Eeephour: Math Question for Cacarulo

[Eeephour]

On the side count of sevens, I have attempted to practice this technique and found it difficult. Specifically, I found the extra multiplication difficult. I thought of a way to simplify this count. Instead of starting at the deck # and multiply by -4. start at the deck # and multiply by -2. for example 2 deck game use ?4. then count each 7 as a 0.5 so your count would go ?4, - 3.5, - 3.0, - 2.5 ECT. If I am correct, then you eliminate all extra calculations. The only difference is to do your true count you would either round up or truncate.

Does this sound right to you?

Furthermore, while I have not tried this, what if you used 9 as a ? 0.5 count. You would then have a balanced approach and would not need to start at a ? count. Again does this sound right? Or am I missing something? While I have not tried this, on the surface this second 0.5 count seems like it adds a lot of complexity. Would this give you even more advantage?

eephour

[Brick]

I've considered Cacarulos 7 count also but decided there is too much margin of error for the head. What about starting the 7 count at 0? When we have deck(s) that are short of the normal ratio we simply add .5 to the true count. In other word 3 decks remaining that are depleted of six 7s will result in an additionaly TC of plus 1 to the existing TC. This seems to be easier with virtually the same gain of ev.

I like the idea of side counting 9s to 7s but I would think learning a level 2 count system would be easier and more efficient.

When using a side count of 7s only,we must assume the 9s are equally porportioned. They have about equal value for the player and dealer,but I believe one of them is slightly more valuable. Does anyone knoe the exact value of the 7 and 9?

thanks,
Brick

> On the side count of sevens, I have
> attempted to practice this technique and
> found it difficult. Specifically, I found
> the extra multiplication difficult. I
> thought of a way to simplify this count.
> Instead of starting at the deck # and
> multiply by -4. start at the deck # and
> multiply by -2. for example 2 deck game use
> ?4. then count each 7 as a 0.5 so your count
> would go ?4, - 3.5, - 3.0, - 2.5 ECT. If I
> am correct, then you eliminate all extra
> calculations. The only difference is to do
> your true count you would either round up or
> truncate.

> Does this sound right to you?

> Furthermore, while I have not tried this,
> what if you used 9 as a ? 0.5 count. You
> would then have a balanced approach and
> would not need to start at a ? count. Again
> does this sound right? Or am I missing
> something? While I have not tried this, on
> the surface this second 0.5 count seems like
> it adds a lot of complexity. Would this give
> you even more advantage?

> eephour

[Cacarulo]

> On the side count of sevens, I have
> attempted to practice this technique and
> found it difficult. Specifically, I found
> the extra multiplication difficult. I
> thought of a way to simplify this count.
> Instead of starting at the deck # and
> multiply by -4. start at the deck # and
> multiply by -2. for example 2 deck game use
> ?4. then count each 7 as a 0.5 so your count
> would go ?4, - 3.5, - 3.0, - 2.5 ECT. If I
> am correct, then you eliminate all extra
> calculations. The only difference is to do
> your true count you would either round up or
> truncate.

> Does this sound right to you?

Yes, it's correct but you would have to work with non-integer RCs. If you find it easier then go ahead.

Esentially the system works as follows:

Count System 1) -1 1 1 1 1 1 0 0 0 -1 (Hi-Lo)
which is used for playing purposes.

Count System 2) -2 2 2 2 2 2 1 0 0 -2 (Level-2 system) which is used for betting.
Note that this system is 2*Hi-Lo + 7s (side-count)

Now, it's possible to eliminate the multiplication by using half the value of the tags. This would be -1 1 1 1 1 1 0.5 0 0 -1.
As you see, this is "simply" Hi-Lo + 7s (but counted as 0.5).

> Furthermore, while I have not tried this,
> what if you used 9 as a ? 0.5 count. You
> would then have a balanced approach and
> would not need to start at a ? count. Again
> does this sound right? Or am I missing
> something? While I have not tried this, on
> the surface this second 0.5 count seems like
> it adds a lot of complexity. Would this give
> you even more advantage?

First, you will have to keep track of two additional cards instead of one and particularly the nine is not good for insurance purposes.
Insurance wouldn't be a problem if you take it using Hi-Lo but this would be a little disadvantage.
More advantage? I don't think so. Perhaps a very tiny advantage which does not justify the increase of complexity.

Sincerely,
Cacarulo

[Cacarulo]

> I've considered Cacarulos 7 count also but
> decided there is too much margin of error
> for the head. What about starting the 7
> count at 0? When we have deck(s) that are
> short of the normal ratio we simply add .5
> to the true count. In other word 3 decks
> remaining that are depleted of six 7s will
> result in an additionaly TC of plus 1 to the
> existing TC. This seems to be easier with
> virtually the same gain of ev.

