Выигрышная ситема ставок в букмекерской конторе

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  I think, all are familiar to that rates in bookmaker offices happen following kinds:
- A normal level
- express
- System
- The conditional rate

About game by normal levels and expresss know everything, game by the conditional rate is a little specific, therefore we will disassemble it later. But here what I have faced recently. What even the people, long enough playing to bookmaker office, badly understand features of game by system. Now I will try to tell about it. In the form of questions and answers.

1) That such system?

Can glance in rules of any bookmaker office, and you will meet there about such inscription. “The system is the set of  express representing full search of variants of express of one size from the fixed set of outcomes. It is characterised by the identical size of the rate on each express (a system variant) and identical quantity of outcomes in each express. At the rate it is necessary to specify total of outcomes and dimension of the express (a system variant) in system. The prize on system is equal to the sum of prizes on the expresss entering into system” (from game rules of one known bookmaker office).

Similar it is possible to meet and in rules of other offices. But here an ill luck. To understand this it is difficult enough. Therefore I will try to explain same the most simple accessible language.

We have solved, on what we will put. That is have selected for game some matches. But here an ill luck. If to play their normal levels what well to win, it is necessary to put the decent sums of money for each match. It is possible to arrive in another way: to generate the express from the pleasant events. To put it is not enough, but with the big factor. But in this case if any rate loses, we absolutely will receive nothing. There is also a third variant. To make of the selected events some expresss. And if any event in one of expresss does not play, we will receive 0 only on this express, and the others will win.

But how to break rates on expresss? Can play doubles (expresss from two events), tees (from three) or still somehow? Further I will write, how it correctly to make. So, for example we have chosen that all of us we will play rates doubles. But what rates to include in what doubles? By and large they after all all are equal (here to me likely will object, after all there is a superrate which 100 % as will win. But remember how many time this your superrate has not played?).

Here to us to the aid game by system also comes. At game by system we speak. Here our events. Make of them expresss. In each express include … (here we speak dimension of system) events. In all possible combination. For example: our events:

1. A Torpedo - Spartak of M: the Victory of 2 commands
2. Krylja Sovetov – the Rotor: the Victory of 1 command
3. An Uralan - the M Locomotive: the Victory of 2 commands

Then the set of doubles for them will be:

1. (A Torpedo - Spartak of M: the Victory of 2 commands) * (Krylja Sovetov – the Rotor: the Victory of 1 command)

2. (A Torpedo - Spartak of M: the Victory of 2 commands) * (an Uralan - the M Locomotive: the Victory of 2 commands)

3. (Krylja Sovetov – the Rotor: the Victory of 1 command) * (an Uralan - the M Locomotive: the Victory of 2 commands)

What does it give to us? That if the rate (a Torpedo - Spartak of M has not played: the Victory of 2 commands) and on 1 and 2 express we have received a prize 0. And the third express has played. And we have won money. That is in a case if we have united all 3 events in one express, we would lose money. And in case of system even in the presence of one not guessed event we will win money.

2) I know that at game on system, it is necessary to name on what system “How many from How many” I play. What should I name?

Name quantity of events which you have selected for rates. And how many events you would like to include in each express from these events. The system 2 of 3 (we have chosen 3 events and in each express have included on 2 events) has been above disassembled. It is called dimension of system.

3) That such quantity of variants in system?

That, how many expresss it is possible to make it, if all events to break into expresss according to the set dimension. For example, in system 2 from 3, it is possible to make 3 expresss (on 2 events in everyone). We assorted this case above.

4) What can be dimension of system? What restrictions are on system?

Yes any. For example, 2 from 3, 7 from 9, 4 from 6 and ect. But here it is necessary to consider the following.
Offices can limit:
Quantity of events included in system.
The Maximum quantity of variants in system.
The Minimum sum on 1 variant of system
The Maximum factor of one express in system

For each bookmaker office these parametres. To find them it is possible in rules.
And here for what they? In my opinion: "and" and are connected with the out-of-date software of offices. In"that players would put more money.“G”– what not to be ruined if to players will carry.

5) What there are systems?

- Simple (what accept all offices, we have disassembled them above)
- Difficult (what accept only the most advanced offices (as I understand offices by which it allows to do their software))
- Invented by various people – a set of ordinary rates grouped in the different ways. wow! And how many their everyones happens!

