What's the perfect amount of players...

while playing and counting on a 6D table?

My theory is that it's not good 1 on 1 because when the pack goes negative, there's nobody but you to grind out the rest of a cold shoe. Wonging out is okay but there is no one left to eat up the bad cards during the bathroom break. Plus, 1 on 1 is not good for cover. Only someone who is not afraid of the game will accept playing 1 on 1 and that�s not the typical ploppy.

If the table is full, we can't spread to 2 hands and we can't get enough hands before the cut card shows up.

I think it's best if there is a total of 3 players plus the D. Does anyone know if my assumption is correct? How can we know what will result in the best EV return?

Stealth B.

According to Ian Andersen, 1 on 1 is still the best because it provides the maximum number of hands per hour. Although during negative counts, the extra players help eat the cards, however, during positive counts, they also eat up your "good" cards to make your EV lower. Comparing the two, the effects at high counts is much more prevalent.

Other factors to consider is that the extra ploppies tend to slow down the game through unnecessary hesitation.

Thus IMO, 1 on 1 still provides a higher EV for the counter.

Yep TJ, I believe you're right. But given all the circumstances, I think it is best in general to play with 1 other player on the table with 6D.

Also, at times I have noticed ploppies get mad at the D for not busting then leave the table just before the shoe was set to unload the big cards, thus leaving me 1 on 1 with a monster C. Sometimes they eat up the bad cards from the top of the pack. Now that's coooool.

Another ?: If a counter was 1 on 1 during 6D and the TC was +6 or higher, wouldn't it make more sense mathematically (heat not consided) to play as many hands as is alowed? My rationale being that the player would get multiple amounts of hands in relation to the number of hands the D received, all while the C was high.

S B

Generally speaking, two hands is the best choice. One hand's SD is higher. Three hands not only draws heats, but also eats precious cards.

When 1 on 1, I always play two hands when max bet out.

...and a tableful of ploppies to take over when the count goes negative. :-)

Your question about multiple hands in positive counts is discussed in detail in Blackjack Attack. You need to consider that if you play two hands in one round you should only bet 75% of your usual bet on each hand, whereas if you play one hand in two rounds you will make your usual bet each round. Thus there is a point where you maximise your earnings per card used. Of course, this assumes that another round will be dealt.

does it make mathematical sense to play as many hands as possible on 6D when there is a very high count? Let's assume this scenario is being played out 1 on 1 and heat is NOT part of the equation.

My theory is to get more hands in play in comparison to what the D gets during the monster C. ???

When you know that the cut card will come out during next round you should play as many hands as possible in order take full advantage of the high count. But playing more than three hands at an earlier stage of the shoe is counterproductive because two many cards will be consumed and the advantage situation will be gone sooner than it came.
Its also bad for practical reasons: You will be so busy performing your plays that you might lose control of the situation. There is a big danger that you get cheated on pay-outs.

Francis Salmon

When you know that the cut card will come out during next round you should play as many hands as possible...

Yes absolutely, for sure, for sure, I do that.

But playing more than three hands at an earlier stage of the shoe is counterproductive because two many cards will be consumed and the advantage situation will be gone sooner than it came.

This is the area I'm baffled by. For instance, let's say the situation is 1 on 1 with the D and let's assume the house allows us 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 us 4 total rounds of most likely + count cards (good cards). 4 hands times 4 rounds is 16 hands total during a + count. Also during which, the D gets only 4 hands. Therefore, 20 hands including the D's will flop on the table. Now, scenario #2, we are 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. Subsequently we now know the cut card is about to come out so we play 4 hands on the last round. That's 21 hands including the D's that flop on the table during the high + portion of the shoe.

With scenario #1, player gets 16 hands during a high + C. The D gets only 4 hands.

With scenario #2, player gets 12 hands during a high + C. The D gets 9 hands.

This is why my poor brain fries on this every time I think about it. I'll be headed for the Xanax if I can't get this one resolved. Heeeellllp meeeee!

Stealth Bomber

When you play 4 hands simultaneously,you cannot bet the same amount per hand as you would with 1 hand because the risk of ruin would be greatly increased.Imagine a situation where you bet maximum on four hands but you get rather lousy hands and the dealer gets the ace.You have to insure.Now the dealer gets the 9 and you lose all your bets including the insurance. You have lost 6 max bets at once.Comparatively,losing 6 single hands in a row is much less likely.
Now I really have to run off but I hope you got the point.I'll be back after Christmas.

Francis Salmon

Your answers are all in Blackjack Attack, but briefly... Let's say the count is high enough that you should be making your max bet (e.g. $100) each round.

With scenario #1, player gets 16 hands during a high + C. The D gets only 4 hands.

To maintain the same ROR you should be bet approx. $75 on each of your two hands. So you would have 2*$75*8 = $1200 in action.

With scenario #2, player gets 12 hands during a high + C. The D gets 9 hands.

You would also have $100*12 = $1200 in action.