Perfect insurance

[JSTAT]

Your "1/4 deck count at 1/2 deck since 3/4 deck hasn't been reached yet" comment is not comprehensible to me, but I understand that you are saying that 4 aces played with 14 cards left is treated as 2 extra aces. It looks like you are saying that whenever you are considering the number of extra aces, always round up.

JSTAT, Can you verify that this is your intention:

In the first quarter deck (1 to 12 cards seen), we expect to see no aces, so any aces seen are extra. This value can be from 0 to 4.

In the second quarter deck (13 to 25 cards seen), we expect 1 ace, so if we see more than 1, those are extra. This value can be from 0 to 3.

In the third quarter deck (26 to 38 cards seen), we expect 2 aces, so the extra aces value can be either 0, 1, or 2.

In the final quarter deck (39 or more cards seen), we expect 3 aces, so the extra aces value can be either 0 or 1.

Right?

At 13 cards, one ace should be seen. +3 is the break even point with one ace. +2 with 2 aces seen add +1 ,+1 with 3 aces add +2, even count with 4 aces seen add +3. At 14-25 cards, the count is rounded up to the first 13 card level.

At 26 cards, 2 aces should be seen. +2 is the break even point with the two aces seen. +1 with 3 aces seen add +1, even count with four aces seen add +2. With only one ace seen, a count of +3 is needed. At 27-38 cards, the count is rounded up to the 26 card level.

At 39 cards, 3 aces should be seen. +1 is the break even point with three aces seen. An even count with 4 aces seen, add +1. With only one ace seen, a count of +3 is needed to account for the extra two aces. With two aces seen, a count of +2 is needed to account for the extra ace. Piece of cake Ken!

 

[k_c]

At 13 cards, one ace should be seen. +3 is the break even point with one ace. +2 with 2 aces seen add +1 ,+1 with 3 aces add +2, even count with 4 aces seen add +3. At 14-25 cards, the count is rounded up to the first 13 card level.

At 26 cards, 2 aces should be seen. +2 is the break even point with the two aces seen. +1 with 3 aces seen add +1, even count with four aces seen add +2. With only one ace seen, a count of +3 is needed. At 27-38 cards, the count is rounded up to the 26 card level.

At 39 cards, 3 aces should be seen. +1 is the break even point with three aces seen. An even count with 4 aces seen, add +1. With only one ace seen, a count of +3 is needed to account for the extra two aces. With two aces seen, a count of +2 is needed to account for the extra ace. Piece of cake Ken!

With all due respect if you have a perfect count of tens by tagging tens = -2 and non-tens (excluding aces) = +1 and also a perfect side count of aces then a perfect insurance decision can be made by simply subtracting aces remaining from running count - insure when this number is greater than 0. This requires no calculation of how many aces per deck remain and works for any number of decks.

Assuming initial running count = 0, this will give perfect insurance decisions without any errors assuming the counter is counting accurately. :grin:

 

[itrack]

With all due respect if you have a perfect count of tens by tagging tens = -2 and non-tens (excluding aces) = +1 and also a perfect side count of aces then a perfect insurance decision can be made by simply subtracting aces remaining from running count - insure when this number is greater than 0. This requires no calculation of how many aces per deck remain and works for any number of decks.

Assuming initial running count = 0, this will give perfect insurance decisions without any errors assuming the counter is counting accurately. :grin:

Exactly.

"At 27-38 cards, the count is rounded up to the 26 card level."

As a side note, if you are counting every single ace in the deck why would you ever need to round numbers? You should KNOW the EXACT number. If you were to just listen to K_C's Above post, there would be no rounding.

 

[KenSmith]

Piece of cake Ken!

OK, so we can go through all this complicated bull, and come up with the right answer, or you can use the simple and perfectly accurate approach that me, k_c, and itrack have been advocating throughout this thread.

Why on earth would you continue to do this the hard way?

To recap... Take your running count. Add the numbers of aces seen. Insure at higher than +4. There. Done. In 15 simple words.

If you're unwilling to admit that your complex system is inferior to this, then I reiterate my belief expressed earlier that you have no interest in being accurate in your articles.

 

[JSTAT]

OK, so we can go through all this complicated bull, and come up with the right answer, or you can use the simple and perfectly accurate approach that me, k_c, and itrack have been advocating throughout this thread.

Why on earth would you continue to do this the hard way?

To recap... Take your running count. Add the numbers of aces seen. Insure at higher than +4. There. Done. In 15 simple words.

If you're unwilling to admit that your complex system is inferior to this, then I reiterate my belief expressed earlier that you have no interest in being accurate in your articles.

Thank you Ken, k_c, itrack, johndoe, and FLASH1296 for your inputs about the article and shortcuts towards perfect insurance. We all agree that the Ten Count is accurate but complexity is the issue. Well guys, I've always done it this way and chose to publish it at http://www.examiner.com/x-18051-San-Francisco-Blackjack-Examiner~y2009m9d27-Casino-blackjack-101-Perfect-insurance this afternoon. Explained how the sausage (perfect insurance) is made with the questions you posed to me Ken. We all came up with the right answers in this thread which enhanced the accuracy of the article.

 

[itrack]

congratulations, you just confused the hell out of 95% of the readers of the examiner.

 

[johndoe]

Use this formula at single deck to calculate perfect insurance, at 13 cards played, one ace should be seen. +3 is the break even point with one ace. +2 with 2 aces seen add +1 ,+1 with 3 aces add +2, even count with 4 aces seen add +3. At 14-25 cards, the count is rounded up to the first 13 card level. At 26 cards played, 2 aces should be seen. +2 is the break even point with the two aces seen. +1 with 3 aces seen add +1, even count with four aces seen add +2. With only one ace seen, a count of +3 is needed. At 27-38 cards, the count is rounded up to the 26 card level. At 39 cards played, 3 aces should be seen. +1 is the break even point with three aces seen. An even count with 4 aces seen, add +1. With only one ace seen, a count of +3 is needed to account for the extra two aces. With two aces seen, a count of +2 is needed to account for the extra ace. At each 1/4 deck, account for the normal amount of aces, then add and subtract as necessary.

or

Take your running count. Add the numbers of aces seen. Insure at higher than +4. There. Done.

:laugh:

 

[golfnut101]

wow

I shouldnt even weigh in, but I just dont get why you would try to impress
experienced people on this site with such nonsense. Your posts are embarrassing to read. If you are sincere about helping people learn the game, leave it to people who know what they are talking about.