Do you want to know which mathematical odds on the success of your bet? Then we have two good news for you. First: to calculate the patency, you do not need to carry out complex calculations and spend a lot of time. It is enough to use simple formulas, which will take a couple of minutes to work with. Second, after reading this article, you will easily be able to calculate the probability of passing any of your trades.
To correctly determine the patency, you need to take three steps:
- Calculate the percentage of the probability of the outcome of an event according to the bookmaker's office;
- Calculate the probability from statistical data yourself;
- Find out the value of a bet given both probabilities.
Let us consider in detail each of the steps, using not only formulas, but also examples.
The first step is to find out with what probability the bookmaker evaluates the chances of a particular outcome. After all, it is clear that bookmakers do not bet odds just like that. For this we use the following formula:
PB=(1/K)*100%,
where P B is the probability of the outcome according to the bookmaker's office;
K - bookmaker odds for the outcome.
Let's say the odds are 4 for the victory of the London Arsenal in a duel against Bayern. This means that the probability of its victory by the BC is regarded as (1/4) * 100% = 25%. Or Djokovic is playing against South. The multiplier for Novak's victory is 1.2, his chances are equal to (1/1.2)*100%=83%.
This is how the bookmaker itself evaluates the chances of success for each player and team. Having completed the first step, we move on to the second.
Calculation of the probability of an event by the player
The second point of our plan is our own assessment of the probability of the event. Since we cannot mathematically take into account such parameters as motivation, game tone, we will use a simplified model and use only the statistics of previous meetings. To calculate the statistical probability of an outcome, we use the formula:
PAND\u003d (UM / M) * 100%,
WherePAND- the probability of the event according to the player;
UM - the number of successful matches in which such an event took place;
M is the total number of matches.
To make it clearer, let's give examples. Andy Murray and Rafael Nadal have played 14 matches. In 6 of them, total under 21 games were recorded, in 8 - total over. It is necessary to find out the probability that the next match will be played for a total over: (8/14)*100=57%. Valencia played 74 matches at the Mestalla against Atlético, in which they scored 29 victories. Probability of Valencia winning: (29/74)*100%=39%.
And we all know this only thanks to the statistics of previous games! Naturally, for some new team or a player, such a probability cannot be calculated, so this betting strategy is only suitable for matches in which opponents meet not for the first time. Now we know how to determine the betting and own probabilities of outcomes, and we have all the knowledge to go to the last step.
Determining the value of a bet
The value (valuability) of the bet and the passability are directly related: the higher the valuation, the higher the chance of a pass. The value is calculated as follows:
V=PAND*K-100%,
where V is the value;
P I - the probability of an outcome according to the better;
K - bookmaker odds for the outcome.
Let's say we want to bet on Milan to win the match against Roma and we calculated that the probability of the Red-Blacks winning is 45%. The bookmaker offers us a coefficient of 2.5 for this outcome. Would such a bet be valuable? We carry out calculations: V \u003d 45% * 2.5-100% \u003d 12.5%. Great, we have a valuable bet with good chances of passing.
Let's take another case. Maria Sharapova plays against Petra Kvitova. We want to make a deal for Maria to win, which, according to our calculations, has a 60% probability. Bookmakers offer a multiplier of 1.5 for this outcome. Determine the value: V=60%*1.5-100=-10%. As you can see, this bet is of no value and should be refrained from.
Bet Pass Probability: Conclusion
When calculating the passability of a bet, we used a simple model that is based only on statistics. When calculating the probability, it is desirable to take into account many different factors that are individual in each sport. It happens that it is not statistical factors that have more influence. Without it, everything would be simple and predictable. By choosing your niche, you will eventually learn to take into account all these nuances and give a more accurate assessment of your own probability of events, including many other influences. The main thing is to love what you do, gradually move forward and improve your skills step by step. Good luck and success in the exciting world of betting!
There will also be tasks for an independent solution, to which you can see the answers.
General statement of the problem: the probabilities of some events are known, but the probabilities of other events that are associated with these events need to be calculated. In these problems, there is a need for such operations on probabilities as addition and multiplication of probabilities.
For example, two shots were fired while hunting. Event A- hitting a duck from the first shot, event B- hit from the second shot. Then the sum of events A And B- hit from the first or second shot or from two shots.
Tasks of a different type. Several events are given, for example, a coin is tossed three times. It is required to find the probability that either all three times the coat of arms will fall out, or that the coat of arms will fall out at least once. This is a multiplication problem.
