# side pot scenario

What happens if we have 4 players at the table with the following bets

-player1 10\$ - all-in - Cards: 9d, 5h

-player2 50\$ - all-in - Cards: 9h, 4h

-player3 100\$ not all-in - Cards:2d,Kc

-player4 100\$ not all-in - Cards:2h,Jh

Table cards:9s, 3d, 3s, Ac, 2c

During flop, player 4 raises to 100\$. player1 and player2 go all-in. Player3 calls the bet. Player1 and player2 have the same formation rank and kickers, so they should win, but how much?

I assume that player1 takes 10\$ from each player, so he wins 40\$ /2 because he splits the first position with another player.

What about player2, player3 and player4?

You have 3 pots
\$40 - player1, player2, player3, player4
\$120 - player2, player3, player4
\$100 - player3, player4

Hand order
1 - Player1 Player2 tie 9933A
2 - Player3 Player4 tie 3322A

Player1 and Player 2 split the \$40 pot
Player2 wins the \$120 pot
Player3 and Player4 spit the \$100 pot

• The cards/hands of each player is written above at the end of each line. Player3 has pair of 2 and pair of 3, with ACE kicker. Player4 has the same as player3.
– NVG
Commented Feb 12, 2016 at 15:32
• Would this assumption be true: the best hand from each side pot wins that pot or splits it with other players from the pot that have the same formation rank
– NVG
Commented Feb 12, 2016 at 15:59
• Yep, if they have equivalent hands and an equal amount of money in the pot, they would split their side pot(s). Commented Feb 12, 2016 at 16:02
• @nvg OK? I assumed the \$100 in text on the river was in addition. I edited the answer. Yes, pots should total bets. Commented Feb 12, 2016 at 17:53
• That action would typically not happen on the river. You might want to just take the rive part out of the question. Commented Feb 12, 2016 at 18:20