Adding multiple hands in +EV count question!! Help!

Ok, so here's your (hypothetical) situation:

You're playing heads up against the dealer and the TC of the shoe is currently at +1. You look at the dealers shoe and notice that the cut card is the next to be dealt so this coming round will be the last before the shuffle. In terms of maximizing EV only(no bankroll restrictions, etc), should you:
A. Spread to three hands at the table maximum
or
B. Play only one hand at the the table maximum?
or
C. ??

Open as many hands as you can, and Put as much as you can on each hand.

P.S. In any positive EV games (doesn't have to be Blackjack game/s).
You should Open as many hands as you can, and Put as much as you can on each hand.

Quote: TiffanyH

Ok, so here's your (hypothetical) situation:

You're playing heads up against the dealer and the TC of the shoe is currently at +1. You look at the dealers shoe and notice that the cut card is the next to be dealt so this coming round will be the last before the shuffle. In terms of maximizing EV only(no bankroll restrictions, etc), should you:
A. Spread to three hands at the table maximum
or
B. Play only one hand at the the table maximum?
or
C. ??

Depending on the blackjack rules at that table, a true count of +1 might have an expected value of 0. When the EV = 0 for the next round, then playing more hands would not increase your EV.

But if a TC of +1 has a positive EV and your only goal is to maximize EV, you should play table maximum on as many spots as allowed.

(However, a more realistic goal than maximizing EV is maximizing bankroll growth, or at least avoid going broke. To achieve maximum bankroll growth, if a TC of +1 has a positive EV and your optimal 1-hand bet size at that count is B, then you on that last round you should bet a total of 2B divided equally among three hands.)

Quote: ChesterDog

Depending on the blackjack rules at that table, a true count of +1 might have an expected value of 0. When the EV = 0 for the next round, then playing more hands would not increase your EV.

But if a TC of +1 has a positive EV and your only goal is to maximize EV, you should play table maximum on as many spots as allowed.

(However, a more realistic goal than maximizing EV is maximizing bankroll growth, or at least avoid going broke. To achieve maximum bankroll growth, if a TC of +1 has a positive EV and your optimal 1-hand bet size at that count is B, then you on that last round you should bet a total of 2B divided equally among three hands.)

ChesterDog,

If optimal 1-hand bet size at that count is B, what is the optimum bet(under same conditions) if spread to :-

1) 2 hands - each hand bet = ?
2) 3 hands - each hand bet = 2B/3 =0.67B.
3) 4 hands - each hand bet = ?
4) 5 hands - each hand bet = ?
5) 6 hands - each hand bet = ?

Thanks in advance.

Quote: ssho88

ChesterDog,

If optimal 1-hand bet size at that count is B, what is the optimum bet(under same conditions) if spread to :-

1) 2 hands - each hand bet = ?
2) 3 hands - each hand bet = 2B/3 =0.67B.
3) 4 hands - each hand bet = ?
4) 5 hands - each hand bet = ?
5) 6 hands - each hand bet = ?

Thanks in advance.

Your question made me recheck my calculations. Now, I see my advice was wrong. When playing 3 hands, the optimal bet per hand would be closer to 0.58 B instead of 0.67 B.

If the optimal bet playing one hand is B, then when playing N hands, the optimal bet on each would be:
B / [1 + (N-1)C/V ] where V is the variance for a hand of blackjack and C is the covariance between two hands.

The Wizard gives the variance and covariance for various sets of rules here.

To complicate the matter further, the variance and covariance would depend on the blackjack count, but suppose V = 1.3 and C = 0.48, then here are the optimal bet sizes:

1) 2 hands - each hand bet = 0.730B
2) 3 hands - each hand bet = 0.575B
3) 4 hands - each hand bet = 0.474B
4) 5 hands - each hand bet = 0.404B
5) 6 hands - each hand bet = 0.351B

(I will show you the derivation of my formula if you wish to check it.)

Quote: ChesterDog

Your question made me recheck my calculations. Now, I see my advice was wrong. When playing 3 hands, the optimal bet per hand would be closer to 0.58 B instead of 0.67 B.

If the optimal bet playing one hand is B, then when playing N hands, the optimal bet on each would be:
B / [1 + (N-1)C/V ] where V is the variance for a hand of blackjack and C is the covariance between two hands.

The Wizard gives the variance and covariance for various sets of rules here.

To complicate the matter further, the variance and covariance would depend on the blackjack count, but suppose V = 1.3 and C = 0.48, then here are the optimal bet sizes:

1) 2 hands - each hand bet = 0.730B
2) 3 hands - each hand bet = 0.575B
3) 4 hands - each hand bet = 0.474B
4) 5 hands - each hand bet = 0.404B
5) 6 hands - each hand bet = 0.351B

(I will show you the derivation of my formula if you wish to check it.)

I try to derive it, please correct me if I am wrong

For single hand BJ with expectation, ev and variance, V, the optimum bet size = B = bankroll * ev/ V * k, where k is kelly ratio.

So for multiple hands in a round, the variance for a hand = V +(N-1) *C. so optimum bet = bankroll * ev/[V+(N-1)C] * k = B* V/[V+(N-1)C] = B / [1 + (N-1)C/V ]

Thanks