Addition of probabilities of incompatible events
Probability addition is used when it is necessary to calculate the probability of a combination or a logical sum of random events.
Sum of events A And B designate A + B or A ∪ B. The sum of two events is an event that occurs if and only if at least one of the events occurs. It means that A + B- an event that occurs if and only if an event occurs during the observation A or event B, or at the same time A And B.
If events A And B are mutually inconsistent and their probabilities are given, then the probability that one of these events will occur as a result of one trial is calculated using the addition of probabilities.
The theorem of addition of probabilities. The probability that one of two mutually incompatible events will occur is equal to the sum of the probabilities of these events:
For example, two shots were fired while hunting. Event A– hitting a duck from the first shot, event IN– hit from the second shot, event ( A+ IN) - hit from the first or second shot or from two shots. So if two events A And IN are incompatible events, then A+ IN- the occurrence of at least one of these events or two events.
Example 1 A box contains 30 balls of the same size: 10 red, 5 blue and 15 white. Calculate the probability that a colored (not white) ball is taken without looking.
Solution. Let's assume that the event A– “the red ball is taken”, and the event IN- "The blue ball is taken." Then the event is “a colored (not white) ball is taken”. Find the probability of an event A:
and events IN:
Events A And IN- mutually incompatible, since if one ball is taken, then balls cannot be taken different colors. Therefore, we use the addition of probabilities:
The theorem of addition of probabilities for several incompatible events. If the events make up the complete set of events, then the sum of their probabilities is equal to 1:
The sum of the probabilities of opposite events is also equal to 1:
Opposite events form a complete set of events, and the probability of a complete set of events is 1.
The probabilities of opposite events are usually denoted in small letters. p And q. In particular,
from which the following formulas for the probability of opposite events follow:
Example 2 The target in the dash is divided into 3 zones. The probability that a certain shooter will shoot at a target in the first zone is 0.15, in the second zone - 0.23, in the third zone - 0.17. Find the probability that the shooter hits the target and the probability that the shooter misses the target.
Solution: Find the probability that the shooter hits the target:
Find the probability that the shooter misses the target:
More difficult tasks in which you need to apply both addition and multiplication of probabilities - on the page "Various tasks for addition and multiplication of probabilities" .
Addition of probabilities of mutually joint events
Two random events are said to be joint if the occurrence of one event does not preclude the occurrence of a second event in the same observation. For example, when throwing a dice, the event A is considered to be the occurrence of the number 4, and the event IN- dropping an even number. Since the number 4 is an even number, the two events are compatible. In practice, there are tasks for calculating the probabilities of the occurrence of one of the mutually joint events.
The theorem of addition of probabilities for joint events. The probability that one of the joint events will occur is equal to the sum of the probabilities of these events, from which the probability of the common occurrence of both events is subtracted, that is, the product of the probabilities. The formula for the probabilities of joint events is as follows:
Because the events A And IN compatible, event A+ IN occurs if one of three possible events occurs: or AB. According to the theorem of addition of incompatible events, we calculate as follows:
Event A occurs if one of two incompatible events occurs: or AB. However, the probability of occurrence of one event from several incompatible events is equal to the sum of the probabilities of all these events:
Similarly:
Substituting expressions (6) and (7) into expression (5), we obtain the probability formula for joint events:
When using formula (8), it should be taken into account that the events A And IN can be:
- mutually independent;
- mutually dependent.
Probability formula for mutually independent events:
Probability formula for mutually dependent events:
If events A And IN are inconsistent, then their coincidence is an impossible case and, thus, P(AB) = 0. The fourth probability formula for incompatible events is as follows:
Example 3 In auto racing, when driving in the first car, the probability of winning, when driving in the second car. Find:
- the probability that both cars will win;
- the probability that at least one car will win;
1) The probability that the first car will win does not depend on the result of the second car, so the events A(first car wins) and IN(second car wins) - independent events. Find the probability that both cars win:
2) Find the probability that one of the two cars will win:
More difficult tasks in which you need to apply both addition and multiplication of probabilities - on the page "Various tasks for addition and multiplication of probabilities" .
Solve the problem of addition of probabilities yourself, and then look at the solution
Example 4 Two coins are thrown. Event A- loss of coat of arms on the first coin. Event B- loss of coat of arms on the second coin. Find the probability of an event C = A + B .
Probability multiplication
Multiplication of probabilities is used when the probability of a logical product of events is to be calculated.
