Question about spreads

I was just rereading Burning the Tables by Ian Anderson. When he was talking about his bet spread he is making it sound as if a 6x6 (6 units on each of 2 hands) is the same as a 1-12 spread. I always thought that a 1-12 spread on a shoe assumed you were only betting one hand. spreading to two hands will get you approximately 2/3 of the cards in good shoes instead of approximately 1/2, but it doesnt double your total action, and therefore they can't be the same thing right?

I agree; two hands of 1 to 6 bet spreads still only gets you a min to max bet spread of 1 to 6 (just twice as often), not 1 to 12.

Interesting point about getting 2/3 vs 1/2 of the good cards in a 1 on 1 gm.

I'd be interested in hearing more input on this. Since you're betting more at a count where high cards are equally likely, whether you take one hand or two, aren't you still betting more with an advantage? I always thought betting 6x6 would offer less risk/variance than betting 12, and be somewhere similar to a bet of ~10 (no math behind that, just my thinking). I've been thinking a lot about spreads lately, as I started that other spread about a hypothetical blatant counter.

I've been thinking about using a $10 - 2x$25 spread (which I know is not very aggressive and may be hard to get a long term edge).

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Quote: Ibeatyouraces

Your spread is the amount of your betting in units, not dollar amounts. Betting $10 - $100 on one hand is a 1-10 spread as your unit is $10. Now betting that same way but now with $100 on two hands, its a 1-20 spread since $100 x 2 = $200 or 20 units.

Oh, right. It took me a day to figure this out, but now I see your point. But for the 1 to 20 spread to hold true in this example, the player would have to be allowed to reduce to 1 spot's minimum bet whenever he wanted, correct?

At my local casino, players can play multiple hands, but there's special rules to prevent them from using multiple spots to circumvent the maximum bet limits. I've never done it, but I think players aren't allowed to change the amount of spots they're playing in the middle of a shoe (they might also have to play at least double the table minimums in each spot?).

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Quote: Ibeatyouraces

Your spread is the amount of your betting in units, not dollar amounts. Betting $10 - $100 on one hand is a 1-10 spread as your unit is $10. Now betting that same way but now with $100 on two hands, its a 1-20 spread since $100 x 2 = $200 or 20 units.

Thanks for answering I just have a follow up question to this. In your example above I understand the logic and see why it is betting 1-20 but are you sacrificing profits and trying to reduce variance by spreading your action into two spots?

My thinking is as follows:

If the deck is rich in high cards the probability of being dealt a 21 should be greater. If that is the case wouldn't you want to have all your money on one spot to maximize the blackjack payoff of 3:2? I am still learning so my logic could be terribly flawed, just trying to get input from those more experienced than me.

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Quote: SlackJawYokel

If the deck is rich in high cards the probability of being dealt a 21 should be greater. If that is the case wouldn't you want to have all your money on one spot to maximize the blackjack payoff of 3:2? I am still learning so my logic could be terribly flawed, just trying to get input from those more experienced than me.

I'm no expert but I wanted to point out that the majority of a counters money does not come from hands of blackjack. The wizard has a table on his card counting appendix that estimates how much (%) benefit each move gets you. Blackjack is only estimated at 7% of the counters benefit. The benefit of two hands (besides circumventing the max bet at a table and getting 2/3 of the good cards, as previously mentioned) would be better EV on doubles, splits and higher cards with the dealer more likely to bust.

Quote: SlackJawYokel

If the deck is rich in high cards the probability of being dealt a 21 should be greater. If that is the case wouldn't you want to have all your money on one spot to maximize the blackjack payoff of 3:2? I am still learning so my logic could be terribly flawed, just trying to get input from those more experienced than me.

Heh, first, cool screen name. Does that make you Cleetus?

Second, good question. I think the difference between maximizing the player's profits by betting one spot or more than one, including the profits from the BJ component, of a "good" remaining shoe might comes down to whether the Player wants to bet more than the table's max bet for one hand. If not, then a single hand's the way to go, since spreading the bet out to more than one spot just increases the amount of cards used per hand without increasing the players' potential profits, but if he wants to bet more than the max bet for 1 spot, it's more complicated.

Let's say it's late, Player's wife recently left him, he's got a snoot full of scotch and he's in a reckless mood. He's playing 1 on 1 with the dealer, no doubt after hitting on her. He decides to count a shoe and gets near the end w/ a really good count, let's say + 5 TC (or at least as close to + 5 as the aforementioned snoot of scotch allows).

Cursing God and the whore he married, he wants to bet a lot. The table limit's $100. If $100's what he wants to bet per hand, he definitely wants only 1 spot, since that minimizes the total cards played per $100 bet, preserving cards for hopefully an extra hand or two before reshuffle. But if he wants to bet $200 or more, he should expand out to as many spots as he needs to get that bet in on this hand, since that would minimize his amount of cards played per $200 (or however much our sad and lonely hero wants to bet per hand).

