progressive martingale

[phineas]

after this post please do not respond to tell me how the mg cna never work, is very dangerous, etc.

start with 16 units.
assume a 4 step capped martingale... 5-10-20-40 = 80 units
after 3 wins you progress to 6-12-24-48= 96
after 3 wins at 6 you prgress to 7-14-28-56= 112
after 3 wins at 7 you progress to 8-16-32-64= 128

if you complete 12 wins before the four loss tring occurs, you won 16 units and start over. you will lose nearly half the time. if you lose at any point you will have lost 80 dollars.

if the odds of losing 4 in a row (in bac) are (.49) to the 4th power = .0576 or 576 times in 10,000, rougly once in every 17 hands. right?
if you play a straight capped martingale you need win 16 units... = 16 bets. if you play the progressive mg you will only need 12 bets. you will win more often than you lose? this doesnt seem right... where is the flaw?
seems like if i keep my average under 17 hands, i should win over the long haul. but i know its not right... is it?

[Nickels_n_Bullets]

start with 16 units.
assume a 4 step capped martingale... 5-10-20-40 = 80 units
after 3 wins you progress to 6-12-24-48= 96
after 3 wins at 6 you prgress to 7-14-28-56= 112
after 3 wins at 7 you progress to 8-16-32-64= 128


The flaw...

5+10+20+40 = 75 units each
6+12+24+48 = 90 units "
7+14+28+56 = 105 units "
8+16+32+64 = 120 units "

[phineas]

i get that... but either way my loss is teh same:75 if i lose at the 6 dollar point i would have already made a 15 profit, making my overall loss the same. same thing at every other point. the question is whether i will win th ewhole thing more often than i lose it. and more importantly, why?

[Ray]

phineas- Here is a crack at your question: If we forget the .5 house
advantage we can figure probabilities without too much trouble.

Starting with the first progression you have a 50% chance on each
hand to win. If you lose the first,second and third, you still have the
same 50% chance to win. So, it's safe to say for one win you have
a 50% chance to win. What happens now is, you must win three in a
row and this is where things change. In probability calc's we can say
that each of the three wins amount to individual events and to figure
the probability it looks like this:

P(A) * P(B) * P(C) = The probability of the first three events which are:

1/2 * 1/2 * 1/2= 1/8 or about 12%( note that this turns out to be the
same as winning 3 in a row in BJ without the HA.....It's between the
published probability to win and lose....as it should be)

The next progression follows the same logic and ends up being:
1/2 * 1/2 * 1/2= 1/8 or 12%..We will call this P(DEF)

Now we have the task of combining the first and second progressions
for the probability to complete the first two steps:
P(ABC) * P(DEF)= The probability to win 6 in a row and complete the first
two steps. This amounts to: 1/8 * 1/8=1/64 or 6.4% which is the win 6
in a row probability.

The other two steps follow the same logic and it ends up that your
probability to win the 16 units in 12 bets ends up being .0132% and
this is the same as winning 12 in a row.
You will lose many more times than you will win as you can see.

One thing to remember: when we say the probability to lose 4 in a row,
we can't nail that down to some specific timeframe as related to the here
and now. Also, when you lose the first bet there is no relation between
the fact that we lost the first bet and what happens next.

Ray

[phineas]

ray wrote: So, it's safe to say for one win you have
a 50% chance to win. What happens now is, you must win three in a
row and this is where things change.

i understand your point about losses not effecting the probability of future wins, except in BJ i think there is a small correlation. losses reflect removal of "bad cards" thereby leaving the deck richer in "good cards" for the upcoming hands. however the correlation is miniscule and not of ant effect in my question.

Im not sure what you are saying when you say i must now win 3 in a row. in the progression, i needn't win more than one in a row to profit. do you mean that i must win 3 loss progressions? i don not understand. also, are you saying that the probability of winning 4 capped progressions is the same as winning 4 straight hands? that doesnt seem right. the probablity f 3 in a row is 1/8... in a 4 step martingale i should win 3 hands to get the same result, but have potentially 12 tries at it, which would be 1/4 of the total hands played. I do not doubt your math, i am just unclear on what you are saying.