In this case, random events must be independent. Two events are said to be mutually independent if the occurrence of one event does not affect the probability of the occurrence of the second event.
Probability multiplication theorem for independent events. The probability of the simultaneous occurrence of two independent events A And IN is equal to the product of the probabilities of these events and is calculated by the formula:
Example 5 The coin is tossed three times in a row. Find the probability that the coat of arms will fall out all three times.
Solution. The probability that the coat of arms will fall on the first toss of a coin, the second time, and the third time. Find the probability that the coat of arms will fall out all three times:
Solve problems for multiplying probabilities yourself, and then look at the solution
Example 6 There is a box with nine new tennis balls. Three balls are taken for the game, after the game they are put back. When choosing balls, they do not distinguish between played and unplayed balls. What is the probability that after three games there will be no unplayed balls in the box?
Example 7 32 letters of the Russian alphabet are written on cut alphabet cards. Five cards are drawn at random, one after the other, and placed on the table in the order in which they appear. Find the probability that the letters will form the word "end".
Example 8 From a full deck of cards (52 sheets), four cards are taken out at once. Find the probability that all four of these cards are of the same suit.
Example 9 The same problem as in example 8, but each card is returned to the deck after being drawn.
More complex tasks, in which you need to apply both addition and multiplication of probabilities, as well as calculate the product of several events - on the page "Various tasks for addition and multiplication of probabilities" .
The probability that at least one of the mutually independent events will occur can be calculated by subtracting the product of the probabilities of opposite events from 1, that is, by the formula.
A union (logical sum) of N events is called an event 
, which is observed every time it occurs at least one of events 
. In particular, the union of events A and B is the event A+
B(some authors
), which is observed when comesor
A,or
Bor
both of these events at the same time(Fig. 7). A sign of intersection in the textual formulations of events is the union "or".
Rice. 7. Combining A+B events
It should be taken into account that the event probability P(A) corresponds as the left part of the shaded in Fig. 7 figures, and its central part, marked as 
. And the outcomes corresponding to event B are located both on the right side of the shaded figure and in the labeled 
central part. Thus, when adding 
And 
area 
actually enters this sum twice, and the exact expression for the area of the shaded figure has the form 
.
So, association probability two events A and B is
For a larger number of events, the general calculation expression becomes extremely cumbersome due to the need to take into account numerous options for the mutual overlap of areas. However, if the combined events are incompatible (see p. 33), then the mutual overlap of areas is impossible, and the favorable zone is determined directly by the sum of the areas corresponding to individual events.
Probability associations arbitrary number incompatible events 
is defined by the expression
|
|
Corollary 1: A complete group of events consists of incompatible events, one of which is necessarily realized in the experiment. As a result, if events
…
,form a complete group, then for them
Thus,
WITHconsequence 3 We take into account that the opposite of the statement “at least one of the events will occur 
…
' is the statement 'none of the events 
…
is not implemented." That is, in other words, “events will be observed in the experience 
, And 
, and …, and 
”, which is already the intersection of events that are opposite to the original set. Hence, taking into account (2 .0), to combine an arbitrary number of events, we obtain
|
|
Corollaries 2, 3 show that in those cases where the direct calculation of the probability of an event is problematic, it is useful to estimate the complexity of studying an event opposite to it. After all, knowing the meaning 
, get from (2 .0) the desired value 
no more work.
Examples of calculating the probabilities of complex events
Example 1 : Two students (Ivanov and Petrov) together Icurled up to defend the laboratory work, having learned the first 8 kontrolling questions for this work out of 10 available. Checking readiness,the teacher asks everyone only onen randomly selected question. Determine the probability of the following events:
A= “Ivanov will defend his laboratory work”;
B= “Petrov will defend his laboratory work”;
C= “both will defend laboratory work”;
D= “at least one of the students will defend the work”;
E= “only one of the students will defend the work”;
F= “none of them will defend the work.”
Solution. Note that the ability to defend the work as Ivanov, tlike Petrov individually is determined only by the number of mastered questions, the poetat. (Note: in this example, the values of the resulting fractions were deliberately not reduced to simplify the comparison of the calculation results.)
EventCcan be formulated differently as "both Ivanov and Petrov will defend the work", i.e. will happenAnd eventA, And eventB. Thus the eventCis the intersection of eventsAAndB, and according to (2 .0)
where the factor “7/9” appears due to the fact that the occurrence of the eventAmeans that Ivanov got a “good” question, which means that out of the remaining 9 questions, Petrov now has only 7 “good” questions.
