
low stakes rookie: handheld games
When using a balanced count like the HiLo or HiOpt I, is it always necessary to keep the exact # of cards in order to calculate tht TC?

GMan: Re: handheld games
> When using a balanced count like the HiLo or HiOpt
> I, is it always necessary to keep the exact # of cards
> in order to calculate tht TC?
No, you divide your running count by the number of decks remaining. Round your estimate to the nearest 1/2 deck for hand held games and you will do fine.

JSTAT: Re: handheld games
> When using a balanced count like the HiLo or HiOpt
> I, is it always necessary to keep the exact # of cards
> in order to calculate tht TC?
I use a balanced ten count and side count the aces at 1/4 decks. Tens are 2, non tens(except aces)are +1. If the count is even at 3/4 of a single deck and no aces have appeared, your odds of getting a blackjack are about one every 15 hands. At the top of the deck, it is one every 20.7 hands. Count at 1/4 decks for perfect insurance and blackjack frequencies. For perfect insurance, at 1/4 deck left, an even count is needed. At 1/2 deck, a combination of aces and/or plus count totalling two is needed. At 3/4, a combination of aces and/or plus count totalling 3 is needed. Anything higher is to the player advantage with insurance.
JSTAT

Aruuba: Re: handheld games
> I use a balanced ten count and side count the aces at
> 1/4 decks. Tens are 2, non tens(except aces)are +1.
> If the count is even at 3/4 of a single deck and no
> aces have appeared, your odds of getting a blackjack
> are about one every 15 hands.
Strictly speaking, I don't see how the count could be even after 13 cards dealt if no aces have appeared?
Count at 1/4 decks for
> perfect insurance and blackjack frequencies. For
> perfect insurance, at 1/4 deck left, an even count is
> needed. At 1/2 deck, a combination of aces and/or plus
> count totalling two is needed. At 3/4, a combination
> of aces and/or plus count totalling 3 is needed.
> Anything higher is to the player advantage with
> insurance.
> JSTAT
Guess I must be missing someting on the insurance angle as I just don't see how any of those scenarios you mention would necessarily make for a "perfect" insurance bet? Maybe it has something to do with how you treat your side count of Aces? Like, in your halfdeck example, there you are sitting with your 2 card hand vs a dealer Ace and 26 cards remaining being all the Spades and Clubs. Count would be even? with still 2 Aces remaining? Would those 2 Aces mean a plus 2 and you'd take insurance?

JSTAT: Re: handheld games
> Strictly speaking, I don't see how the count could be
> even after 13 cards dealt if no aces have appeared?
On average an ace should appear every 13 cards. If 4 tens and 8 non tens(except aces)appear, we have an even count.
> Guess I must be missing someting on the insurance
> angle as I just don't see how any of those scenarios
> you mention would necessarily make for a
> "perfect" insurance bet? Maybe it has
> something to do with how you treat your side count of
> Aces? Like, in your halfdeck example, there you are
> sitting with your 2 card hand vs a dealer Ace and 26
> cards remaining being all the Spades and Clubs. Count
> would be even? with still 2 Aces remaining? Would
> those 2 Aces mean a plus 2 and you'd take insurance?
At halfdeck, 18 non tens and 8 tens are used in theory. Take two non tens out, we have 16. 16 non tens and 8 tens makes insurances an even bet. This can be +1 with ace showing. Or 1 ace gone with ace up on even count. Or 1 with 2 aces gone and ace showing.
JSTAT

GMan: Re: handheld games
> I use a balanced ten count and side count the aces at
> 1/4 decks. Tens are 2, non tens(except aces)are +1.
> If the count is even at 3/4 of a single deck and no
> aces have appeared, your odds of getting a blackjack
> are about one every 15 hands. At the top of the deck,
> it is one every 20.7 hands. Count at 1/4 decks for
> perfect insurance and blackjack frequencies. For
> perfect insurance, at 1/4 deck left, an even count is
> needed. At 1/2 deck, a combination of aces and/or plus
> count totalling two is needed. At 3/4, a combination
> of aces and/or plus count totalling 3 is needed.
> Anything higher is to the player advantage with
> insurance.
> JSTAT
Are you 10% sure you answered the question ?

