# Do previous bets count towards All-In raises?

Let's say 3 players are playing poker. Player A has $11, B has$10, C has $9 Player A opens by betting$3, leaving A with $8 Player B re-raises A "All-In" and C calls, going "All-In" as well. Player A then put his remaining$8 into the pot.

My question is, is player A considered the "short stack" in this situation, does he go all-in with $8 (after betting) or$11 (before betting)? If player A loses, will he lose all his money or is he eligible to win from side pots?

• Player A only has to bet as much as the maximum previous bet. That is player B, who bet $10. Since A has$11, he only has to call the amount up to $10 to call the "all-in" thus he still keeps his extra$1 despite calling the all in. – Kenshin Jan 10 '13 at 12:13

Specific to the example, player A can only call the All-In for $7, although his total amount staked is$10, equaling that of player B. Opposing players still to act can only call the difference between their current staked amount and what opponents have bet, assuming they are closing the action. Otherwise, they can bet/raise up to their total stack assuming another player still to act can match it. You mention "before betting" and "after betting" and i'm not certain of what you mean, but it seems irrelevant as the game is sequential. Further action only exists after any action state. The term short stack is only a description and has no real use in game mechanics.
• players A, B, & C contest the main pot. ($27) • players A & B contest the only side pot. ($2)
In a little more detail, player C can only win a multiple of his total staked amount ($9), thus the$27 main pot, or 3 players in that pot for $9 each. Players A & B can win the main pot and the side pot, where the side pot consists of the extra money staked above what the lowest amount staked was (i.e. above$9). Player A only has to match this extra amount of $1 (i.e. difference between what player B & C staked), thus 2 players in the side pot totaling$2.