One big bet vs two hands

the majestic star in gary, indiana (and probably a lot of other casinos too) have a rule that you can play 2 hands at the table minimum, but if the table is full, you must bet double the table min on each hand (i would assume the chances are decent that somebody who wants to join the table are planning on playing less than or equal to twice the table min, thats why they do that)
revive a thread or start another one on the same subject? its a double edged sword, so who cares.. people that like to bitch will..

Sonny said:
If there is no covariance then the bets should be the same size since there is no increased risk.
same size? what do you mean by that? i thot you were supposed to do 150% of your bet into 2 hands to keep the risk the same

Sonny said:
That's exactly right. The optimal bet will be somewhere between 1 bet and 2. We can use the Kelly formula to find out the optimal bet for any number of hands we want to play. This will ensure that we are not adding any extra risk to our betting system by spreading our bets. For a $13,300 bankroll and a 1% edge:

(Bankroll * Advantage) / Variance = Bet

($13,300 * 0.01) / 1.33 = 133 / 1.33 = $100
i thot the variance on 1 hand was 1.1? and i also thot you were supposed to bet 1% of your bankroll at 1% advantage, thus the above would = $133.. could you explain this formula a little more please?

Sonny said:
If we want to play two hands at the same table then our variance will change. As you pointed out, it will be greater than one hand but less than double. Each hand you spread to (at the same table) will increase your variance by about 0.5. That gives us:
but isnt variance supposed to go down? im confused.. unless of course going up by .5 is actually going down, since its 1.1 per hand, so it should be 2.2 but only goes up to 1.6 thus its actually less?

jee_pack said:
From what I read in this post, I made a few calculations and came up with this: To avoid detection as a counter, lower your spreads. Well you can make a 1-4 spread look like a 1-2 spread. This is how, lets say you take 28$ as a Unit bet. Under TC+2, bet one hand at table minimum, lets say the min is 15$. At TC+2 to +3, bet 1 unit (28$ rounded up to 25$) for 1 hand; total bet: 25$. At TC+3 to +4, bet 40$ on 2 hands; total bet is 80$. TC+4 to +5, bet the same thing, 40$, but on 3 hands; total bet: 120$. And for TC+5 to +6, bet 55$ on 3 hands; total bet: 165$. So this is even more optimal than playing just 1 hand when the count is over +3, with the same risk. And instead of spreading from 25$ to 100$ (75$ increase or 300% increase), you are spreading from 25$ to 55$ (30$ increase or 220% increase). But you are actually increasing your profits......

1: Does this make sens, did I make do the math correctly (I rounded up to the closes multiple of 5$ on each calculation and used the 50% and 75% rule, although I think I should have went with the exact number... but I can do that later)...

2: So your max bet with this is 165$ total on the table, and on each hand, 55$. The other way, always playing 1 hand, your max bet is lower: 100$ total on the table instead of 165$. But your max bet on 1 hand is much higher, 100$ instead of 55$. So which of these 2 situations do you think would bring less heat, taking out the fact that one of them is more optimal than the other and also taking out all the other variables that can influence heat...

Thanks in advance...
why did you start with something odd like $28, then round? why not just start with $30? also, how does 28 round UP to 25?

ScottH said:
Well if you only spread to multiple hands in high positive counts like most counters, that isn't usually a problem. I've heard that the general rule is if you go 2 spots you have to be 3X the table min on each spot.
i thot the general rule was 2x

Sonny said:
One of the drawbacks is that you will be using up more cards in positive situations and that might cause you to get less rounds, which will hurt your win rate. Blackjack Attack by Don Schlesinger has a good chapter on playing multiple hands.
but you can just play longer right? its confusing, which do you want, higher EV or more rounds? i would think the former.. besides, if the table is gonna be full anyways..

