Accuracy of Illustrious 18.

[chessplayer]

I have a problem with two entries of the illustrious 18 as listed in http://jay.purplewire.com/blackjack/ill18.html (Archive copy)

1) You shoulds insure against A when the TC is +3 .

The payout is 2 to1. Simple math tells me then I need 10 or Q or K or J to occur in AT LEAST one third of the time for it to be worth while.

This means the TC should be + 4 for this to be advisable. in a pack of 52, only the removal of 4 low cards can make the tens one third, unless you are talking about the 7 , 8 , 9 which are not counted in Hi Lo counting,which in this case may mean the guide is not so reliable.

2)We should Double when having 11 vs A when TC is >=+1

To me A is better than 11. Blackjack wins 21. if 11 takes on other values, A is likely to not bust and win too. Further, you take away your option to take addtional cards with Double.

I wonder if any1 else wonder along with me

 

[Sucker]

I have a problem with two entries of the illustrious 18 as listed in http://jay.purplewire.com/blackjack/ill18.html (Archive copy)

1) You shoulds insure against A when the TC is +3 .

The payout is 2 to1. Simple math tells me then I need 10 or Q or K or J to occur in AT LEAST one third of the time for it to be worth while.

This means the TC should be + 4 for this to be advisable. in a pack of 52, only the removal of 4 low cards can make the tens one third, unless you are talking about the 7 , 8 , 9 which are not counted in Hi Lo counting,which in this case may mean the guide is not so reliable.

I believe you may be confusing True Count with RUNNING Count. Perhaps someone with intimate knowledge of Hi-Opt can explain this fully.

2)We should Double when having 11 vs A when TC is >=+1

To me A is better than 11. Blackjack wins 21. if 11 takes on other values, A is likely to not bust and win too. Further, you take away your option to take addtional cards with Double.

I wonder if any1 else wonder along with me

Ace IS better than 11, unless the dealer has already checked and does not have BJ. This means that when it's time to play the dealers' hand, the next card she turns over (which is her hole card) can NOT give her 21. If she makes 21, she has to WORK at it. You, on the other hand; CAN get a ten on your double down card. If you think about it, you WILL get it.

 

[chessplayer]

I am not. TC is TC which is the count for an average deck left.

As to your second point, I rechecked the BS and found the Illustrious 18 to be accurate. However the first point is unresolved

I believe you may be confusing True Count with RUNNING Count. Perhaps someone with intimate knowledge of Hi-Opt can explain this fully.



Ace IS better than 11, unless the dealer has already checked and does not have BJ. This means that when it's time to play the dealers' hand, the next card she turns over (which is her hole card) can NOT give her 21. If she makes 21, she has to WORK at it. You, on the other hand; CAN get a ten on your double down card. If you think about it, you WILL get it.

 

[kewljason]

I have a problem with two entries of the illustrious 18 as listed in http://jay.purplewire.com/blackjack/ill18.html (Archive copy)

1) You shoulds insure against A when the TC is +3 .

The payout is 2 to1. Simple math tells me then I need 10 or Q or K or J to occur in AT LEAST one third of the time for it to be worth while.

This means the TC should be + 4 for this to be advisable. in a pack of 52, only the removal of 4 low cards can make the tens one third, unless you are talking about the 7 , 8 , 9 which are not counted in Hi Lo counting,which in this case may mean the guide is not so reliable.

In single deck game (pack of 52 cards) the insurance index is actually 1.4 complicating your question even further. :laugh: The +3 index is for multideck games. (actually 3.0 for 6 deck and 3.1 for 8 decks but generally rounded to 3)

 

[Seaclusion]

This means the TC should be + 4 for this to be advisable. in a pack of 52, only the removal of 4 low cards can make the tens one third, unless you are talking about the 7 , 8 , 9 which are not counted in Hi Lo counting,which in this case may mean the guide is not so reliable.

In order to have a +3 count with the dealer showing an ace, at least five cards have to have been removed from the deck. The dealers card which is known to be an ace and therefore makes the count -1. Your two cards which are low and now makes the count +1, and the other player at the table whose two cards have to be low to make the count now a +3.

That means there are 47 cards left in the deck with only one high card removed. 16 ten cards (not including aces) left divided by 47 cards left in the deck = 34% of the cards left are ten cards. More than 1/3.

 

[SleightOfHand]

In order to have a +3 count with the dealer showing an ace, at least five cards have to have been removed from the deck. The dealers card which is known to be an ace and therefore makes the count -1. Your two cards which are low and now makes the count +1, and the other player at the table whose two cards have to be low to make the count now a +3.

