advantage due to knowing your first card cannot be a five

[chucky baby]

I have had answers from Norm and Don when I previously asked this question but maybe the question was not well-articulated by me in my previous post.
To clarify...
In a 6-deck game, if I am 100% certain that my first card will not be a five but my second card could be anything, including fives, and the dealer's first card and/or second card could be anything, including fives..and all hit cards could be anything, including fives...
What advantage does this knowledge offer to my game?

Thank you...
Chucky Baby

[DSchles]

I have had answers from Norm and Don when I previously asked this question but maybe the question was not well-articulated by me in my previous post.
To clarify...
In a 6-deck game, if I am 100% certain that my first card will not be a five but my second card could be anything, including fives, and the dealer's first card and/or second card could be anything, including fives..and all hit cards could be anything, including fives...
What advantage does this knowledge offer to my game?

Thank you...
Chucky Baby

What did I answer the first time?

I have to think about this, and I may be missing something, but I think the answer may not be so difficult. When a player hand starts with a five in 6-deck, the disadvantage is just a shade under 20%. If you guarantee that the hand can't start with a five, then you eliminate HALF of all such hands that contain a five. That is, the second card in the hand can be a five, but the first card can't be. So, I think that this means you eliminate as a possibility half of the 20% disadvantage, which would mean that knowledge of no five as the first card is worth 10% to the player.

But, I'm not sure ... yet. :-)

Don

[Joe Mama]

I think it will be the average of player's expectation for the twelve cards besides the five.

Theory of Black Jack page 146 lists the player's advantage/disadvantage -- A(52%), 2(-12%), 3(-14%), 4(-16%),5(-19%), 6(-18%),7(-17%),8(-9%),9(0%), Ten Value(+13%)

(52-12-14-16-18-17-9+0+13+13+13+13)/12 = 1.5%

[DSchles]

I think it will be the average of player's expectation for the twelve cards besides the five.

Theory of Black Jack page 146 lists the player's advantage/disadvantage -- A(52%), 2(-12%), 3(-14%), 4(-16%),5(-19%), 6(-18%),7(-17%),8(-9%),9(0%), Ten Value(+13%)

(52-12-14-16-18-17-9+0+13+13+13+13)/12 = 1.5%

Most times, when I write that I'm not sure if my answer is correct, it turns out that it isn't! Thankfully, I don't write that too often. :-)

In this instance, I now recall vaguely the first time I answered this, and I think I actually did it your way, which I now believe is the right approach. But, there's even a simpler way to do it without having to add and average. Say that the disadvantage from having a five as the first card is 19.5%. By stipulating that you won't get a five first, you are stating that you have a 1/13 chance of not having a 19.5% disadvantage. 19.5%/13 = 1.5%, which now becomes your edge, since this negative event is NOT happening.

Don

[Joe Mama]

By stipulating that you won't get a five first, you are stating that you have a 1/13 chance of not having a 19.5% disadvantage. 19.5%/13 = 1.5%, which now becomes your edge, since this negative event is NOT happening.

I think this is true if all the possibilities add up to zero advantage ( they actually add up to -1). Try both ways for the ace, and you get -4% and -4.4% -- Yikes, I didn't realize not getting an ace up created such a disadvantage.

[DSchles]

I think this is true if all the possibilities add up to zero advantage ( they actually add up to -1). Try both ways for the ace, and you get -4% and -4.4% -- Yikes, I didn't realize not getting an ace up created such a disadvantage.

If they don't add to zero, why is one way different than the other?

Don

[Joe Mama]

If they don't add to zero, why is one way different than the other?

Don

Because that is the only case where the sum of the remaining cases add up to the negative value of the removed case. What if all possibilities were negative (some more so than others)? Your premise would yield a positive probable outcome which is not possible if all outcomes are negative probability.

[chucky baby]

I like the simplicity of Joe Mama's solution for the first card not being a five. Can we assume it is halved because the second card is random and so could be a five?
BTW Joe, most ace sequencers know that the ace-depletion syndrome has to be factored in if card counting on the first box.

Thanks for your efforts,
Chucky Baby

[DSchles]

Because that is the only case where the sum of the remaining cases add up to the negative value of the removed case. What if all possibilities were negative (some more so than others)? Your premise would yield a positive probable outcome which is not possible if all outcomes are negative probability.

OK, agreed. But then, if we know the overall starting imbalance (positive or negative), we should be able to factor that in to my suggestion and still use just the one value, no?

Don

[Joe Mama]

OK, agreed. But then, if we know the overall starting imbalance (positive or negative), we should be able to factor that in to my suggestion and still use just the one value, no?

Don

Yes, (Initial overall imbalance minus card effect)/12.

In the OP case => (-1 -(-19))/12 = 1.5

[DSchles]

Yes, (Initial overall imbalance minus card effect)/12.

In the OP case => (-1 -(-19))/12 = 1.5

There you go! :-)

Don

[Three]

In a 6-deck game, if I am 100% certain that my first card will not be a five but my second card could be anything, including fives, and the dealer's first card and/or second card could be anything, including fives..and all hit cards could be anything, including fives...
What advantage does this knowledge offer to my game?

Just out of curiosity, how would you know this information? It seems like an odd piece of information to be certain of. I could see knowing there are no fives left, or the next card is paint. But knowing the next card is not a 5 and the cards after could be a 5, it is hard to figure out how you could possibly have such information. Don't answer if it is sensitive information.

[peterlee]

I have a quite similar question.
6 decks
The decler's up card is from a segment with TC-1 ,
All the other cards deal are from a segment with TC+1 ,including all player's cards, dealer's second card and the hitting cards.
How should I calculate my EV?