# Thread: AsZehn: KO vs Red 7

1. ## John Auston: Jumping in here

> UBZII should outperform K-O under any
> setting.

Just of the top of my head, it seems to me . . .

Except a situation where the counts encountered are forced to be disproportionately on the high positive side, no?

An unbalanced count performs better the more the counts are near it's pivot. UBZII is near a hi-lo +2, KO near a hi-lo +4.

In strict back-counting, a disproportionate number of higher-bet counts would favor KO's better accuracy at hi-lo +3, +4, and beyond. Conversely, KO's poorer performance at negative counts ( and UBZII's better) are ruled out of the equation.

Further, if you are only going to be playing positive counts, then the indices used for unbalanced counts in RC-mode could be recomputed to reflect the decks-left composition of wong situations.

-JA

2. ## Don Schlesinger: Re: Wrong chart

> I've looked at that page and the game
> described there is 5/6,S17,DAS,LS, PLAY-ALL
> ,I18+F4. My game is 5/6,S17,DAS, BACK-COUNT
> ,C22.

No, the chart is for both play-all and back-counting (at the bottom).

> Besides, if you look at the bottom of p. 297
> you'll find that 5/6,S17,DAS,BACK-COUNT,I18
> gives a SCORE of \$63.54 which is close to
> what I've posted: \$65.30. The difference is
> that I'm using C22 instead of I18.

The point is not to determine the absolutes, but rather to determine the relatives. I'm questioning the outperformance of K-O over UBZII; I'm not concerned with the differences in raw SCOREs, which will vary because we're simming different games.

Don

3. ## Don Schlesinger: Re: Jumping in here

An interesting point, John, but the fact that K-O is weaker away from the pivot is K-O's problem. It isn't UBZII's problem, which should be happy no matter what index it is at.

So, whereas I understand that providing a situation that plays into K-O "strong point" disproportinately HELPS K-O, I don't see why that's the same thing as "favoring" K-O. I would have still thought that UBZII would continue to outperform, but maybe I'm wrong.

Don

> Just of the top of my head, it seems to me .
> . .

> Except a situation where the counts
> encountered are forced to be
> disproportionately on the high positive
> side, no?

> An unbalanced count performs better the more
> the counts are near it's pivot. UBZII is
> near a hi-lo +2, KO near a hi-lo +4.

> In strict back-counting, a disproportionate
> number of higher-bet counts would favor KO's
> better accuracy at hi-lo +3, +4, and beyond.
> Conversely, KO's poorer performance at
> negative counts ( and UBZII's better) are
> ruled out of the equation.

> Further, if you are only going to be playing
> positive counts, then the indices used for
> unbalanced counts in RC-mode could be
> recomputed to reflect the decks-left
> composition of wong situations.

> -JA

4. ## John Auston: Re: Jumping in here

> An interesting point, John, but the fact
> that K-O is weaker away from the pivot is
> K-O's problem. It isn't UBZII's problem,
> which should be happy no matter what index
> it is at.

Does this sound reasonable?:

Given: unbalanced counts are less accurate at departures and amount-to-bet, the further they are
from pivot.

Consider the following hi-lo trues:

<-5 UBZII always 2 "better" than KO
-5: -7 below UBZII pivot, -9 below KO
-4: -6 below UBZII pivot, -8 below KO
-3: -5 below UBZII pivot, -7 below KO
-2: -4 below UBZII pivot, -6 below KO
-1: -3 below UBZII pivot, -5 below KO
0: -2 below UBZII pivot, -4 below KO
+1: -1 below UBZII pivot, -3 below KO
+2: right at UBZII pivot, -2 below KO
+3: 1 above UBZII pivot, -1 below KO
+4: 2 above UBZII pivot, right at KO
+5: 3 above UBZII pivot, 1 above KO
>5: UBZII always 2 "worse" than KO

Play-all, it all evens out, and UBZII level 2 has the edge.

But restrict to Wong only:

+1: -1 below UBZII pivot, -3 below KO
+2: at UBZII pivot, -2 below KO
+3: 1 above UBZII pivot, -1 below KO
+4: 2 above UBZII pivot, at KO
+5: 3 above UBZII pivot, 1 above KO
>5: UBZII always 2 "worse" than KO

UBZII only has the "accuracy advantage" at +1 and +2,
KO has the rest, which is probably just enough to overcome the
level 1 versus level 2.

Another observation. KO's disadvantage at negative counts would hurt it more than it does were it not for the fact that only around 1 unit is bet there anyway.

The reverse happens to UBZII at high counts, since we are at max anyway.

So, when we Wong, the key counts are +2, +3, +4, and +5, and KO has the accuracy edge there.

Or so it seems to me, just trying to "thought experiment" it.

Also, if what I am thinking is right, then it should hold that the larger the Wonging spread, the more KO should inch ahead, and the reverse.

But I could be wrong. Just trying to think what could explain those results.

John

5. ## Don Schlesinger: Re: Jumping in here

> Does this sound reasonable?:

> Given: unbalanced counts are less accurate
> at departures and amount-to-bet, the further
> they are
> from pivot.

