots no matter how much he's invested in them??

Thanks again.

Yuval.

[golddog]

Hell there...
So I'm almost done with our game, last stages, and I've come to this situation I don't know what to do with...

Player A goes all in with 50$.
Player B plays normally and have 1000$ on the table.
Player C plays normally and have 1000$ on the table.

So far we have 2 pots, one with 150$ and one with 1900$.

In this point, for unknown reasons we don't care about, player B and C both fold....

My logic tells me that player A, anyway, can win just the 150$ pot... what happens to the 1900$ pot?? or when a player is left alone at the table he takes all the pots no matter how much he's invested in them??

Thanks again.

Yuval.

Correct on player A's pot.

For the other pot, the second player to fold is taken out and beaten to death for being so stupid (hopefully before they've reproduced), so I guess it would go to the first by default.

Good on you to think of the corner cases, but that's not a real-world situation. C (if he was after B) would never fold and would pick up the side pot.

To allow for a mis-click, I suppose the side would be split between B and C?