In Texas Hold'em, who is awarded second and third place in this scenario?

Question

Consider the scenario where there are three players remaining:

• Player A has 1,000 chips
• Player B has 200 chips
• Player C has 50 chips and is the dealer

The hand starts and Player C goes all in. Player A calls and Player B goes all in. Player A calls Player B's all in. At this point everyone is all in and turns their cards over.

In the end, Player A's hand beats both Player B and Player C. Clearly, Player A has won first place. And if Player B's hand beat Player C's hand then clearly Player B would have won second place, and Player C third.

But what if Player C's hand beats Player B's hand - who gets second place and who gets third place in that scenario? I wager one of the following three outcomes is what should happen, but I'm not familiar with the rules to know what the "official" outcome should be:

1. Player C wins second place because his hand beat Player B's hand, even though Player B had more chips at the start of the hand.
2. Player B wins second place since she had more chips at the start of the hand than Player C.
3. Player A wins first place and gets to go off and roll around in the piles of money he has won. Player C gets 50 of Player B's chips, since he went all in with his 50 and won the hand and the two stay at the table and play heads up to determine second and third place.

Thanks

Answer 1

Here, the important factor is that Player B and Player C both lost to, and were "busted" by Player A's hand. Not that Player C's hand beats Player B's.

If two players "bust out" on the same hand, then the prizes are awarded in DESCENDING chip order as of BEFORE the mutual bust. That is Player B comes in ahead of Player C.

If Player C had sat out with 50 chips, and only Player B was busted, Player C would continue. (And probably win second place unless he were able to stage a complete reversal.)