Insurance index numbers

[Running Count]

I just spent the last 1/2 hour in bjmath.com and, after taking advil for the headache the formulas gave me, I still don't know the answer to my question:

Why should the index number for insurance increase so dramatically as the number of decks increases? The proportions of high to low cards remain the same as the TC increases regardless of the number of decks. Also, the index numbers for most other playing decisions in the Ill18 remain mostly stable as the number of decks increases. What's the math going on here?

More broadly, how can I explain to my brother why, mathematically, single deck gives a better advantage for the BS player than 6D. The best I've managed is a weak "is has to do with the lower variation with less cards in the deck."

Running Count

 

[Rob McGarvey]

Simply put, most counts do not include 789 in the count, and as the # of decks goes up, the number of non counted cards does too. This is compensated for by upping the number from +1.4 for 1 deck to +3 for 6 and 8. If you want perfect Insurance bets you can use Thorps 10 count, but again, we have to sacrifice certain things to gain advantage in the rest of the game.

 

[Theef]

A big reason why SD is best for BS players

To get a blackjack, you need a 10 and an Ace - two different-valued cards. You can never get a blackjack when you draw two of the same-valued cards.

The fewer decks you use, the less chance you'll get two same-valued cards. Suppose you get an Ace as your first card in SD. Already the number of Aces left in the deck has decreased by 25% because there are only three left. This decreases your chances of getting another Ace. Put another way, it increases your chances of getting a non-Ace. This is good because you want a certain type of non-Ace - a 10.

Now suppose you have 8 decks. There are 32 Aces in there. When you draw that first Ace, you've decreased the supply of Aces by only 3.125%. So your chances of getting a non-Ace didn't increase as much as they would have if there were only one deck.

It works the other way, too - if you get a 10 (or anything else) your chances of getting a non-whatever-it-was increase more with one deck than with multiple decks.

So fewer decks means fewer pairs, which means more non-pairs, which means more A-10 non-pairs.

Of course, the dealer gets more blackjacks too, but that's okay, because when the dealer gets blackjack, you only lose one bet, whereas if you get blackjack, you win a bet and a half. The more blackjacks, the better, even though the dealer gets as many as you get.

 

[T-Hopper]

Totally wrong. The correct answer is...

The insurance index only changes based on the number of decks for counts that include the ace. When you count the dealer's ace as -1, you are introducing an error in your true count that varies according to the number of decks remaining at the time. If you use High-Low, I would suggest using a +3 index in any number of decks, but DON'T count the dealer's ace before you make your decision.

 

[Running Count]

Really?...

T-Hopper wrote:

> insurance index only changes based on the number
> of decks for counts that include the ace

Sounds good, but according to the system indices on the menu to the left, the numbers increase for HiOpt1 (a non-ace count) just as they do for HiLo (an ace-reckoned count). In fact they are identical. So your explanation (which makes intuitive sense) seems not borne out by the numbers (unless Eliot mis-typed).

Robbie Mack's explanation seems not to make sense, either, since the proportion of 7s, 8s, and 9s theoretically stays the same as the number of decks increases.

My non-expert understanding was that the increase in the insurance count had to do with the fact that SD play more closely aligned to the expected outcome -- less variation is expected. But I can't translate that into a logical explanation.

RC

 

[Rob McGarvey]

That Too

Use Hi Opt II and you'll never go wong, I mean wrong.

 

[T-Hopper]

Look like a cut & paste from High-Low

See the Hi-Opt II chart on the same page, where the index is 3 in 1 deck and 4 in 2+ decks. That's a difference of just one point between 1 and 6 decks (and probably just a fraction of a point) with an ace-neutral multi-level count.

 

[Running Count]

A Typo in the Indices, Mr. Mayor? Your constituents want to know!

 

[The Mayor]

Re: A Typo in the Indices, Mr. Mayor? Your constituents want to know!

I don't understand what the question is, or else I would answer.

If the question is are my numbers correct, the answer is yes.

If the question is "why are my numbers correct", the answer is because that is what the simulator showed them to be.

Sometimes that's the only argument I have.

--Mayor

 

[Running Count]

So if the nu,bers for hiopt1 are right...

...then how does that affect your theory, T-Hopper?

RC

 

[T-Hopper]

Some theory

The only time the number of decks makes a difference in the outcome of a hand is when you are drawing more than one card - not for insurance. A given (exact) TC represents the same ratio of 10s no matter how many cards remaining.

Since we use not exact TCs but TCs bundled into groups, you will see minor fluctuations in the index due to things like # of players or level of penetration, since these things affect the distribution of exact counts within each interval n..n+1 or n-.5 to n+.5. Usually these factors won't change the number by more than a single point.

Humble and Cooper give +2 for single deck and +3 for 2 or more decks for Hi-Opt I.

 

[illustrious bakedbean]

great answer

that was wonderfully clear.

and it tracks the best explanation i could come up with during a conversation with a bj friend of mine.

well done. i wholeheartedly concur.

 

[Stephen Bauer]

Re: A big reason why SD is best for BS players

Nice explanation *NM