What happens when 4 players all in?

lets say there are 4 players, at which player 1 and player 2 goes all in with the same amount and player 3 and player 4 all in at lower different amounts...what happens then...
you gotta make 3 seperate pots?for player1&2, player 3 and player 4 :S
i dont understand what happens then.

The lowest stacked player's chips go in with an equal amount of chips from the other 3. This is the main pot -- whoever wins of the 4 wins this pot. Then the 2nd lowest stack's remaining chips go into a side pot, which is matched by the other 2 players. Only those 3 can win this side pot. The last side pot is the remaining chips (rest of the 3rd lowest stack, matched by the big stack) and is won by one of the 2 big stacks.

and if player 1 or 2 wins thats the end of the FREEROLL for the other 3

lol yeah, thanks guys

Probably easiest to demonstrate with an example:

Player 1: $1000 at the start of the hand, holding AcKc
Player 2: $1000 at the start of the hand, holding QsQd
Player 3: $500 at the start of the hand, holding AsJd
Player 4: $250 at the start of the hand, holding ThTd

They all get all-in preflop. To make life easy we'll assume Players 1 and 2 were in the blinds and there are no antes. The pots work out as follows:

Main Pot: $1000 (Player 4's $250 and $250 from every other player), everybody is eligible to win this one.
Side Pot 1: $750 (Player 3's remaining $250, and $250 apiece from Players 1 and 2), only Players 1, 2 and 3 can win this one.
Side Pot 2: $1000 (the remaining $500 in each of Player 1 and 2's stacks), only Players 1 and 2 are eligible for this one.

The bit people often get wrong is that the pots need to be awarded in the reverse order from what they were created in. So let's say after the pots have been made right the board comes down like this:

2h 4s Tc 9s Ah

We award Side Pot 2 first. Player 1's pair of aces beat Player 2's queens, so Player 1 is awarded the $1000. Player 2 is felted, and eliminated if it's a freezeout tournament.

Next we award Side Pot 1. Player 1's pair of aces with a king kicker beat Player 3's aces with a jack kicker, so Player 1 takes the $750 from that side pot and Player 3 is also eliminated.

Then we come to the Main Pot. Player 4's set of tens beats Player 1's aces, so Player 4 is awarded the $1000 main pot.

Hand done, move on. Note that if this hand happened in a tournament, Player 2 would be awarded the higher finishing position because they started the hand with more chips than Player 3.

There are two ways of making the pots right BTW. You can just do the math if you're confident but if you're not then the best way is to always start with the player with the smallest stack. Go clockwise around the table and take an amount equal to the smallest stack from each of the other players and use that to make your main pot (ensure any blinds and antes are included in this pot as well, if they're present and assuming the smallest stack is bigger than the big blind).

Once that's done repeat the process with the next smallest stack and so on until there's only one (or even none, like in the above example) player left with chips. That's their change, which they keep, and then you can proceed with the hand, remembering to always award the pots in reverse order.

Thank you guys, now i have a question that is related to this one :
Lets say Player1 calls, Players2 raise, Players3 calls and player4 calls
then player1 re-raises and player2 folds, player 3 all in(because he doesnt match the amount of money that was rised and so creates a side pot) and player 4 calls.

Since player2 folds and has money in the pot...where does the money go...in the side pot or in the main pot?

Follow this rule:

The main pot is the lowest stack that is in the hand (if that player is all in) multiplied by the amount of players remaining in the hand.

For Instance:
P1: 1500 chips
P2: 750 chips
P3: 450 chips
P4: 2050 chips

If raises > 450 when P2 folds: Main POT: 450*4=1800 chips
If raises < 450 when P2 folds it goes this way:

SB 50, BB 100
P1 calls: 100
P2 raise: 250
P3 calls: 250
P4 calls: 250
(blinds fold)
P1 Raises: 600
P2 fold
P3 All in
P4 Calls

Main POT: 150 (SB&BB) + 250 (p2) + 450*3 (Remaining players) = 1750 chips
Side POT: 150*2 = 300 chips

I dont get it,
I mean i get the main pot but i dont get why the side pot is 150*2...
And in your example's, is the blinds like p5 and p6 ?

You work it out the exact same way that I detailed above, by starting with the smallest stack that still has cards. If the player who raised then folded bet more than the size of the shortest stack before folding then the remainder goes into the next pot up and so on - the golden rule is that the shortest stack cannot win more from any one player than they started the hand with, even if that player that has since folded.

Let's do an example using the same stack sizes I had in the example above, and in this case assuming for simplicity's sake that Players 3 and 4 are in the blinds and there are no antes:

Player 1: $1000 at the start of the hand, holding AcKc
Player 2: $1000 at the start of the hand, holding QsQd
Player 3: $500 at the start of the hand, holding AsJd
Player 4: $250 at the start of the hand, holding ThTd

Player 1 open limps, Player 2 raises to $200, Players 3 and 4 both flat call, Player 1 then shoves, Player 2 folds because he's a nit and Players 3 and 4 call off the remainder of their stacks. You end up with two pots, as follows:

Main Pot: $950 (Player 4's $250, $250 each from Players 1 and 3 and the $200 that Player 2 bet before folding). Everybody (other than Player 2, who has folded) is eligible to win this one.
Side Pot: $500 (Player 3's remaining $250 and a matching $250 from Player 1). Only Players 1 and 3 can win this pot, and Player 1 gets the other $500 in his stack back since nobody matched it.