EventDimplies that “the work will be protectedor Ivanov,or Petrov,or they are both together”, i.e. at least one of the events will occurAAndB. So the eventDis a union of eventsAAndB, and according to (2 .0)
which is in line with expectations, because even for each of the students individually, the chances of success are quite high.
WITHevent E means that “either the work will be defended by Ivanoc, and Petrov "ncollapses",or Ivanov will get unsuccessful inpros, and Petrov will cope with the defense. The two alternatives are mutually exclusive (incompatible), so
Finally, the statementFwill only be true ifAnd Ivanov,And Petrov with protectionNot cope." So,
This completes the solution of the problem, but it is useful to note the following points:
1. Each of the obtained probabilities satisfies the condition (1 .0), no if for
And 
get conflictwith(1 .0) is impossible in principle, then for
try andusing (2 .0) instead of (2 .0) would result in a clearly incorrectproject value
. It is important to remember that such a probability value is fundamentally impossible, and when such a paradoxical result is obtained, immediately begin to search for an error.
2. The found probabilities satisfy the relationsm
.
Ethen it is quite expected, because eventsC, EAndFform a completeth group, and eventsDAndFare opposite to each other. Accounting for theseratios on the one hand can be usedvan for rechecking calculations, and in another situation it can serve as the basis for an alternative way to solve the problem.
P note : Don't neglect writingexact wording of the event, otherwise, in the course of solving the problem, you may involuntarily switch to a different interpretation of the meaning of this event, which will lead to errors in reasoning.
Example 2 : In a large batch of microcircuits that did not pass the output quality control, 30% of the products are defective.If any two microcircuits are chosen at random from this batch, then what is thethe probability that among them:
A= “both fit”;
B= “exactly 1 good chip”;
C= “both defective”.
Let us analyze the following variant of reasoning (careful, contains an error):
Since we are talking about a large batch of products, the removal of several microcircuits from it practically does not affect the ratio of the number of good and defective products, which means that by choosing some microcircuits from this batch several times in a row, we can assume that in each of the cases there are unchanged probabilities
=
P(a defective product is selected) = 0.3 and
=
P(good product selected) = 0.7.
For an event to occurAit is necessary thatAnd at first,And for the second time, a suitable product was chosen, and therefore (taking into account the independence of the success of choosing the first and second microcircuit from each other), for the intersection of events we have
Similarly, for the event C to occur, both products must be defective, and to obtain B, you need to select a good product once and a defective product once.
Error sign. Xalthough all the probabilities obtained aboveand look plausible, when they are analyzed together, it is easy tonote that .However, casesA, BAndCform a completegroup of events for which the .This contradiction indicates the presence of some error in reasoning.
WITH ut errors. Let us introduce two auxiliaryevents:
= “the first chip is good, the second is defective”;
= “the first chip is defective, the second one is good”.
It is obvious that , however, just such a calculation option was used above to obtain the probability of the eventB, although the eventsBAnd
are not eequivalent. In fact,
, because wordingeventsBrequires that among the microcircuits exactlyone
, but completelynot necessarily the first
was good (and the other was defective). Therefore, although
event 
is not a duplicate event 
, but should be taken into accounthang out independently. Given the inconsistency of events 
And 
,
the probability of their logical sum will be equal to
After this correction of the calculations, we have
which indirectly confirms the correctness of the found probabilities.
Note : Pay special attention to the difference in wording of events like “onlyfirst of the listed elements must…” and “onlyone of the items listedents must…”. The last event is clearly broader and includesTinto its composition the first as one of (possibly numerousx) options. These alternatives (even if their probabilities coincide) should be taken into account independently of each other.
P note : The word “percentage” comes from “per cent”, i.e."a hundred". The representation of frequencies and probabilities as a percentage allows you to operate with larger values, which sometimes simplifies the perception of values “by ear”. However, using multiplication or division by “100%” in calculations for correct normalization is cumbersome and inefficient. In this regard, notAvoid using values by mentioningas a percentage, substitute them in the calculated expressions foror as fractions of a unit (for example, 35% in the calculation is writteni as “0.35”) to minimize the risk of erroneous normalization of the results.
Example 3 : Resistor set contains one resistor nnominal value of 4 kOhm, three resistors of 8 kOhm and six resistorsorov with a resistance of 15 kOhm. Three resistors chosen at random are connected in parallel. Determine the probability of obtaining a final resistance not exceeding 4 kOhm.
