Let's assume heat is a non issue for the moment and the count goes high. Isn't it mathematically better to play as many hands as is possible so as to obtain more of the good cards into our hands during the high C?
This is something I'm especially interested in for shoe games where when the C goes high in the beginning, it will on average keep the same TC to the end.
Without heat and bankroll considerations you should play as many spots as there is on the table. It increases the total hands you get and decreases variance.
somewhere in my past I read that it's best to play only one hand when playing one on one. That has never made sense to me.
Playing one hand does enable you to vary your bets more often. If you were playing single deck and getting 6 rounds for 1 spot or 1 round for 7 spots then clearly you would not want to play 7 spots as you would only be depth charging. Once the count is high you want to spread to as many spots as you can in order to raise the ratio of player vs dealer bjs. If heat is not a factor the only reason to keep from spreading to max hands is to vary your bet more often which would not really be a factor in the shoe game. More hands is almost always mathematically better.
Once the count is high you want to spread to as many spots as you can in order to raise the ratio of player vs dealer bjs.
I also believe this to be true.
Let's assume for instance the game is 6D, 80% pen, +15TC, two aces extra on average and 4.? decks have already gone to the discard tray. Let's also assume in this situation that there is on average 16 play positions left in the pack to be played before the shoe is over and the pit crew doesn't have a care in the world about us being AP's or how many hands we play. Now let's also assume we will be playing 1 on 1 in any of the following manners to finish out the shoe:
1) play one position with a total of eight hands to the D and eight hands to the player -or-
2) play three positions with a total of four hands to the D and twelve to the player -or-
3) play seven positions with a total of two hands to the D and fourteen to the player.
Which option would mathematically provide the highest EV?
Without question the more money you can get on the table the more EV you have. In this case 3) gives you 14 hands, which is more money and clearly more EV.