Knox said:
Let's take the example of head up play. Count gets very high/positive. Lots of tens and aces. If you have 2 hands out to the dealers one, you have twice the odds of getting a blackjack. You want to suck up as many of those good cards as you can.
it would seem that way, but somebody else in another post explained it.. if its just your 1 hand vs the dealer, you will get 50% of the blackjacks right? well if its your 2 hands vs the dealers, will you get twice as many? no, you will get 67% of the blackjacks, thus your getting 33% more blackjacks (66.7/50), not 100% more, so you could say in a weird way that your 67% wrong on that one!

asiafever said:
When the dealer gets a blackjack, all your regular hands loose and all your blackjacks push. When you get a blackjack, you get paid 3:2 on a single hand if the dealer don't have a blackjack too, therefore even if the dealer is less likely to get a blackjack every times he will get one it will damage you more.
but when the dealer busts you win both hands, altho you still have to not bust.. it all evens out to it being better for you.. its all about averages

jee_pack said:
thanks for all the input!

So the only "ewwww mommy its icky!" is that I still don't know if they would allow it, actually, before being a counter, I played twice at my local casino, and once, at the 15$ min table, I saw someone decide to play 2 hands at one point, so later I did the same thing, at the table min on both hands, seemed to pass like a normal thing. Though I have never seen someone play 3 hands so I don't know if there is a 2 hand limit or not.

Question to sunny? So I think I get the math now... One thing I wish I knew though is BY HOW MUCH does it increase your EV or by how much does it decrease your RISK? If there is a formula behind this, I'd love to see it....

thanks in advance
also, everybody knows that 150% into 2 raises EV (by how much?), and that 100% into 2 lowers risk (by how much?), but what is the in between? would it be 125% into 2?

Renzey said:
Betting 1 x $100 won't produce the same overall EV as betting 2 x $50 due to the extra cards used by the extra hands.

Just think about betting 1 x $300 in a positive count with four other players at the table as opposed to 3 x $100. You'll have the same EV on that round with a much lower variance, but will get in fewer rounds with each positive count. Thus, EV suffers. However, 2 x 73% or 3 x 57% improves EV with the same variance providing enough other players are at the table.
dont you mean win rate suffers? other players at the table cant effect your EV (im pretty sure on this one)

Knox said:
Let's make it even simpler. Assume I play two hands against the dealer's one in a highly favorable count.

In three rounds, three blackjacks come out. They should be evenly distributed, so say I get one on each of my hands and the dealer gets one. Assume further $50 bets and 50% win rate on each hand.

Round 1:
Hand 1: blackjack pays $75
Hand 2: loses $50

Round 2:
Hand 1: wins $50
Hand 2: blackjack pays $75

Round 3 dealer blackjack both hands lose total of $100 :(

In this case we win $150 on the first two hands and lose $100 to the blackjacking dealer. That looks like a win to me of $50.

The key here is that we win 1.5x the bet on blackjack, but only lose 1:1 when we lose, even if it happens to be two hands.

Am I missing something here, this looks like pretty simple math?
im confused what 1. a win rate of 50 PERCENT means, and 2. where is win rate calculated into your example?
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Well-Known Member
When playing 2 hands, your betting .72% ( .75% for Simplicity.) of your BU on each hand multiplied by the advantage. And since we are only playing with an advantage.....

(edit ) I just saw Freds post.. He beat me to t..
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Well-Known Member
SilentBob420BMFJ said:
i thot the variance on 1 hand was 1.1?
The standard deviation for 1 hand is about 1.16. The variance is about 1.33.

SilentBob420BMFJ said:
and i also thot you were supposed to bet 1% of your bankroll at 1% advantage, thus the above would = $133.. could you explain this formula a little more please?
It’s the Kelly formula. You multiply your advantage by your bankroll and adjust for the riskiness of the wager. The result is the optimal Full Kelly wager.

SilentBob420BMFJ said:
but isnt variance supposed to go down? im confused.. unless of course going up by .5 is actually going down, since its 1.1 per hand, so it should be 2.2 but only goes up to 1.6 thus its actually less?
You got it.

SilentBob420BMFJ said:
but you can just play longer right? its confusing, which do you want, higher EV or more rounds?
You want both. Sometimes playing only one hand will give you more rounds, which gives you more chances to get money on the table. Other times you will be able to get more money on the table by playing multiple hands (even if there are fewer rounds). You have to look at each situation and decide what the best plan is. You only get a few +EV hands per hour so you want to make the most of them.

Sure, you could just play longer to make up for the lost EV, but if you know how to make more money sooner why would you opt not to? If you knew how to earn $20/hr instead of $17/hr, why wouldn’t you? Also, you might not be able to play longer if you catch too much heat.

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