> Consider the following hi-lo trues:

> -5: -7 below UBZII pivot, -9 below KO
> -4: -6 below UBZII pivot, -8 below KO
> -3: -5 below UBZII pivot, -7 below KO
> -2: -4 below UBZII pivot, -6 below KO
> -1: -3 below UBZII pivot, -5 below KO
> 0: -2 below UBZII pivot, -4 below KO
> +1: -1 below UBZII pivot, -3 below KO
> +2: right at UBZII pivot, -2 below KO
> +3: 1 above UBZII pivot, -1 below KO
> +4: 2 above UBZII pivot, right at KO
> +5: 3 above UBZII pivot, 1 above KO

> Play-all, it all evens out, and UBZII level
> 2 has the edge.

> But restrict to Wong only:

> +1: -1 below UBZII pivot, -3 below KO
> +2: at UBZII pivot, -2 below KO
> +3: 1 above UBZII pivot, -1 below KO
> +4: 2 above UBZII pivot, at KO
> +5: 3 above UBZII pivot, 1 above KO

> UBZII only has the "accuracy
> advantage" at +1 and +2,
> KO has the rest, which is probably just
> enough to overcome the
> level 1 versus level 2.

> Another observation. KO's disadvantage at
> negative counts would hurt it more than it
> does were it not for the fact that only
> around 1 unit is bet there anyway.

> The reverse happens to UBZII at high counts,
> since we are at max anyway.

> So, when we Wong, the key counts are +2, +3,
> +4, and +5, and KO has the accuracy edge
> there.

> Or so it seems to me, just trying to
> "thought experiment" it.

> Also, if what I am thinking is right, then
> it should hold that the larger the Wonging
> the reverse.

> But I could be wrong. Just trying to think
> what could explain those results.

It's my thinking that's probably wrong here. I sometimes confuse UBZII with Zen. I couldn't see any sense that K-O would outperform a level-two balanced count, but all the while we were speaking about two unbalanced counts. I think you're right.

Don

6. ## Cacarulo: Re: Jumping in here

> UBZII only has the "accuracy
> advantage" at +1 and +2,
> KO has the rest, which is probably just
> enough to overcome the
> level 1 versus level 2.

Correct.

> Also, if what I am thinking is right, then
> it should hold that the larger the Wonging
> the reverse.

You don't gain that much with a larger spread.

1-12:
KO = \$65.30 (-6)
UBZII = \$63.85 (-5)

1-20:
KO = \$65.46 (-6)
UBZII = \$63.93 (-5)

Even with a smaller spread KO outperforms UBZ2.

1-6:
KO = \$64.69 (-5)
UBZII = \$63.30 (-3)

Sincerely,
Cacarulo

7. ## Don Schlesinger: BJRM 2002

I just can't seem to get K-O to outperform UBZII with any of the canned BJRM sims.

Cac, why do you think using four more indices would change anything?

Try optimal Wonging 1-6, 5/6, das, s17, LS, and let me know which count wins. I'm getting UBZII by about \$2 or so.

Don

8. ## Cacarulo: Re: BJRM 2002

> I just can't seem to get K-O to outperform
> UBZII with any of the canned BJRM sims.

Try 1-6 with S17,DAS,5/6 and you will find that KO is a little better than UBZII. BTW, I have BJRM2000.

KO = \$62.88 (-1)
UBZII = \$62.73 (-2)

> Cac, why do you think using four more
> indices would change anything?

I don't think. It just happens Those four indices are very important for KO. Besides, in BJRM, KO is not using the I18 but the KO preferred. An apple to apple comparison would require the same set of indices which is what I do (C22 against C22).

> Try optimal Wonging 1-6, 5/6, das, s17, LS,
> and let me know which count wins. I'm
> getting UBZII by about \$2 or so.

LS is another thing. In that scenario I agree that
UBZII is better than KO.

Sincerely,
Cacarulo

9. ## Don Schlesinger: Re: BJRM 2002

> Try 1-6 with S17,DAS,5/6 and you will find
> that KO is a little better than UBZII. BTW,
> I have BJRM2000.

> KO = \$62.88 (-1)
> UBZII = \$62.73 (-2)

The latest version, BJRM 2002, gives KO (-1) = \$62.88, and UBZII (-2) = \$62.73. So, there is no change, and you are correct. However, for K-O, it says BOTH K-O preferred AN I18 and Fab4. What does that mean??

> I don't think. It just happens Those four
> indices are very important for KO. Besides,
> in BJRM, KO is not using the I18 but the KO
> preferred. An apple to apple comparison
> would require the same set of indices which
> is what I do (C22 against C22).

Why would those four indices be any more important to K-O than to UBZII? Most of them are actually closer to the UBZII pivot than to the K-O pivot, no?

> LS is another thing. In that scenario I
> agree that UBZII is better than KO.

So it seems.

Don

10. ## Cacarulo: Re: BJRM 2002

> = \$62.88, and UBZII (-2) = \$62.73. So, there
> is no change, and you are correct. However,
> for K-O, it says BOTH K-O preferred AN I18
> and Fab4. What does that mean??

I don't know. John? Is it KO preferred or I18?

> Why would those four indices be any more
> important to K-O than to UBZII? Most of them
> are actually closer to the UBZII pivot than
> to the K-O pivot, no?

You're right. The only index in KO that is exactly at the pivot is 8v5. The others are closer in UBZII. But, insurance is closer in KO as well as ten-splittings. I think the difference comes from these indices. Also, my guess is that KO using I18 should be more powerful than UBZII using I18. What do you think?

Sincerely,
Cacarulo

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