Resh ion. Parallel connection resistance reshistories can be calculated by the formula
.
This allows you to consider events such as
A= “three 15 kΩ resistors selected” = “
”;
B= "intwo resistors of 15 kOhm and one with resistancem 8 kOhm” =“
”…
The full group of events corresponding to the condition of the problem includes a number of options, and it is precisely those thatwhich correspond to the advanced requirement to obtain a resistance of not more than 4 kOhm. However, although the “direct” solution path, involving the calculation (and subsequent summationing) of the probabilities that characterize all these events, and is correct, it is not advisable to act in this way.
Note that in order to obtain a final resistance of less than 4 kOhm dit remains that the used set includes at least one resistor with a resistanceeat less than 15 kOhm. Thus, only in the caseAtask requirement is not fulfilled, i.e. eventAisopposite researched. However,
.
Thus, .
P
ri
tossing
: Calculating the probability of some eventA, do not forget to analyze the complexity of determiningI probabilities of an event opposite to it. If rassread
easy, then it is with this that we must begin.other tasks, completing it by applying the relation (2 .0).
P example 4 : There arenwhite,mblacks andkred balls. Balls are drawn one at a time from the box.and returned after each extraction. Determine ProbabilityeventsA= “white ballwill be extracted before black” .
Resh ion. Consider the following set of events
= “the white ball was removed at the first attempt”;
= “first a red ball was taken out, and then a white one”;
= “a red ball was taken out twice, and a white one the third time”…
So toas the balls return, then the sequence of eventsytiy
can be formally infinitely extended.
These events are incompatible and together constitute the set of situations in which the event occurs.A. Thus,
It is easy to see that the terms included in the sum formgeometric progression
with initial element
and denominator 
. But sumsand elements of an infinite geometric progression is equal to
|
|
.
Thus, . LIt is curious that this probability (as follows from the obtainedexpression) does not depend on the number of red balls in the box.
How to convert the probability of an event into a coefficient? How to find a value (valuable or inflated) coefficient on the outcome of an event?
To increase your chances of winning, the player must understand how the bookmaker works.
Bookmaker odds represent the probability of an event with a certain percentage of markup (margin), which varies between 1.5-10% in different offices. If the margin did not exist, all bookmakers would go bankrupt in a matter of hours.
The player must understand what the coefficients are and bet only on favorable prices. Therefore, he needs to be able to convert coefficients into probabilities and vice versa.
The formula for converting the coefficient into a percentage of the probability of an event:
V=1/coff*100%
The conversion of probability to coefficients is calculated by the formula:
K=100%/probability
Example
Bookmaker quotes for the match between Real Madrid and Liverpool are:
2.25 (L1) – 3.7 (draw) – 3.09 (L2)
Convert coefficients of probability
V(P1) = 1/2.25*100%= 44.4%
V(draw) = 1/3.7*100%= 27%
V(P2) = 1/3.09*100%= 32.4%
We add the probabilities of this match and get the total probability
V = 44.4%+27%+32.4%= 103.8%
Many will wonder why the probability is more than one hundred percent. The answer is tritely simple, everything over 100% is the bookmaker's margin. In our case, it is 3.8%.
Odds for equally likely events should ideally be K(P1) = K(P2) = 2.0 (50%), however, due to the betting margin, they will be underestimated. For example, if the bookmaker markup is 7%, then the odds will be 1.86, if 2%, then the odds will be 1.96.
The key to the success of a successful player is to always bet at the best odds. Bookmakers employ traders who can also make mistakes in their calculations. Skilled players make a good living by such miscalculations.
For example, the bookmaker estimates the victory of Juventus over Roma with a probability of 60% (1.66), and you, having carefully analyzed the match, calculated the probability of 67% (1.49). If your calculations are correct, then the bookmaker gives an overestimated (valuable) odds for this outcome of this event. The player should definitely take advantage of this opportunity by placing a bet on the victory of Juventus. Such coefficients are called value and in the long run they will certainly bring profit to the player.
If your probability was less than 60%, it would mean that the bookmaker underestimated the odds for this outcome. It is strictly forbidden to bet on clearly underestimated odds!
To find value bets, a player needs to be able to correctly analyze the probability of an outcome, although there are many reputable services that provide such services for a fee.
