Risk of Ruin

I have a question about how to determine the risk of ruin for the following strategy.

I start with 10 units.
My goal is to win 2 units and then quit.
I am playing a 6 deck game S17, ndas, double on 10&11 only

If I achieve my goal of winning 2 units, therefore having a total of 12 units before the first shoe is finished being dealt, I will quit the game regardless of the count or anything else.

If I have not achieved my goal before the shoe is finished I will start the following shoe. However I now reduce my goal to winning 1 unit for a total of 11 units. If I achieve this goal at any point during the shoe I quit the game.

If I am unable to achieve the goal, I start the next show, but now my goal is just to have a total of 10 units. This continues for each shoe, reducing my expectation by one unit, until I either achieve that goal, or run out of units completely.

I am flat betting 1 unit for all bets, but using count to determine strategy.

Any ideas on how to figure this out?
Thanks!

If you are unable to achieve the goal within 4 shoes you leave anyway.

Otherwise if your goal would eventually be to have just 1 unit, which must be about 99.9999999999%! :)

Why are you doing this? It makes no sense.

How can it be 100%?

If I win my first two hands, then I've achieved my goal. I'm experimenting with this using CVBJ, and it seems to work far more often than not.

So what I'm interested in is just how effective (or uneffective) it is.

You are flat betting regardless of the count. The count-based play variations you can make at this horrible game will not overcome the house advantage, without the use of a large bet spread or wonging out.

You said nothing to indicate that you are playing with an advantage. Artificial manipulation of session results will not change the inevitable. You said you will play until you lose all your units. Eventually this will happen. The house edge will grind you down and eventually out, regardless of the meaningless "stopping points" you set for yourself when you get ahead a couple of units.

You said: "This continues for each shoe, reducing my expectation by one unit, until I either achieve that goal, or run out of units completely."

Eventually you will be down to just one unit. When you lose a hand, you've reached expectation, which is to have zero units.

yes that system sux i believe you would be better to use galan's theorory bring all the bankroll put it on one hand of black jack with rat hole money for a good split or double leave if you win or lose if you are smart you will just back count on a 5 dollar table to less than one deck then bet it the casinos near where i live have 5 10 15 dollar games all with no sign saying "no mid shoe entry"

Why will I be down to just one unit? In my experince with this in CVBJ I have won about 80% of the time.

My concept is to use the standard deviation and quit when it is slightly up. I realize that if I had tons of money and played forever I would lose. But I will not play forever.

As an example:
I have 10 units, lose 1 hand, noe I have 9
Say I win on a DD now I have 11,
If I lose then next I have 10
then win two times in a row I Have 12 and I'm done.

I do not care that if I do this 10,000 times (or even 1,000 or 100) I may lose. I am interested in what happens if I do this 4 or 5 times. It seems to me that in this case I have a possibility of achieving my goal.

As for the continuing until I run out of units statement, I did qualify that by saying that I will only play 4 shoes. So basically I am playing four games using ranges of 0-12; 0-11; 0-10 and 0-9.

I don't have the ROR answer for your question. But if the question is modified as below there might be a possible answer.
You quit the game if you've won two units or you've lost two units. You have to keep playing until the above condition is met. Assuming the house advantage is p and the player advantage is q (so that p+q = 1), then your ROR is p^2/(p^2 + q^2). Your chance of winning is 1 - ROR or q^2/(p^2 + q^2).
So if the house advantage is 0.51, then the ROR = (0.51^2)/(0.51^2 + 0.49^2) = 0.52 or 52%.
If the question is: You quit the game if you've won 3 units or you've lost 3 units. Then your ROR is p^3/(p^3 + q^3). Your chance of winning is 1 - ROR or q^3/(p^3 + q^3).
So if the house advantage is 0.51, then the ROR = (0.51^3)/(0.51^3 + 0.49^3) = 0.53 or 53%.

How do I figure out the formula? I don't know now. That's from my memory long time ago. If somebody can find the way to prove the formula, please post it here or email me. Thanks.