Flop comes down as above (2h 4s Tc 9s Ah). We award the side pot first, where Player 1's aces with a king beats Player 3's aces with a jack, and Player 1 takes the $500 side pot. We then move on to the main pot, where Player 4's set of tens beats Player 1's aces and Player 4 takes the $950 main pot.

If Player 2 bets more, however... here's another example, again assuming that Players 3 and 4 are in the blinds and there are no antes:

Player 1: $1000 at the start of the hand, holding AcKc
Player 2: $1000 at the start of the hand, holding QsQd
Player 3: $500 at the start of the hand, holding AsJd
Player 4: $250 at the start of the hand, holding ThTd

Player 1 open limps, Player 2 raises to $600, Players 3 and 4 call off their stacks, Player 1 then shoves, Player 2 folds because he's a really stupid nit and Players 3 and 4 are already all in. You still end up with two pots, as follows:

Main Pot: $1000 (Player 4's $250 and $250 each from Players 1, 2 and 3). Everybody (other than Player 2, who has folded) is eligible to win this one.
Side Pot: $750 (Player 3's remaining $250 and a matching $250 from Players 1 and 2). Only Players 1 and 3 can win this pot.

Player 1 gets his remaining $500 back, plus the remaining $100 from Player 2's bet which nobody is left to challenge him for.

Then repeat the above process, 2h 4s Tc 9s Ah board, Player 1 wins the $750 side pot but Player 4 wins the $1000 main pot.

Does that cover it?

It covers it but now i am also a little confused because i though that when someone folds he's out of the game...that he cant win anything till the next hand :S

I am talking about Texas holdem poker :/

That's absolutely right. If a player folds they can't win anything in the current hand.

In both of the examples above Player 2 folds and therefore isn't eligible to win anything in the hand. I don't see where I've implied that they could win anything though?

This is one of the main reasons we like online poker....... Computers do all the nasty math ....never a documented mistake......

I am sorry, somehow i readed it wrong...i went and get a sleep and come back and it seems i readed the wrong thing.
Thanks

Seems like i cant edit my last post...anyways..i havent understood a part tho


According to your example i have trained it a little...i will show you it, let me know if its correctly.
(This example excludes blinds)

Player 1: calls $100
Player 2: calls($100)
Player 3: raises to $200
Player 4: calls $200
Player 1: All ins at $150(creating a side pot)
Player 2: calls $100(to match the $200)

Main pot : $150*4=$600(everyone can win this pot)
Side pot : $50*3=$150(the remains of the players, everyone can win this pot except player1)


And here is another example i have practiced using the same numbers.
(This example also excludes blinds)

Player 1: calls $100
Player 2: calls($100)
Player 3: raises to $200
Player 4: All in's with $150
Player 1: Folds(because hes a nit :P )
Player 2: calls $100(to match the $200)

Main pot : $150*3=$450(everyone can win this pot except player1)
Side pot : $50*2=$100+$100(player1 remains)...(only player 3 and 2 can win this pot)



Am i counting it the right way?, pleas let me know.

The last example is wrong:

P1: Calls 100$
P2: Calls 100$
P3: Raises 200$
P4: All in 150$
P1: Fold
P2: Calls 100$

Main Pot: 100+150*3=550 $ (Money from P1 + Lowest stack * Remaining players) (Pot for P2,P3 and P4)
Side Pot: The rest: 50*2 = 100 $ (Only for P2 and P3)

P4 Can win P1 money before folding because he has a bigger stack (150$ > 100$)

^ this. All of Player 1's $100 goes into the main pot that all three remaining players can win, because he bet less than the amount of the shortest stack's all-in bet before folding.

Note also that in this example Players 2 and 3 can continue to bet (because we haven't specified that they're all in, I'm assuming they still have chips behind). If they continue to bet, you add that money to the $100 side pot, not the main pot.

It's like I said above, the golden rule of side pots is that when you're working them out you always always ALWAYS start with the smallest stack first, remember that player is entitled to win up to the amount of their full stack off every player, even those that have folded, make your main pot right based on that and then move up to the next smallest stack. If you think about it any other way, say by starting with the two biggest stacks, the pots will probably end up wrong.

FWIW working out multiple side pots is one of the most difficult things for a beginning dealer to learn. When I was being taught, often my boss would dump eight random piles of chips on the table and say "These are eight different players and all of them are all-in. Make the pots right." It was a pain in the arse at first but give it a try - once you can get something like that right, situations like the above will seem like a piece of cake in comparison.

Thanks guys, i finally got it

And if your playing a self dealt game make sure when you deal you don't let other players stick their hands in the pot to try and help you if you know how to do it.. it usually only causes more confusion when everyone is trying to help, at least for me.

By that logic you've gotta love casinos too, because there's a dealer to do all the math for you LOL

I though i got it but it seems i have one more question about this to ask.

Lets say there is a situation like this :
Player 1: calls $100
Player 2: calls($100)
Player 3: raises to $200
Player 4: All in's with $150
Player 1: all in's with $30(a total of $130)
Player 2: fold



Which one is the correct way of counting this?

Way #1 :
or way#2

This. I've gotten into many arguments with drunk friends that have forgotten how to count lol.

Just count up the shortest stack and match it from everyone that has them covered, then do the same with the next shortest stack and so on until you have made side pots for everyone that is all in.

From your example, way # 2 is the right way. The extra fifty is returned to player 3 because it is an uncalled bet.

It's way two, except there is no third pot - the $50 is simply an uncalled bet that gets returned to the player before you even bother to deal another card, player 3 doesn't need to "win" it even if he's the only one eligible. Then you've got a main pot of $490 that players 1, 3 and 4 can win and a side pot of $40 that only players 3 and 4 can win.