5) That such difficult system?

The most ridiculous in this question that to me set it not only simple players, but also some bookmaker offices. Even one very big known bookmaker office has set. Can because I think, what they do not aspire to learn and enter something new? So after all big …

The difficult system is a system each event in which can consist of expresss. That is, for example, modifying earlier considered example, earlier we had 3 events:

1. A Torpedo - Spartak of M: the Victory of 2 commands
2. Krylja Sovetov – the Rotor: the Victory of 1 command
3. An Uralan - the M Locomotive: the Victory of 2 commands

And we have wanted that the first event at us would be not
“A Torpedo - Spartak of M: the Victory of 2 commands”, and the express
1. (A Torpedo - Spartak) * (CSKA – the Ruby the Victory of 1 command)

Then the system 2 of 3 will be:

1. ((A Torpedo - Spartak * (CSKA – the Ruby the Victory of 1 command)) * (Krylja Sovetov – the Rotor: the Victory of 1 command)

2. ((A Torpedo - Spartak * (CSKA – the Ruby the Victory of 1 command)) * (an Uralan - the M Locomotive: the Victory of 2 commands)

3. (Krylja Sovetov – the Rotor: the Victory of 1 command) * (an Uralan - the M Locomotive: the Victory of 2 commands)

6) And for what these most difficult systems in general are necessary? What to us for game of simple systems will not suffice?

Difficult systems are convenient at game with small factors. That is, for example, we have 2 events, at which big factors. And some events, at which very small factors. The optimal way in this case to make difficult system 2 of 3. As first 2 events events with the big factors will act. And the third event will be the express from other events (those which with small factors).

Let's continue to consider systems further. In general, what purpose of game in bookmaker office? The majority will tell – to win money! Also will be not right. The overall objective, this money not to lose! As it is known on statisticans lose 90. 92 % of players. At the expense of them the others also live of 10 % of players which win, and bookmaker offices.

Therefore today we will disassemble. As it is possible not to lose, using systems.
We take such situation. We play system 2 of 3. Also have guessed 2 outcomes. A question: We will get profit? Or there will be at us losses. The answer: all depends on factors which we played. If enough big (for system 2 of 3 = "> 1.73)" that we have won average factor of advantageous events of system, to us will return money back. If it is a bit less, we will a little lose.

Therefore for each system there is a so-called point of break-even. We will result the table in which these factors are described:

2 from 3 = ">" 1.73
3 from 4 = ">" 1.58
4 from 5 = ">" 1.49
5 from 6 = ">" 1.43
6 from 7 = ">" 1.38
7 from 8 = ">" 1.34
8 from 9 = ">" 1.31

That is, for example, playing system 8 of 9 and guessing 8 events, we will get profit if the won factors in system are not less than 1.31.

Now. As to other systems.

2 from 4 = ">" (3 1.4(2) 2.45

3 from 5 = ">" (4 1.35(3) 2.15
2 from 5 = ">" (4 1.29(3) 1.8(2) 3.15

4 from 6 = ">" (5 1.32(4) 1.97
3 from 6 = ">" (5 1.26(4) 1.71(3) 2.71
2 from 6 = ">" (5 1.22(4) 1.58(3) 2.23(2) 3.87

5 from 7 = ">" (6 1.27(5) 1.85
4 from 7 = ">" (6 1.23(5) 1.63(4) 2.44
3 from 7 = ">" (6 1.20(5) 1.52(4) 2.06(3) 3.27
2 from 7 = ">" (6 1.21(5) 1.48(4) 1.91(3) 2.7(2) 4.7

6 from 8 = ">" (7 1.26(6) 1.74
5 from 8 = ">" (7 1.22(6) 1.56(5) 2.23
4 from 8 = ">" (7 1.19(6) 1.47(5) 1.93(4) 2.9
3 from 8 = ">" (7 1.17(6) 1.41(5) 1.77(4) 2.43(3) 3.83
2 from 8 = ">" (7 1.15(6) 1.35(5) 1.67(4) 2.15(3) 3.05(2) 5.3

What for figures stand in brackets? In brackets there is a quantity of events guessed by us. That is 3 from 8 = ">" (5 1.77, it is necessary to understand that at game in system 3 from 8 at guessing of 5 events we will be returned by money if the average factor of the won events is not less than 1.777

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