[Ray]

phineas- If I understand your original post correctly, you will be limited
to 12 wins which will include all four progressions. Any loss means you
have lost 4 in a row. If this is the case you must win 3 in a row for every
progression, is that correct? I understand that just one win in the 4 steps
is a win, but don't you need to do that all the time to equal 12 wins?

[Ray]

-Maybe I can explain: To win $5 you can take up to 4 tries.
-To win $15 you can take up to 12 tries.
-Each win($5) is an individual event...not affected by the preceeding win.
-Each win must be in sequence 1,2,3=$15

Probabilities:

The probability to win $5 is 1/2 or 50%; the same is true for the next two
events/wins. To determine the probability of all the events/wins in seq:

P(A,B,C)= 1/2 * 1/2 * 1/2=1/8 or 12%

The rule for the probabilities that must occur in seq. is to multiplfy them
together. They must also be independent of each other.


Everything that I've said applies to the other three progressions. They
all end up with a probability of 1/8 or 12%

Now we know the probabilities for all four progressions is 1/8 each and
we know that they must occur in seq and that they are independent.

Therefore the probability looks like this: P(T)= 1/8 * 1/8 * 1/8 * 1/8=.013
or about 1 in 10,000 hands.

The final prob of .013 is taken from the BJ prob for 12 in a row and will
be slightly different from the above.

[phineas]

The probability to win $5 is 1/2 or 50%; the same is true for the next two
events/wins. To determine the probability of all the events/wins in seq:

P(A,B,C)= 1/2 * 1/2 * 1/2=1/8 or 12%

it seems like the 12% should be on the other side... the event is occuring 3 times is a loss... the probablity of three losses is 1/2 x 1/2 x 1/2...
i dont know, i am confused. i need to take a mth course i think.

[Ray]

phineas- it is the same on the other side...remember we are talking
about 50/50.

Example: the out-come of the hand can be a win or loss and from there,
there are a total of 30 possibilities. 15 ways to win and 15 ways to lose.

P=15/30=1/2=50%(win or lose)

Draw a tree chart starting with the first hand:

Four Step Progression

If a L then W/L then W/L,W/L then W/L,W/L,W/L,W/L

If a W then W/L then W/L,W/L then W/L,W/L,W/L,W/L

There are 30 posibilities in the configuration space and 15 each win/loss
possibilities.

Don't mix this with the probability to lose 4 in a row they are not the same.

[phineas]

are you saying that this will lose 12 times for every one win? bottom line, how many times will it lose, compared with it winning?

[Ray]

phineas- You ask tuff questions...I don't believe that we can say how
standard deviation would play out for such a short run of cards, but
we can look at it from an equal/equal point of view:

1st Progression:

(1),2,3,4
1,2,3,(4)
(1),2,3,4 Total hands = 6 as highlighted(win 2 at first hand/ 1 at 4th)

2nd Progression

1,2,3,(4)
1,(2),3,4
1,(2),3,4 Total hands = 8(win one 4th/2 at 2nd)

3rd Progression

1,2,(3),4
1,2,(3),4
1,(2),3,4 Total hands = 8(win 2 at 3rd/1 at 2nd)

4th Progression

1,2,(3),4
(1),2,3,4
1,2,3,(4) Total hands = 8(win 1 at 3rd/1 at 1st/1 at 4th)



What we have done is give equal chance for all possible outcomes thru
every step of the progression and end up with a total of 30 hands.
Now if on average we lose 4 in a row every 17 hands we can see that
we have played too many hands. 13/17=80% or 13 hands greater than
our limit of 17%

The question then becomes when do we start counting hands. By having
13 more hands than we would like we are sure to lose the 4 in a row
more times than not. I think the alternative to this kind of evaluation
is a very long simulation.

Ray