Ace Side Count For Insurance In Double Deck Blackjack

[MJS]

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.

[Dog Hand]

<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

[MJS]

MJS,

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

Hope this helps!

Dog Hand

Thank you

[DSchles]

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.

Still looks wrong to me. Not sure how you are attempting to determine deck composition simply from the running count. Maybe you can give a specific example with RC = 0 and six aces seen (instead of the expected two), to tell me why you think you need to have seen that many aces to make insurance profitable. I would contend that if you had seen five aces (three extra), insurance would be correct.

There are similar arguments in other areas.

Don