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Thread: Ace Side Count For Insurance In Double Deck Blackjack

  1. #1


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    Ace Side Count For Insurance In Double Deck Blackjack

    I am considering side counting aces in double deck games for insurance purposes only. Is this table accurate?


    Hi Low Running Count 0 +1 +2 +3 +4 +5 +6




    1.5 Decks Remaining 4 3 2 1 0 0 0




    1 Deck Remaining 6 5 4 3 2 1 0




    .5 Decks Remaining 7 6 5 4 3 2 1




    The numbers in the table are the minimum number of Aces that must have been played and counted at a specific running count and deck penetration, which would indicate that the remaining pack contains one third or more 10's. An example would be a 0 running count with 1.5 decks (78 cards) remaining with 4 Aces played, adding an extra 2 10's to the 24 that Hi Low Count would indicate. Dividing the 26 10's by the 78 remaining cards yields a ratio of one third, which would be the break even point to take insurance. Is this a worthwhile effort in an attempt to increase the overall edge?

  2. #2


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    What do you think is the gain from side counting Aces with HiLo and simply using the index for insurance (as Aces are factored into the count)? What % of gain makes it worthwhile to take that extra step? 15%?

  3. #3


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    Quote Originally Posted by ZeeBabar View Post
    What do you think is the gain from side counting Aces with HiLo and simply using the index for insurance (as Aces are factored into the count)? What % of gain makes it worthwhile to take that extra step? 15%?
    He is not side counting for betting purposes. He is specifically doing it to alter his playing (insurance) decisions). Whereas you are using 2.4 and 3.0 true count hi lo to take insurance at double deck and 6 deck respectively, he wants to alter the threshold based On the surplus and deficit of aces.

    For example, true 3.5 with a large surplus of aces would cause him to decline insurance, and a large deficit of aces would cause him to 8nsure at true 2.5. He would adjust other true count decisions based again on that surplus deficit Situation.

    You asked about gain? Think about a drastic improvement in results on your most profitable and also potentially costly most important index. This is something that would improve your game. Consider the regaled FBM ASC for that purpose.

    To the OP
    I’ll look at your numbers later. For best results though, consider the impact of intermediate density on your calculations.
    Last edited by Freightman; 06-28-2020 at 06:22 AM. Reason: To add that 8mportant word - regaled

  4. #4


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    Quote Originally Posted by MJS View Post
    I am considering side counting aces in double deck games for insurance purposes only. Is this table accurate?


    Hi Low Running Count 0 +1 +2 +3 +4 +5 +6




    1.5 Decks Remaining 4 3 2 1 0 0 0




    1 Deck Remaining 6 5 4 3 2 1 0




    .5 Decks Remaining 7 6 5 4 3 2 1




    The numbers in the table are the minimum number of Aces that must have been played and counted at a specific running count and deck penetration, which would indicate that the remaining pack contains one third or more 10's. An example would be a 0 running count with 1.5 decks (78 cards) remaining with 4 Aces played, adding an extra 2 10's to the 24 that Hi Low Count would indicate. Dividing the 26 10's by the 78 remaining cards yields a ratio of one third, which would be the break even point to take insurance. Is this a worthwhile effort in an attempt to increase the overall edge?
    Sorry but you aren't doing it correctly. Do you have the older ("bible") version of Wong's pro BJ in which he explains how to side count aces for Hi-Lo?

    Don

  5. #5


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    Quote Originally Posted by DSchles View Post
    Sorry but you aren't doing it correctly. Do you have the older ("bible") version of Wong's pro BJ in which he explains how to side count aces for Hi-Lo?

    Don
    I have both the 1977 softback and 1981 hardback editions of Professional Blackjack; Wong shows on pages 110 - 111 in the 1977 edition and 119 - 121 in the 1981 edition how to count aces to improve both insurance and playing strategy decisions. I want to simplify the procedure and only improve my insurance decisions. If you check the numbers that were calculated, I think that you will find that they correspond with depleted packs containing one third or more 10's. This method is quite a bit simpler than what Wong explains because it only tries to improve the insurance decision. Thank you for your response.

  6. #6


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    If you try to follow Wong's explanation, your numbers simply don't equate. When you side count aces in Hi-Lo, you need 2.9 for insurance (not to be confused with 2.4 if you don't count aces).

    To take but a single example, you state that, with a +4 RC, the minimum number of aces that have to have been played to take insurance is 0, with 1.5 decks left. Since two aces should have been played, there are an extra two aces remaining over normal. As they count -2 each, your RC of +4 gets reduced to 0, and so you should NOT insure. Many of the other examples don't work either.

    Don
    Last edited by DSchles; 06-29-2020 at 11:44 AM.

  7. #7


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    Ive often thought about a secondary count to use with hi-lo..It would entail counting the ace at +2 and the 7 and 8 at +1 each vs. tens at -1 each.

    Hi-Lo: 11111000-1-1

    W/Side-Count: 111111110-2
    ins(.9813)

    Im sure some work would be involved but it would also be very efficient for 11 12's 13's and 14's..Now just surrender your 15's and 16's
    http://bjstrat.net/cgi-bin/cdca.cgi

  8. #8


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    Quote Originally Posted by DSchles View Post
    If you try to follow Wong's explanation, your numbers simply don't equate. When you side count aces in Hi-Lo, you need 2.9 for insurance (not to be confused with 2.4 if you don't count aces).

    To take but a single example, you state that, with a +4 RC, the minimum number of aces that have to have been played to take insurance is 0, with 1.5 decks left. Since two aces should have been played, there are an extra two aces remaining over normal. As they count -2 each, your RC of +4 gets reduced to 0, and so you should NOT insure. Many of the other examples don't work either.

    Don
    Thank you for pointing out the errors that I made in the original post. This will explain the attempted methodology; I am simply trying to identify situations when the ratio of 10's to all cards is equal to or greater than 1 to 3. With 1.5 decks remaining, 26 of the remaining 78 cards need to be 10's, this yields the decimal equivalent of .333. With 1 deck remaining, 18 of the remaining 52 cards need to be 10's, this yields a decimal equivalent of .346. With .5 decks remaining, 9 of the remaining 26 cards need to be 10's, this also yields a decimal equivalent of .346. Hopefully, the following table corrects my previous mistakes.

    Running Count 0 +1 +2 +3 +4 +5 +6

    1.5 decks 6 5 4 3 2 1 0

    1.0 deck 7 6 5 4 3 2 1

    .5 decks 7 6 5 4 3 2 1

    Again, the numbers in the table indicate the minimum number of aces that need to be counted at various penetrations to make insurance profitable. Thank you for your time and patience.
    Last edited by MJS; 06-30-2020 at 10:10 PM.

  9. #9


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    Quote Originally Posted by MJS View Post
    <snip>the ratio of 10's to non 10's is equal to or greater than 1 to 3.<snip>
    MJS,

    That should say "the ratio of 10's to all cards..."

    Hope this helps!

    Dog Hand

  10. #10


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    Quote Originally Posted by Dog Hand View Post
    MJS,

    That should say "the ratio of 10's to all cards..."

    Hope this helps!

    Dog Hand
    Thank you

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