Game beat-ability confusion

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gothic

Well-Known Member
So I was reading the "Money Management 1-3" on the Blackjack School and noticed how the example game had an advantage of .40%, meaning a true count of one would mean the player has a positive .1% advantage, a true count of two would have a .6%, yadda yadda and, on either the second or last Money Management lesson, an 'average advantage' is figured to be 1.25.

The game I play has a house advantage of .66%, so this would mean I'd still be losing at a true count of one and just barely winning at a true count of two. Basically, this means I'm playing a LOT of cold action to the very tiny bit of hot play, or at least this is how I understand it. How can I calculate my 'average advantage?' The section also said that, for the described game on the page, a player bets 76% of their advantage to get their optimum bet, but since my game has different conditions, would this change how I determine my best bet? If so, how? And would the game be worth playing?

Another thing... I read somewhere that the 'average variance for a hand of blackjack is 1.33' and that the 'standard deviation is equal to the square root of variance (1.153)' I understand what EV is and basically how variance relates to losing/winning streaks, but I'm not 100% on understanding the formulae for figuring out what to expect as far as variance and standard deviation is concerned for any particular session, obviously by starting with 'hands played' as units of measurement for time. Now I feel like I'm confusing myself more. I did a few searches on the subject, but didn't find what I was looking for. Thanks for your patience, I'm learning a lot from this site.

Gothic
 
Ferretnparrot

Ferretnparrot

Well-Known Member
You are correct, if the game has a higher house edge, you will see fewer hands dealt at a player advantage and overall you will make less money per period of time because you are playing fewer hands.

You cannot just pull out a formula and calculate the player edge because factors of
-how much you bet on those hands in contrast to other -ev hands
-Penetration
-surrender and other rules
All weigh in and play a role.

You would want to use simulation programs to simulate the play and see what happens over the span of a googolplex of hands
 
C

Canceler

Well-Known Member
gothic said:
Another thing... I read somewhere that the 'average variance for a hand of blackjack is 1.33' and that the 'standard deviation is equal to the square root of variance (1.153)' I understand what EV is and basically how variance relates to losing/winning streaks, but I'm not 100% on understanding the formulae for figuring out what to expect as far as variance and standard deviation is concerned for any particular session, obviously by starting with 'hands played' as units of measurement for time.
Click to expand...
In the Voodoo forum there is a sticky thread called “The Gambler’s Fallacy”. Sonny’s reply to my question in post number 8 helped me understand this a lot better. (Be sure to look at that link included in my quote of Sonny.)
 
FLASH1296

FLASH1296

Well-Known Member
gothic said:
So I was reading the "Money Management 1-3" on the Blackjack School and noticed how the example game had an advantage of .40%, meaning a true count of one would mean the player has a positive .1% advantage, a true count of two would have a .6%, yadda yadda and, on either the second or last Money Management lesson, an 'average advantage' is figured to be 1.25.

The game I play has a house advantage of .66%, so this would mean I'd still be losing at a true count of one and just barely winning at a true count of two. Basically, this means I'm playing a LOT of cold action to the very tiny bit of hot play, or at least this is how I understand it. How can I calculate my 'average advantage?' The section also said that, for the described game on the page, a player bets 76% of their advantage to get their optimum bet, but since my game has different conditions, would this change how I determine my best bet? If so, how? And would the game be worth playing?

Another thing... I read somewhere that the 'average variance for a hand of blackjack is 1.33' and that the 'standard deviation is equal to the square root of variance (1.153)' I understand what EV is and basically how variance relates to losing/winning streaks, but I'm not 100% on understanding the formulae for figuring out what to expect as far as variance and standard deviation is concerned for any particular session, obviously by starting with 'hands played' as units of measurement for time. Now I feel like I'm confusing myself more. I did a few searches on the subject, but didn't find what I was looking for. Thanks for your patience, I'm learning a lot from this site.
Click to expand...
I think that your 'average variance for a hand of blackjack is 1.33' is too high.

Your overall advantage can be computed by using a table of True Count frequencies
and seeing what your advantage is at each True Count to calculate a weighted average.
This is sensitive to penetration. Doh' !
This assumes perfect play and it assumes playing without cover — like splitting 10's and doubling A9 where appropriate.

The penetration and the rules with YOUR bet spread and your count will determine your expectation.

You are playing a weak game and are using a weak count.
With a typical bet spread your actual advantage is < 1%.
 
G

gothic

Well-Known Member
hmmm

If the 6 deck has 70%-75% penetration, can knowing this allow me to calculate how frequently the TC's occur, or is this one of those things that requires running a sim? It seems like there would be an equation.

So I've considered the information above and came to two ideas:

1) The single deck is still playable with < 0.5% house edge at a count of zero, so why not meander over there until I...

2) sharpen up my rudimentary-but-working shuffle tracking technique I figured out by learning the house shuffle, where decks 1-6 stack, which ones combine, yadda (1,3,5 combine, for example, averaging counts), so the six deck splits into two regions of three decks, with a TC usually around +/- one or two, the other obviously being opposite. Okay, so now lets say you just have a three deck region with a TC of one... that's a running count of three in a separated three-deck sub-scenario within this 6 deck piece of work where you virtually have 100% deck penetration. Maybe there's four decks left in the shoe, but with this tactic, there's only the next ONE to consider and that remains so until I can multiply whatever RC I have by three or even four (having a third or quarter of a deck to divide by, being the same as multiplying) to get my TC as I get to the very end of the third deck.

I understand that I might not get to cut the cards to my advantage and a lot of people chop right through the middle, so I'd simply figure in only the portion coming to or currently in action. (eg if they sent a good deck of the three-deck chunk to the back where it won't be played, then I'll still have two decks to work with)

The frequency of high counts gained from the above strategy should make a six deck shoe at 0.66% normal disadvantage all of the sudden playable, unless I totally missed something. Thanks for your input so far, you guys are pros.

Gothic
 
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