Questions on streaks in baccarat

[cho]

in any baccarat shoe, we see runs of 3's or 4's and even 4+ for either banker or player. does anyone have information on the probability of the frequency these runs would occur in any given shoe? or the average probability out of all possible combination of shoes?

 

Any ideas anyones?

thanks

Edited January 25, 2016 by cho
typo error

[avion]

here ya go Cho.

FrequenceChart.pub

[cho]

Thanks for the table chart. However i think thats not quite the information im looking for or that miss read the chart.

To make my question clearer i put the following example:

Say This is the shoe I was playing,

B-1112111231321111111211316212131141121222411

Above we see 3's 4 times(4/43), 4's 2 times(2/43) and 4's+ 1 times(1/43). That is 7 times total out of 43 columns, so 7/43 or roughly 16.3%.

the question would be how to calculate the average percentage for all the shoes.

i.e. Percentage of 3's, 4's, 4's+ in any shoes.

thanks again

 

 

[avion]

Go by the chart I sent you, for the average number of times a 1,2, 3, 4, 5 etc  occurs in a shoe.  Then, accordingly, when you multiply each number on the chart from your shoe, to what is average, above average or below average.  If you are wanting the exact percentage of the total of the 1's in the shoe you just played, you have to get the total number of hands for the shoe, and then get the total number of the 1's in the shoe, to figure out the percentage of the 1's for that shoe, same for 2's, 3's etc.

I believe I answered your first question, from the chart I sent you, as you asked for   "the average probability out of all possible combination of shoes?"

[cho]

does that mean we cant work out the probablity of the shoe we havent played?

because it read from ZuanXIN Baccarat, in non biased shoe, average hands played is 72(ignoring tie), banker and player should be 50%/50%. Then from those bankers and players, Halves should be single(1iers) and the other halves having some repeats being (2's, 3's, 4's and so on). I wanted to know if there is a way to calculate the frequency of those repeats appearing in shoes we have not played.

[avion]

okay, on AVERAGE, the chart I sent you will tell you what each is.  The percentage average for a "1" is one time every 4 hands on AVERAGE.  Most every shoe you play, for example, the 1's will be below average, close to average, average, or above average.  There is no set "LAW" for a any number that it is going occur in any shoe.  

The shoe you share, if I am correct, has 26 1's in it.  So, on average a 1 happens 1 in every 4 hands, 2 in every 8 hands, 3 in every 12 hands, etc.  So you must take the number of 1's in a shoe that takes place, and then multiply the total number by 4, to see if the 1's are average, below average, above average, etc, according to the number of total decisions in a shoe.  

If another member here, reads this and can offer a better explanation for what cho is asking, please feel free to help in this matter.  

[Pando]

We know the theoretical average number or 1's, 2's, 3's 4's etc in a perfect shoe but shoes like that seldom appear. (If the did they would be perfect MDB+ shoes).

The bias in any shoes is what works in our favour providing we are able to identify it, and play it before it changes.

Knowing the theoretical average assist us in working out what kind of bias the shoes has - there will almost always be some bias that we can exploit with the tools of NOR or MDB+. How we play the shoe depends on what the bias is, how strong it is, and whether we catch it in time.

 

 

[Action Mike]

The Stats for Profits system specifically analyzes this, and reports back on a shoe by shoe basis, casino by casino basis, or any number of metrics.  We can see the results on random computer generated data as well as live user entered data.  Coming soon, we will have the ability to enter shoes real time as they are being played which will be a big help for our online players. 

@cho I recommend you look at the videos for the stats for profits system and consider joining as a premium member.