Examples of system bets in a bookmaker's office: sports betting system 2 out of 3, 3 out of 5.

Let's consider examples of system bets:

Sports betting system 2 out of 3. The bookmaker offers three sporting events with odds: 1.90, 2.00, 2.5.

The player placed $600. If the player correctly predicts all three events, the following results are obtained:

odds 1.90 x odds 2.00 = odds 3.80, then multiply the odds by a third of the amount – 3.80 x $200 = $760

odds 1.90 x odds 2.50 = odds 4.75, then multiply the odds by a third of the amount – 4.75 x $200 = $950

odds 2.50 x odds 2.00 = odds 5.00, then multiply the odds by a third of the amount – 5.00 x $200 = $1,000

In total, provided that all outcomes win, the player earns:

$760 + $950 + $1,000 = $2,710

In the event that one outcome does not win, the income will be either $760, $950, or $1,000.

Sports betting using the "3 out of 5" system. The bookmaker offers five sporting events with odds of: 1.90, 2.00, 2.5, 2.8, and 3.0. This system will consist of 10 accumulators, with three events in each accumulator. The player has staked $1000. The total stake is evenly distributed across the 10 accumulators, meaning $100 per accumulator. If the player correctly predicts all five events, the result is as follows:

odds 1.90 х odds 2.00 х odds 2,50 = odds 9.50 х 100 руб. = $950

odds 1.90 х odds 2.00 х odds 2.80 = odds 10.64 х 100 руб. = $1064

odds 1.90 х odds 2.00 х odds 3.00 = odds 11.40 х 100 руб. = $1140

odds 2.00 х odds 2.50 х odds 2.80 = odds 14.00 х 100 руб. = $1400

odds 2.00 х odds 2.50 х odds 3.00 = odds 15.00 х 100 руб. = $1500

odds 2.00 х odds 2.80 х odds 3.00 = odds 16.80 х 100 руб. = $1680

odds 2.50 х odds 2.80 х odds 3.00 = odds 21.00 х 100 руб. = $2100

odds 2.50 х odds 2.80 х odds 1.90 = odds 13.30 х 100 руб. = $1330

odds 2.50 х odds 3.00 х odds 1.90 = odds 14.25 х 100 руб. = $1425

odds 2.80 х odds 3.00 х odds 1.90 = odds 15.96 х 100 руб. = $1596

In total, if all outcomes are successful, the player earns $14,185. If one or two outcomes do not succeed, the player's income will range from $950 to $6,680. The odds in this system are lower than those of an accumulator with the same number of events. In each accumulator included in the system, the odds are multiplied together, and then the resulting odds from the accumulators are summed up.

How many accumulators are in the system? How can a player in a bookmaker's office independently calculate the number of accumulators in the system? This information will be useful for the player to double-check the bookmaker, primarily for their own peace of mind. Any bookmaker provides players with calculators where they can accurately find out all the stake values. In each accumulator, the number of events equals the first number of the system's dimension. For example, in the "2 out of 3" system – 2 events in the accumulator, "3 out of 5" – 3 events, "5 out of 7" – 5 events.

However, the number of accumulators in the system is calculated more complexly, and there is a special formula for this:

m! / (n! x (m – n)!)

m – number of events in the system (the second number specified in the system dimension: «2 out of 3» – 3, «3 out of 5» – 5, «5 out of 7» – 7 etc.)

n – number of events for the bet to pass. (the first number specified in the system dimension: «2 out of 3» – 2, «3 out of 5» – 3, «5 out of 7» – 5 etc.)

! – factorial. The product of all natural numbers from 1 to a given number (the factorial of the number 5 will be equal to 1 х 2 х 3 х 4 х 5 = 120, the factorial of the number 7 will be equal to 1 х 2 х 3 х 4 х 5 х 6 х 7 = 5040, the factorial of the number 2 will be equal to 1 х 2 = 2)

There will be 10 express trains in the “3 out of 5” system:

5! / (3! х (5 – 3)!) = 120 / (6 х 2!) = 120 / 12 = 10

In the system "5 out of 7" there are 21 express:

7! / (5! х (7 – 5)!) = 5040 / (120 х 2!) = 5040 / 240 = 21