Standing of all-in losers in a tournament

Question

When multiple players go all-in during a hand in a tournament, and lose, how is their relative standing (5th, 6th, etc.) determined?

Adam Dymitruk · Accepted Answer · 2012-01-26 09:30:51Z

11

It is determined by their stack size prior to the hand starting.

answered Jan 26, 2012 at 9:30

Adam Dymitruk

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1

This is also true when playing multiple tables on the bubble in a hand-for-hand situation. So if there are two tables and player on each goes all in on the same hand, and they both bust, then the player with the shortest stack at the start of the hand is deemed to be the first person out of the two.
– Lee Oades
Jan 26, 2012 at 11:32
@Adam Dymitruk: so what happens in the hypothetical case where the players had the same stack size prior to the hand? : )
– TacticalCoder
Jan 26, 2012 at 23:15
2

The universe implodes, I'm guessing.
– Adam Dymitruk
Jan 26, 2012 at 23:23
2

If all players have the same amount of chips exactly on the bubble, they split the last prize according to the number of players with the same number of chips
– amigal
Jan 31, 2012 at 16:35

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