Stealth Bomber
Well-Known Member
I would assume the answer is 'as many as is possible'. I've heard other AP's disagree. This is an area on 6D, which baffles me.
For instance: (any issues other than math are aside for the moment) Let's say the situation is 1 on 1 with the D and the player is about mid pack with a monster C. Let's assume the house allows the player to play up to 4 hands and we also assume the shoe will be able to produce 3 rounds before the cut card shows up. So that would give 4 total rounds of most likely + cards (good cards); 4 hands times 4 rounds is 16 hands total during a high C. Also during which, the D gets only 4 hands. Therefore, 20 hands including the D's will flop on the table.
Now, scenario #2: The player is able to play this same high C situation 1 on 1 with the D but this time the player gets only 1 hand per round until 16 total hands are played including the D's. That's 8 for the P and 8 for the D. Subsequently, the player now knows the cut card is about to come out so he/she plays 4 hands on the last round, the D of course gets 1. That's 21 hands including the D's that flop on the table during the high C portion of the shoe.
With scenario #1, player gets 16 hands during the high C. The D gets only 4 hands. (20 hands total)
With scenario #2, player gets 12 hands during the high C. The D gets 9 hands. (21 hands total)
Based on scenario #1, I can't see why there is any reason NOT to play as many hands as is allowed when the C is high (all other issues aside i.e. 'heat').
Am I correct to believe that scenario #1 provides the best EV or am I missing something?
Stealth
For instance: (any issues other than math are aside for the moment) Let's say the situation is 1 on 1 with the D and the player is about mid pack with a monster C. Let's assume the house allows the player to play up to 4 hands and we also assume the shoe will be able to produce 3 rounds before the cut card shows up. So that would give 4 total rounds of most likely + cards (good cards); 4 hands times 4 rounds is 16 hands total during a high C. Also during which, the D gets only 4 hands. Therefore, 20 hands including the D's will flop on the table.
Now, scenario #2: The player is able to play this same high C situation 1 on 1 with the D but this time the player gets only 1 hand per round until 16 total hands are played including the D's. That's 8 for the P and 8 for the D. Subsequently, the player now knows the cut card is about to come out so he/she plays 4 hands on the last round, the D of course gets 1. That's 21 hands including the D's that flop on the table during the high C portion of the shoe.
With scenario #1, player gets 16 hands during the high C. The D gets only 4 hands. (20 hands total)
With scenario #2, player gets 12 hands during the high C. The D gets 9 hands. (21 hands total)
Based on scenario #1, I can't see why there is any reason NOT to play as many hands as is allowed when the C is high (all other issues aside i.e. 'heat').
Am I correct to believe that scenario #1 provides the best EV or am I missing something?
Stealth