JSTAT: Re: handheld games
> Are you 10% sure you answered the question ?
I answered low stakes rookie's question about using a balanced count to calculate a true count perfectly. I suggested that he count with 1/4 decks for true count conversions. My Ten Count(JSTATII) is superior to hilo for the reasons I explained to him. How can we get blackjacks with only tens left? Hilo makes the mistake of including aces WITH tens as a minus count.
JSTAT

Aruuba: Re: handheld games
> On average an ace should appear every 13 cards. If 4
> tens and 8 non tens(except aces)appear, we have an
> even count.
But that's only 12 cards, not 13, which is what I assumed 3/4 decks meant and also since you specified no aces had yet appeared. No big deal at all.
> At halfdeck, 18 non tens and 8 tens are used in
> theory. Take two non tens out, we have 16. 16 non tens
> and 8 tens makes insurances an even bet. This can be
> +1 with ace showing. Or 1 ace gone with ace up on even
> count. Or 1 with 2 aces gone and ace showing.
> JSTAT
So, for a halfdeck remaining, as far as insurance goes, what is your count if 8 tens and 17 nontens (only 2 thru 9 I assume) have been seen (count is now +1?) and the 26th card is a dealer ace. Would you take insurance now since the count is +1 with only one Ace, the dealer's, seen so far and 3 Aces still remaining in the remaining 26 unseen cards?
I'm probably still missing something.
Just curious, is this also a counting system you use for betting purposes?

JSTAT: Re: handheld games
> But that's only 12 cards, not 13, which is what I
> assumed 3/4 decks meant and also since you specified
> no aces had yet appeared. No big deal at all.
On average, an ace will appear every 13 cards. That's how I calculate perfect insurance and perfect blackjack frequencies.
> So, for a halfdeck remaining, as far as insurance
> goes, what is your count if 8 tens and 17 nontens
> (only 2 thru 9 I assume) have been seen (count is now
> +1?) and the 26th card is a dealer ace. Would you take
> insurance now since the count is +1 with only one Ace,
> the dealer's, seen so far and 3 Aces still remaining
> in the remaining 26 unseen cards?
The dealer ace is added to the +1, thus insurance would be an even bet. At half deck, insurance needs to be a total of two of the Ten Count and/or ace side count.
> I'm probably still missing something.
> Just curious, is this also a counting system you use
> for betting purposes?
I call this count JSTATII. This count is on You Tube. It is called "Card counting at 21" by moviemakerjjcasino. 21,000 views so far. There are a few mistakes(we only took one take), but overall it shows how JSTATII works. It shows how to bet when aces are few and the ten count is good. Perfect insurance is also used in this video.
JSTAT

Aruuba: Re: handheld games
> On average, an ace will appear every 13 cards. That's
> how I calculate perfect insurance and perfect
> blackjack frequencies.
Here's the way I look at it. If you want to say 4 tens and 8 nontens have been dealt in the first 12 cards, then the count is even but the prob of a BJ is less than 1 in 16, not 1 in 15 like you first said. If you want to say that 4 tens and 9 nontens have been dealt then BJ's are indeed 1 in 15+. But then the count is not even, it's positive. Anyway, that's all I was trying to say.
> The dealer ace is added to the +1, thus insurance
> would be an even bet. At half deck, insurance needs to
> be a total of two of the Ten Count and/or ace side
> count.
Yet, clearly, 8 tens are still in the remaining 26 cards. You will be being paid 21 for something that happens less than onethird of the time. It is not an even bet at this point, it is a losing bet. So, just pointing out it seems to me less than "perfect" for insurance decisions in case you thought it was. Nothing wrong with a count that is less than perfect for insurance though.
But your count could be made to be perfect for insurance so maybe, after all, it is just that you slightly misunderstand exactly how to use the perfect insurance count. All you'd have to do is count seen Aces as +1, which you'd know from your sidecount of Aces, and, if at any time, and you don't even have to worry in the slightest what's remaining to be dealt, if you ever get to +4 running count, insurance is an even bet.
> I call this count JSTATII. This count is on You Tube.
> It is called "Card counting at 21" by
> moviemakerjjcasino. 21,000 views so far. There are a
> few mistakes(we only took one take), but overall it
> shows how JSTATII works. It shows how to bet when aces
> are few and the ten count is good. Perfect insurance
> is also used in this video.
> JSTAT
Thanks. Maybe I'll check it out.
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