[Spaceman Spiff]

All right, I should know the answer to this. I think I handled it right, but I got some skeptical looks.

Down to 3 handed. Say Player A is on the button, 'B' is the SB and 'C' is BB. The button busts. For the next hand, the button should move to player B, but then that would force player C to pay the BB two hands in a row. I said that you couldn't do that and the button jumps to player C to keep the rotation of blinds correct.

Should it be done differently?

[CaptLego]

I like the way you handled it. According to Robert's rules, you have a choice:
http://www.chiptalk.net/roberts_rules_of_poker.php

I never really comletely understood the "moving button" method, so I don't know how that would apply in this situation. But your solution seems to be a good application of the "dead button" rule.

I like to advance the BB. If the BB busts out, then the SB ends up on an empty seat (so nobody has to post the SB twice in a row). But in this case, the person to the left of the BB busted out, so just moving the BB and SB ahead to the next available players seems to work out fairly.

[dad604]

I like it where the blinds is posted like it should have been and the button moves to the SB. I think that is the fairest way.

[Spaceman Spiff]

Originally Posted by dad604

I like it where the blinds is posted like it should have been and the button moves to the SB. I think that is the fairest way.

I guess that's basically what I did. I use the Dead Button rule during the rest of the tournament. It just always seems to get tricky when it goes to heads up.

You bring up an interesting situation Lego where the BB busts. I'm sure we've had this happen also and you're right, the same player has probably ended up paying the SB 2 hands in a row. I guess the proper way to do it for hand 1 would be no SB. Button would move as normal, but the button would have no blind for that first hand. The heads up opponent would be the BB. On hand 2, you'd have a dead button and the 2 normal blinds. For hand 3, dammit, that doesn't work either. (Can you tell I'm thinking this through as I type. You'd have the same blinds 2 hands in a row in this case and that's what I want to avoid. I guess the only way around it would be for the button in this case to post the SB again even though he would have just done it.

[Niagara_All_In]

At our home games if that situation came up player B would have been BB and C would have been dealer/SB.

Once its heads up we kill the SB button and just have BB and Dealer with the dealer posting the SB.

there are way to many ways you could have handled it and you did the right thing.

[CaptLego]

Originally Posted by Spaceman Spiff

I guess that's basically what I did. I use the Dead Button rule during the rest of the tournament. It just always seems to get tricky when it goes to heads up.

You bring up an interesting situation Lego where the BB busts. I'm sure we've had this happen also and you're right, the same player has probably ended up paying the SB 2 hands in a row. I guess the proper way to do it for hand 1 would be no SB. Button would move as normal, but the button would have no blind for that first hand. The heads up opponent would be the BB. On hand 2, you'd have a dead button and the 2 normal blinds. For hand 3, dammit, that doesn't work either. (Can you tell I'm thinking this through as I type. You'd have the same blinds 2 hands in a row in this case and that's what I want to avoid. I guess the only way around it would be for the button in this case to post the SB again even though he would have just done it.

Try this:

Hand 1: A=button, B=SB, C=BB (C busts out)
Hand 2: A=BB (it's his turn), no SB (empty seat at C), B = button
Hand 3: A=SB & button, B=BB

carry on

[tomb1]

Just finished a tourney online:

We were down 3 players.

Small blind goes all in and loses.

Next hand -- now heads up for the first time -- button moves to the empty seat where previous SB was, and remaining two players get SB and BB correctly.

Next hand the button moves to the player who got the SB on the first heads-up hand, and continues alternating after that of course. (This means he got SB and first action twice in a row.)

[Fins]

Originally Posted by CaptLego

Try this:

Hand 1: A=button, B=SB, C=BB (C busts out)
Hand 2: A=BB (it's his turn), no SB (empty seat at C), B = button
Hand 3: A=SB & button, B=BB

carry on

That's the way I'd handle it... anytime the BB busts there is no SB next hand... an easier thing to remember in addition to the button not moving till blinds are back on track.

[Captn_All_In]

How about the guy in first position?? What do you do when he busts and then its heads up??

[whitepotatoe]

I didn't read all the posts, so this might be in there already, but I think everyone should have the rule written out to show to people so they don't get their panties in a bunch. I wrote this up because I got sick of settling disputes about it. SInce then I have posted it online a few times and a cardroom actually uses it in their official rules now, and it cites "WhitePotatoe" as a reference, which I think is pretty funny.

Dead button – The big blind is posted by the player due for it, and the small blind and button are positioned accordingly, even if this means the small blind or the button is placed in front of an empty seat, giving the same player the privilege of last action on consecutive hands.
All specific situations are outlined below:


When the action is at least 4 handed:

1. When the small blind busts
- the UTG becomes the big blind
- the big blind becomes the small blind
- the button stays the button

2. When the big blind busts
- the UTG becomes the big blind
- the small blind becomes the button
- there is no small blind posted
- the button stays in the same spot for the next hand when blinds return to normal structure


When the action is 3 handed:

1. When the button busts
- the big blind becomes the small blind and gets the button
- the small blind becomes the big blind

2. When the small blind busts
- the big blind becomes the small blind and gets the button
- the button becomes the big blind

3. When the big blind busts
- the small blind stays the small blind and gets the button
- the button becomes the big blind


When both blinds bust:

The first hand
- the UTG becomes the big blind
- the button stays the button
- there is no small blind posted
The next hand
- the current UTG becomes the big blind
- the current big blind (who was the UTG when the blinds busted) becomes the small blind
- the button stays the button until the next hand


When both blinds bust when the action is 4 handed:
- the UTG becomes the big blind
- the button stays the button and becomes the small blind


So you did it correctly, by having B becone the BB and C become the SB.