I start with 10 units.
My goal is to win 2 units and then quit.
I am playing a 6 deck game S17, ndas, double on 10&11 only

According to http://www.blackjackinfo.com/bjbse.php, the house edge for this game is 0.55%. This is truly a bad game. You would be better off not playing it at all and saving your money for a trip to Las Vegas.

If I achieve my goal of winning 2 units, therefore having a total of 12 units before the first shoe is finished being dealt, I will quit the game regardless of the count or anything else.

If I have not achieved my goal before the shoe is finished I will start the following shoe. However I now reduce my goal to winning 1 unit for a total of 11 units. If I achieve this goal at any point during the shoe I quit the game.

If I am unable to achieve the goal, I start the next show, but now my goal is just to have a total of 10 units. This continues for each shoe, reducing my expectation by one unit, until I either achieve that goal, or run out of units completely.

How does change anything? What is the difference between playing 2 hour sessions or 8 15-minute sessions?

I am flat betting 1 unit for all bets, but using count to determine strategy.

Using strategy variation might decrease the house advantage to 0.40%. This means, over the long run, you will lose 40 cents for every $100 of bets you make. It does not matter that you are stopping and starting all the time. You will still lose at about 0.40%.

Any ideas on how to figure this out?

Since you are playing with a disadvantage, your ROR over the long run is 100%. If you have $1000 bankroll, expect it to be $0 after $250,000 worth of bets. At $10/bet that amounts to 25,000 bets or about 250 hours of play. Variance will affect the actual amount of time, but the result is inevitable. You will lose it all over the long run. You can't make a negative expectation into a positive one simply by arbitrarily stopping and starting anymore than you can by shaking beads or rubbing rabbit's feet.

Now, if you start with the true count is >=+2 and stop when it's <+1, then you'll be playing a winning game.

Thanks!

Glad I could be of help. Now read Blackjack Attack for a better understanding of these concepts.

What you have basically done is what the designers of all "systems" do: define an unlikely series of events, and then bet against it.

The "small Martingale" is the archtypical example, whereunder you keep doubling until you win. Yes, you will PROBABLY win any given start, but you will DEFINITELY lose over the long run.

Although it is concievable that someone could define a series of events SO unlikely one would be unlikely to experience it in a lifetime of play, I have not yet seen that done and doubt it ever could be.

Here is the rule to remember: Probabilities can NEVER overcome odds!

The concepts you propose are dangerous and self-defeating. If you are flat-betting, then you need to wong the shoe to have any chance of coming out ahead in the long run. Otherwise you should be sizing your bets as a reflection of the count. Remember, short-term results don't prove anything except they have a tendency to reinforce an ab initio faulty premise.

I have a computer program based the expectations of a 6D game with much more liberal rules than yours (DAS, DA2, RSA, NS, 75% pen).

This program was originally designed to test every possible sequence of events and compare betting progressions.

Using a 1-1 progression with a goal of 2 units and a "stop loss" of 10 units, the expectation is you will win just under 80% of the time. "Stop Loss" means you will quit once you have dropped 10 units from the peak. Also, betting stops when you do not have enough $ left to make a 1-unit bet (e.g., you hit a blackjack which put you an extra � up, but then drew down to where you only have 9� units left).

The stop loss aspect of the program will not have very much effect at all in this case, since the win goal is only 2 units.

Using a 1-1 progression with $100 units, you will win an average of ~$228 79.86% of the time, and lose an average of ~$940 20.14% of the time; for an average result of a loss of ~$6.94.

If you change to a 2-1 progression with $50 units, you will win an average of ~$218 80.09% of the time, and lose an average of ~$914 19.91% of the time; for an average result of a loss of ~$7.47.

Looking at that ~80% win rate, it means you can win twice in a row ~64% of the time, 3 x in a row 51% of the time, 4 x in a row 41% of the time, etc. You only have about a 1/3 chance of winning 5 times in a row, which means 2/3rd of the time you will lose most, if not all, of your $1000 stake.

I hope and trust this is the kind of info you were looking for.