That means there are 47 cards left in the deck with only one high card removed. 16 ten cards (not including aces) left divided by 47 cards left in the deck = 34% of the cards left are ten cards. More than 1/3.

Hmmm I wonder... Lets say you are playing a face up 6D game with 4 other players at the table and you are at first base playing HiLo. The cards have been dealt and the dealer has an A showing. 3 decks have been played and the RC is -5, including what is currently dealt. Normally, this means that you wouldn't take insurance. However, what if EVERY single card that is shown, aside from the dealer's, is a neutral card? Can you temporarily add these cards to your RC for your insurance decision? This would create a temporary RC of +10, TC +3.33, which would dictate you to take it. For insurance, neutral cards are just as valueble as small cards, so would it make sense to use this info?

 

[Seaclusion]

Well, in the count I use, I would still consider those cards low cards when making decisions about the ratio of tens left in the deck. So for me, yes, they are counted

 

[chessplayer]

Your post seem logical. But it does not answer the question for multidecks.

I have a great mathematics background but am average in Blackjack.

Anyways, your post answer perhaps for single deck but not others.

Basically my point is With a TC of +3, it means the average(I am not saying only 1 deck left, it can be any number left) deck left has had 3 low cards taken from it.

You need at least 4 to make the probability of Insurance to be equal.

Once again, I, also, not the illustration table, am not talking about the single deck only. I listed the single deck as an average deck merely to illustrate the point

In order to have a +3 count with the dealer showing an ace, at least five cards have to have been removed from the deck. The dealers card which is known to be an ace and therefore makes the count -1. Your two cards which are low and now makes the count +1, and the other player at the table whose two cards have to be low to make the count now a +3.

That means there are 47 cards left in the deck with only one high card removed. 16 ten cards (not including aces) left divided by 47 cards left in the deck = 34% of the cards left are ten cards. More than 1/3.

 

[DMMx3]

OP: I'm confused by the way you are wording the issue with the insurance, but I believe you are calculating TC incorrectly. If you "take away" 3 low cards from an average deck, the TC is not +3. The RC is +3, but you have to divide that by the number of decks left (in this case 49/52), so your TC is actually 3.18...

 

[Sharky]

simple math on insurance:

You are betting that the dealer HAS BJ:

Say you have a $20 bet:

you can ins 1/2, or $10, at 2-1 odds

4 of the 13 cards (10, J, Q, & K) you win => 4 x 10 x 2 (2:1 odds) => $80

9 of the 13 cards (2-9, & A) you lose => 9 x 10 => -$90

So, clearly, if you do not count, ins is a losing prop.

Now if you count and the TC is +3, 7 cards win, which is over 1/2 of the cards (7/13) and you get paid 2:1 which makes it a winning prop.

 

[SleightOfHand]

Your post seem logical. But it does not answer the question for multidecks.

I have a great mathematics background but am average in Blackjack.

Anyways, your post answer perhaps for single deck but not others.

Basically my point is With a TC of +3, it means the average(I am not saying only 1 deck left, it can be any number left) deck left has had 3 low cards taken from it.

You need at least 4 to make the probability of Insurance to be equal.

Once again, I, also, not the illustration table, am not talking about the single deck only. I listed the single deck as an average deck merely to illustrate the point

A count of +3 does NOT mean 3 small cards were removed. There are many different scenarios that can generate a +3 count i.e. 3 high 6 small removed, 20 neutral 5 low 2 aces removed, etc. Considering all the probabilities that can produce different counts, a +3 TC is the first point where insurance becomes profitable.

 

[SleightOfHand]

simple math on insurance:

You are betting that the dealer HAS BJ:

Say you have a $20 bet:

you can ins 1/2, or $10, at 2-1 odds

4 of the 13 cards (10, J, Q, & K) you win => 4 x 10 x 2 (2:1 odds) => $80

9 of the 13 cards (2-9, & A) you lose => 9 x 10 => -$90

So, clearly, if you do not count, ins is a losing prop.

Now if you count and the TC is +3, 7 cards win, which is over 1/2 of the cards (7/13) and you get paid 2:1 which makes it a winning prop.

1/2 of the cards to be tens is not required for the 2:1 to be profitable, its 1/3.

 

[Sharky]

1/2 of the cards to be tens is not required for the 2:1 to be profitable, its 1/3.

1/3 x $20 = $6.60
2/3 x $10 = $6.67
------
-$0.07

 

[chessplayer]

DMMx and Sharky, When the low cards are taken out, they are not replaced by high cards, so your math are wrong. Rather, the remaining cards are stretched out to make 52.

