3 ms·

It depends what you mean. As people have noted, the bet sizes before the river are much larger than what would happen in a well-played game today. I believe ev

It depends what you mean. As people have noted, the bet sizes before the river are much larger than what would happen in a well-played game today. I believe every action by both players before the river is a huge mistake relative to theoretically correct poker. So something like this exact hand would never play out between two competent players today. However, the core problem is that one player has shoved the river and the other player beats all missed draws and no value, which is still a common situation. Since the prizes are $100k or $0, and the stacks will be roughly even if Ungar folds on the river, he needs to win more than 50% of the time to profitably call. Do top players ever bluff too much in this situation, and have their opponents profitably call them off? Sure, that happens a lot even at high stakes. Doug Polk recently had a video where he described how he lost a high stakes heads up pot to Jungleman in a similar situation.

confidantlake 7y ago
I may be misunderstanding what you are saying. Are you saying Ungar needs to to win more than 50% to profitably call? That seems incorrect. There is no situation between 2 players where you need to be right more than 50% of the time. Even if there was only 1 chip in the middle and your opponent bet 100 chips, you would still only need to win slightly less than 50% of the time. If there are 100 chips in the middle and your opponent bets 100 chips, you only need to be right 33% of the time.
- yellowstuff 7y ago
  You're correct that I made a dumb mistake. In a tournament we want to calculate our EV in dollars of expected payout, not in chips. Here's the math I should have done: 1) Ungar folds, he is left with about half the chips in play, so about 50% odds of winning the tournament, and EV of ~$50K. 2) Ungar calls and is right, winning $100k. 3) Ungar calls and is wrong. I forgot that he'd still have chips left and assumed that this scenario was worth $0. In fact he'll have ~20% of the chips in play left, which is worth $20K intuitively but can be confirmed here: https://www.icmpoker.com/icmcalculator/ https://www.icmpoker.com/icmcalculator/ So if Ungar calls he needs to be right X% such that $50K = $100K * X% + $20K * (1-X%) X ~= 38%

confidantlake 7y ago

I may be misunderstanding what you are saying. Are you saying Ungar needs to to win more than 50% to profitably call? That seems incorrect. There is no situation between 2 players where you need to be right more than 50% of the time. Even if there was only 1 chip in the middle and your opponent bet 100 chips, you would still only need to win slightly less than 50% of the time. If there are 100 chips in the middle and your opponent bets 100 chips, you only need to be right 33% of the time.

yellowstuff 7y ago
You're correct that I made a dumb mistake. In a tournament we want to calculate our EV in dollars of expected payout, not in chips. Here's the math I should have done: 1) Ungar folds, he is left with about half the chips in play, so about 50% odds of winning the tournament, and EV of ~$50K. 2) Ungar calls and is right, winning $100k. 3) Ungar calls and is wrong. I forgot that he'd still have chips left and assumed that this scenario was worth $0. In fact he'll have ~20% of the chips in play left, which is worth $20K intuitively but can be confirmed here: https://www.icmpoker.com/icmcalculator/ https://www.icmpoker.com/icmcalculator/ So if Ungar calls he needs to be right X% such that $50K = $100K * X% + $20K * (1-X%) X ~= 38%