Knowing how to estimate the probability of an event based on odds is essential to choosing the right bet. If you don't understand how to translate betting odds into odds, you'll never be able to determine how betting odds compare to the actual odds that an event will take place. It should be understood that if the probability of an event according to the bookmakers is lower than the probability of the same event according to your own version, a bet on this event will be valuable. You can compare odds for different events on the Odds.ru website.
1.1. Coefficient types
Bookmakers usually offer three types of odds – decimal, fractional and American. Let's take a look at each of the varieties.
1.2. Decimal Odds
Decimal odds, when multiplied by the size of the bet, allow you to calculate the entire amount that you will receive in your hand if you win. For example, if you bet $1 at odds of 1.80, if you win, you will receive $1.80 ($1 is the returned amount of the bet, $0.80 is the winnings on the bet, which is also your net profit).
That is, the probability of an outcome, according to the bookmakers, is 55%.
1.3. Fractional Odds
Fractional odds are the most traditional kind of odds. The numerator shows the potential amount of net winnings. The denominator is the amount of bet that needs to be made in order to get this same win. For example, odds of 7/2 means that in order to get a net win of $7, you need to wager $2.
In order to calculate the probability of an event based on a decimal coefficient, a simple calculation should be made - the denominator is divided by the sum of the numerator and denominator. For the above coefficient 7/2, the calculation will be as follows:
2 / (7+2) = 2 / 9 = 0,22
That is, the probability of an outcome, according to the bookmakers, is 22%.
1.4. American odds
This type of odds is popular in North America. At first glance, they seem rather complicated and incomprehensible, but do not be afraid. Understanding American odds can be useful, for example, when playing in American casinos, to understand quotes shown in North American sports broadcasts. Let's figure out how to evaluate the probability of an outcome based on American odds.
First of all, you need to understand that American odds are positive and negative. Negative American odds are always in the format, for example, "-150". This means that in order to receive $100 in net profit (winning), you need to wager $150.
A positive American coefficient is calculated in reverse. For example, we have a coefficient of "+120". This means that in order to receive $120 net profit (winning), you need to wager $100.
The probability calculation based on negative American odds is done using the following formula:
(-(negative US odds)) / ((-(negative US odds)) + 100)
(-(-150)) / ((-(-150)) + 100) = 150 / (150 + 100) = 150 / 250 = 0,6
That is, the probability of an event for which a negative American coefficient of “-150” is given is 60%.
Now consider similar calculations for a positive American coefficient. The probability in this case is calculated using the following formula:
100 / (positive US odds + 100)
100 / (120 + 100) = 100 / 220 = 0.45
That is, the probability of an event for which a positive American coefficient of “+120” is given is 45%.
1.5. How to convert coefficients from one format to another?
The ability to translate odds from one format to another can serve you well later on. Oddly enough, there are still bookmakers in which odds are not converted and are shown in only one format, which is unusual for us. Let's look at examples of how to do this. But first, we need to learn how to calculate the probability of an outcome based on the coefficient given to us.
1.6. How to calculate a decimal coefficient based on probability?
Everything is very simple here. It is necessary to divide 100 by the probability of the event as a percentage. That is, if the estimated probability of an event is 60%, you need to:
With an estimated probability of an event of 60%, the decimal odds would be 1.66.
1.7. How to calculate a fractional coefficient based on probability?
In this case, it is necessary to divide 100 by the probability of an event and subtract one from the result obtained. For example, the probability of an event is 40%:
(100 / 40) — 1 = 2,5 — 1 = 1,5
That is, we get a fractional coefficient of 1.5/1 or, for the convenience of counting, - 3/2.
1.8. How to calculate the American coefficient based on the probable outcome?
Here, much will depend on the probability of the event - whether it will be more than 50% or less. If the probability of an event is more than 50%, then the calculation will be made according to the following formula:
- ((probability) / (100 - probability)) * 100
For example, if the probability of an event is 80%, then:
— (80 / (100 — 80)) * 100 = — (80 / 20) * 100 = -4 * 100 = (-400)
With an estimated probability of an event of 80%, we got a negative American coefficient of "-400".
If the probability of an event is less than 50 percent, then the formula will be as follows:
((100 - probability) / probability) * 100
For example, if the probability of an event is 40%, then:
((100-40) / 40) * 100 = (60 / 40) * 100 = 1,5 * 100 = 150
With an estimated probability of an event of 40%, we got a positive American coefficient of "+150".
These calculations will help you better understand the concept of bets and odds, learn how to evaluate the true value of a particular bet.