We have to go back to 4 deck then to take a look.

Say you are playing, 7 players altogether . Dealer gets an Ace while you get 2 and 4. The rest of the players get low cards too.

Now the TC is more than +3. RC is +2(Per Player) x 7 Players -1=+13.

There are 4 or 3.8 Deck left. 13/4 or 13/3.8 Gives you more than what you are asking for,

Now, there are 64 K,Q,J,10 left. This is out of 208-15= 193

64/193 = 0.331......, Just short of a third to make it worthwhile.

OP: I'm confused by the way you are wording the issue with the insurance, but I believe you are calculating TC incorrectly. If you "take away" 3 low cards from an average deck, the TC is not +3. The RC is +3, but you have to divide that by the number of decks left (in this case 49/52), so your TC is actually 3.18...

But, when you have 1 deck remaining with a TC of 3, that means 3 low cards were replaced by high cards. So your 52-card deck which typically has 20 low cards, 12 neutral cards, 16 tens, and 4 aces, now has 17 low cards, 12 neutral cards, 18.4 tens, and 4.6 aces.

18.4/52 > 1/3, so insurance is profitable.

 

[Sharky]

DMMx and Sharky, When the low cards are taken out, they are not replaced by high cards, so your math are wrong. Rather, the remaining cards are stretched out to make 52.

We have to go back to 4 deck then to take a look.

Say you are playing, 7 players altogether . Dealer gets an Ace while you get 2 and 4. The rest of the players get low cards too.

Now the TC is more than +3. RC is +2(Per Player) x 7 Players -1=+13.

There are 4 or 3.8 Deck left. 13/4 or 13/3.8 Gives you more than what you are asking for,

Now, there are 64 K,Q,J,10 left. This is out of 208-15= 193

64/193 = 0.331......, Just short of a third to make it worthwhile.

That's why it's just simple math:laugh:

 

[Sharky]

DMMx and Sharky, When the low cards are taken out, they are not replaced by high cards, so your math are wrong. Rather, the remaining cards are stretched out to make 52.

We have to go back to 4 deck then to take a look.

Say you are playing, 7 players altogether . Dealer gets an Ace while you get 2 and 4. The rest of the players get low cards too.

Now the TC is more than +3. RC is +2(Per Player) x 7 Players -1=+13.

There are 4 or 3.8 Deck left. 13/4 or 13/3.8 Gives you more than what you are asking for,

Now, there are 64 K,Q,J,10 left. This is out of 208-15= 193

64/193 = 0.331......, Just short of a third to make it worthwhile.

Although I was just commenting on OP +3, I do not deal with %ages of remaining high cards, as I find it MUCH simpler to use TC.

1/0.331 = TC of 3.02 => TAKE IT!!!!:whip::whip::whip::whip::whip::whip:

 

[chessplayer]

Ermm.. you mean?

I have Just showed, or rather proved, that EVEN with a TC of 13/4=3.25 or 13/3.8= about 3.4, you STILL do NOT break even. Let alone the 3 proposed by the table.

That's why it's just simple math:laugh:

 

[Sharky]

Ermm.. you mean?

I have Just showed, or rather proved, that EVEN with a TC of 13/4=3.25 or 13/3.8= about 3.4, you STILL do NOT break even. Let alone the 3 proposed by the table.

You are correct and I'm sure you can distinguish .33 decks from .35 :laugh::laugh::laugh::laugh:

PS: the indices are ROUNDED for those of us without you KEEN eyesite:laugh::laugh:

 

[Deathclutch]

DMMx and Sharky, When the low cards are taken out, they are not replaced by high cards, so your math are wrong. Rather, the remaining cards are stretched out to make 52.

We have to go back to 4 deck then to take a look.

Say you are playing, 7 players altogether . Dealer gets an Ace while you get 2 and 4. The rest of the players get low cards too.

Now the TC is more than +3. RC is +2(Per Player) x 7 Players -1=+13.

There are 4 or 3.8 Deck left. 13/4 or 13/3.8 Gives you more than what you are asking for,

Now, there are 64 K,Q,J,10 left. This is out of 208-15= 193

64/193 = 0.331......, Just short of a third to make it worthwhile.

Sounds like you need a count with a higher insurance correlation. Your insurance calls won't be perfect with any count, but you can get some better ones than hi-lo.

 

[Kasi]

They will be with a count only designed for insurance lol.

For single deck Ace thru 9 +1, 10 thru King -2. A Running Count of +4 with this count in a single deck game is a perfect insurance count.

An Insurance bet is profitable with this count.

Adjust accordingly for multiple decks.