I think your idea was discussed with Zenfighter (see Theory & Math or Software & Simulations) and we came to the conclusion that the system was not so equivalent. In fact, I don't find it easier but I understand that there are people capable of doing it your way.

Sincerely,
Cacarulo

[Eeephour]

> Yes, it's correct but you would have to work
> with non-integer RCs. If you find it easier
> then go ahead.

> Esentially the system works as follows:

> Count System 1) -1 1 1 1 1 1 0 0 0 -1
> (Hi-Lo)
> which is used for playing purposes.

> Count System 2) -2 2 2 2 2 2 1 0 0 -2
> (Level-2 system) which is used for betting.
> Note that this system is 2*Hi-Lo + 7s
> (side-count)

> Now, it's possible to eliminate the
> multiplication by using half the value of
> the tags. This would be -1 1 1 1 1 1 0.5 0 0
> -1.
> As you see, this is "simply" Hi-Lo
> + 7s (but counted as 0.5).

> First, you will have to keep track of two
> additional cards instead of one and
> particularly the nine is not good for
> insurance purposes.
> Insurance wouldn't be a problem if you take
> it using Hi-Lo but this would be a little
> disadvantage.
> More advantage? I don't think so. Perhaps a
> very tiny advantage which does not justify
> the increase of complexity.

> Sincerely,
> Cacarulo

thanks for your help

eephour

[Eeephour]

> Yes, it's correct but you would have to work
> with non-integer RCs. If you find it easier
> then go ahead.

> Esentially the system works as follows:

> Count System 1) -1 1 1 1 1 1 0 0 0 -1
> (Hi-Lo)
> which is used for playing purposes.

> Count System 2) -2 2 2 2 2 2 1 0 0 -2
> (Level-2 system) which is used for betting.
> Note that this system is 2*Hi-Lo + 7s
> (side-count)

> Now, it's possible to eliminate the
> multiplication by using half the value of
> the tags. This would be -1 1 1 1 1 1 0.5 0 0
> -1.
> As you see, this is "simply" Hi-Lo
> + 7s (but counted as 0.5).

> First, you will have to keep track of two
> additional cards instead of one and
> particularly the nine is not good for
> insurance purposes.
> Insurance wouldn't be a problem if you take
> it using Hi-Lo but this would be a little
> disadvantage.
> More advantage? I don't think so. Perhaps a
> very tiny advantage which does not justify
> the increase of complexity.

> Sincerely,
> Cacarulo

After my post of last week I decided to take a look at halves counting. Wong claims this method gives a more precise count. However, according to wong?s numbers it seems to give only a slight advantage over hi/lo. I think however, your point about insurance may be applicable to this situation. Since, according to don, insurance is the top play. Maybe if you eliminate the half count for the 9, you would improve the performance. What do you think? Have you ever run any numbers on this?

eephour

[Cacarulo]

> After my post of last week I decided to take
> a look at halves counting. Wong claims this
> method gives a more precise count. However,
> according to wong?s numbers it seems to give
> only a slight advantage over hi/lo.

The advantage of Halves over Hi-Lo is not so slight. Here are some SCOREs for different bet spreads (6D,S17,DOA,DAS,SPA1,SPL3,NS,5/6,Catch-22 floored):

             1-4     1-8     1-12    1-16    1-20 

Halves      11.16   28.69   38.33   44.25   48.27 

Hi-Lo        9.61   25.83   34.84   40.43   44.24

If we choose 1-12 we have that Halves is 10% better than Hi-Lo.

> I think
> however, your point about insurance may be
> applicable to this situation. Since,
> according to don, insurance is the top play.
> Maybe if you eliminate the half count for
> the 9, you would improve the performance.
> What do you think? Have you ever run any
> numbers on this?

Yes, I think you'd have a better performance although the system would be unbalanced.

Sincerely,
Cacarulo

[Zenfighter]

> The advantage of Halves over Hi-Lo is not so
> slight. Here are some SCOREs for different
> bet spreads
> (6D,S17,DOA,DAS,SPA1,SPL3,NS,5/6,Catch-22
> floored):
> 1-4 1-8 1-12 1-16
> 1-20
> Halves 11.16 28.69 38.33 44.25
> 48.27
> Hi-Lo 9.61 25.83 34.84 40.43
> 44.24
> If we choose 1-12 we have that Halves is 10%
> better than Hi-Lo.
>
> Yes, I think you'd have a better performance
> although the system would be unbalanced.
> Sincerely,
> Cacarulo

Absolutely right, but did you have notice, that
a Hi-Lo player who NEVER tips will win aprox. the
same than his counterpartner the Halves player who suffers from generous tipping? :-)
After an exhausted 45 days "Halves-trip", my mind
is flirting with the idea of returning to "Eden".

Nice to be at home again and in contact with all
the friends round here.

Best regards
Z