If he wants to bet somewhere between $100 and $200 per hand, then math needs to be done by someone brighter than me to determine if his profit potential's maximized with one spot at $100 or two spots at $100 and $25 / $50 / $75 / whatever.

Basically, I think he should play every spot he wants to max bet on and there's math to do for deciding whether the last spot, the one with his sub $100 remainder bet, should be laid or dropped.

I think i remember stanford song saying in professional blackjack that if there is one or more others at the table, that the same bet amount across two hands increases EV.

Quote: surrender88s

I think i remember stanford song saying in professional blackjack that if there is one or more others at the table, that the same bet amount across two hands increases EV.

Wong, not Song.

Makes sense to me, a greater proportion of cards during a positive count is going to you than there would be if you were only betting one spot.

Quote: Mission146

Makes sense to me, a greater proportion of cards during a positive count is going to you than there would be if you were only betting one spot.

Unless the bets in the multiple places are max bets, I don't see how spreading out and "using up" cards could be good for a player.

Let's say Counter and one other guy are at a table. A count's reached that dictates a $100 bet under his rules. He can either bet $100 on one spot or spread his $100 out over his and the 5 open spots.

If he spreads out, he'll have to kill 8 hands worth of cards to get to $100 bet. If he just uses his spot and ups his bet to $100, he'll only kill 3 hands worth of cards to get to the same bet amount.

By the time one-spot-Counter's gone through the cards required to make the 8 hands multi-spot-Counter used up on his first hand, he'll have gotten an average of $267 bet at the Advantageous count. How can that not be better than $100 bet at the Advantageous count?

mathematically, as in if you did this infinite times, I think you'd be right. But if you're doing 100 trials (a lot for this unique opportunity), spreading $100 over 2 $50 bets will give less variance, and a better chance (in the short term) of getting doubledown and split opportunities.

Another way to look at this problem would be to assume a max bet of $100, and instead of $50 and $50, two bets of $75 or $100. Might add EV there.

I assumed we were talking about Max betting.

If you played 2 hands instead of 1, you'd get more hands played before the shoe runs out, and more wagered at the positive count. Sorry if I'm missing something, mayhave misread your posts.

Spreading to more hands in positive counts, and decreasing your per hand bet (but increasing your total bet) is a good thing. Your expectation goes up, variance goes down. Wong suggests the trouble of spreading to 3 or more hands is not worth the slight decrease in variance. There is too much covariance in BJ between hands to reduce the variance much further.

Conclusion: spread to 2 hands in +ve counts. Wong's calculations showed that for standard rules, the optimal bet size on each hand is 75% of what you would bet on one hand. So, $200 on one hand or $150 each on 2 hands.

Quote: surrender88s

mathematically, as in if you did this infinite times, I think you'd be right. But if you're doing 100 trials (a lot for this unique opportunity), spreading $100 over 2 $50 bets will give less variance, and a better chance (in the short term) of getting doubledown and split opportunities.

Another way to look at this problem would be to assume a max bet of $100, and instead of $50 and $50, two bets of $75 or $100. Might add EV there.

Heh, I was actually hoping to move the spots discussion to Lonely's thread, but you make a good point.

I think what matters here is how we define the term "100 trials." If we define it to mean 100 independent times the player's facing that Advantage situation, then I think his increased profitability from more total dollars bet using a single spot is likely to overpower any advantages of lower variance / added double and split possibilities / etc.

On the other hand, if we define it to mean 100 hands, each with a total bet amount at the Advantage of $100, then I'm totally with you, the spread out method makes more sense.

Quote: dwheatley

Wong's calculations showed that for standard rules, the optimal bet size on each hand is 75% of what you would bet on one hand. So, $200 on one hand or $150 each on 2 hands.

The math's beyond me, but 75% does "feel" right. Just to make sure I understand the concept, though, betting $200 each on two hands is a more optimal solution than either of the other two, correct?

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Quote: dwheatley

Spreading to more hands in positive counts, and decreasing your per hand bet (but increasing your total bet) is a good thing. Your expectation goes up, variance goes down. Wong suggests the trouble of spreading to 3 or more hands is not worth the slight decrease in variance. There is too much covariance in BJ between hands to reduce the variance much further.

Conclusion: spread to 2 hands in +ve counts. Wong's calculations showed that for standard rules, the optimal bet size on each hand is 75% of what you would bet on one hand. So, $200 on one hand or $150 each on 2 hands.

Nicely said. Agree with your conclusion and playing 2 hands at 75% in positive counts gets you more income and less losses. Chance to split and doubles will quickly creates income. A few such hands, get up and move off, do not look back, do not pass GO, do not collect $200 (Monopoly).

BJ can be profitable if played this way, couple with some Basic S. Only way I see it. Cut House Edge to as low and minimal as possible with less hands and less rounds played. Make some bucks and disappear from that table. Wait for next round. We are not addicted to gambling..we just want to make some money legally. Fair and square.