Do you want to know which mathematical odds on the success of your bet? Then there are two good news for you. First: to calculate the cross-country ability, you do not need to carry out complex calculations and spend a large number 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.
Fast passage
Calculation of the probability embedded in the betting odds
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.
Whether we like it or not, our life is full of all kinds of accidents, both pleasant and not very. Therefore, each of us would do well to know how to find the probability of an event. This will help you make the right decisions under any circumstances that are associated with uncertainty. For example, such knowledge will be very useful when choosing investment options, evaluating the possibility of winning a stock or lottery, determining the reality of achieving personal goals, etc., etc.
Probability Formula
In principle, the study of this topic does not take too much time. In order to get an answer to the question: "How to find the probability of a phenomenon?", you need to deal with key concepts and remember the basic principles on which the calculation is based. So, according to statistics, the events under study are denoted by A1, A2,..., An. Each of them has both favorable outcomes (m) and the total number of elementary outcomes. For example, we are interested in how to find the probability that an even number of points will be on the top face of the cube. Then A is roll m - rolling 2, 4, or 6 (three favorable choices), and n is all six possible choices.
The calculation formula itself is as follows:
With one outcome, everything is extremely easy. But how to find the probability if the events go one after the other? Consider this example: one card is shown from a deck of cards (36 pieces), then it is hidden back into the deck, and after shuffling, the next one is pulled out. How to find the probability that at least in one case the Queen of Spades was drawn? Exists next rule: if considered complex event, which can be divided into several incompatible simple events, then you can first calculate the result for each of them, and then add them together. In our case, it will look like this: 1/36 + 1/36 = 1/18. But what about when several occur at the same time? Then we multiply the results! For example, the probability that when two coins are tossed at the same time, two tails will fall out will be equal to: ½ * ½ = 0.25.
Now let's take even more complex example. Suppose we enter a book lottery in which ten out of thirty tickets are winning. It is required to determine:
- The probability that both will win.
- At least one of them will bring a prize.
- Both will be losers.
So let's consider the first case. It can be broken down into two events: the first ticket will be lucky, and the second one will also be lucky. Let's take into account that the events are dependent, since after each pulling out the total number of options decreases. We get:
10 / 30 * 9 / 29 = 0,1034.
In the second case, you need to determine the probability of a losing ticket and take into account that it can be both the first in a row and the second one: 10 / 30 * 20 / 29 + 20 / 29 * 10 / 30 = 0.4598.
Finally, the third case, when even one book cannot be obtained from the lottery: 20 / 30 * 19 / 29 = 0.4368.
So, let's talk about a topic that interests a lot of people. In this article, I will answer the question of how to calculate the probability of an event. I will give formulas for such a calculation and a few examples to make it clearer how this is done.
What is probability
Let's start with the fact that the probability that this or that event will occur is a certain amount of confidence in the final occurrence of some result. For this calculation, a total probability formula has been developed that allows you to determine whether an event of interest to you will occur or not, through the so-called conditional probabilities. This formula looks like this: P \u003d n / m, the letters can change, but this does not affect the very essence.
Probability Examples
On the simplest example, we will analyze this formula and apply it. Let's say you have some event (P), let it be a throw dice, that is, an equilateral cube. And we need to calculate what is the probability of getting 2 points on it. To do this, you need the number of positive events (n), in our case - the loss of 2 points, on total number events (m). The loss of 2 points can be only in one case, if there are 2 points on the die, since otherwise, the amount will be larger, it follows that n = 1. Next, we calculate the number of any other numbers falling on the dice, per 1 dice - these are 1, 2, 3, 4, 5 and 6, therefore, there are 6 favorable cases, that is, m \u003d 6. Now, according to the formula, we do a simple calculation P \u003d 1/6 and we get that the loss of 2 points on the dice is 1/6, that is, the probability of an event is very small.
Let's also consider an example on the colored balls that are in the box: 50 white, 40 black and 30 green. You need to determine what is the probability of drawing a green ball. And so, since there are 30 balls of this color, that is, there can be only 30 positive events (n = 30), the number of all events is 120, m = 120 (according to total all balls), according to the formula, we calculate that the probability of drawing a green ball will be P = 30/120 = 0.25, that is, 25% out of 100. In the same way, we can calculate the probability of drawing a ball of a different color (black, it will be 33% , white 42%).
A professional better should be well versed in odds, quickly and correctly evaluate the probability of an event by a coefficient and, if necessary, be able convert odds from one format to another. In this manual, we will talk about what types of coefficients are, as well as using examples, we will analyze how you can calculate the probability from a known coefficient and vice versa.
What are the types of coefficients?
There are three main types of odds offered by bookmakers: decimal odds, fractional odds(English) and american odds. The most common odds in Europe are decimal. IN North America American odds are popular. Fractional odds - most traditional look, they immediately reflect information about how much you need to bet in order to get a certain amount.
Decimal Odds
Decimals or else they are called European odds is the usual number format represented by decimal accurate to hundredths, and sometimes even to thousandths. An example of a decimal odd is 1.91. Calculating your profit with decimal odds is very easy, just multiply your bet amount by that odd. For example, in the match "Manchester United" - "Arsenal", the victory of "MU" is set with a coefficient - 2.05, a draw is estimated with a coefficient - 3.9, and the victory of "Arsenal" is equal to - 2.95. Let's say we're confident United will win and bet $1,000 on them. Then our possible income calculated as follows:
2.05 * $1000 = $2050;
Isn't it really that difficult? In the same way, the possible income is calculated when betting on a draw and the victory of Arsenal.
Draw: 3.9 * $1000 = $3900;
Arsenal win: 2.95 * $1000 = $2950;
How to calculate the probability of an event by decimal odds?
Imagine now that we need to determine the probability of an event by the decimal odds set by the bookmaker. This is also very easy to do. To do this, we divide the unit by this coefficient.
Let's take the data we already have and calculate the probability of each event:
Manchester United win: 1 / 2.05 = 0,487 = 48,7%;
Draw: 1 / 3.9 = 0,256 = 25,6%;
Arsenal win: 1 / 2.95 = 0,338 = 33,8%;
Fractional Odds (English)
As the name implies fractional coefficient presented common fraction. An example of an English odd is 5/2. The numerator of the fraction contains a number that is the potential amount of net winnings, and the denominator contains a number indicating the amount that must be wagered in order to receive this winnings. Simply put, we have to wager $2 dollars to win $5. Odds of 3/2 means that in order to get $3 of net winnings, we will have to bet $2.
How to calculate the probability of an event by fractional odds?
It is also not difficult to calculate the probability of an event by fractional coefficients, you just need to divide the denominator by the sum of the numerator and denominator.
For the fraction 5/2, we calculate the probability: 2 / (5+2) = 2 / 7 = 0,28 = 28%;
For the fraction 3/2, we calculate the probability:
American odds
American odds unpopular in Europe, but very unpopular in North America. Perhaps, this species coefficients is the most difficult, but it is only at first glance. In fact, there is nothing complicated in this type of coefficients. Now let's take a look at everything in order.
The main feature of American odds is that they can be either positive, and negative. An example of American odds is (+150), (-120). The American odds (+150) means that in order to earn $150 we need to bet $100. In other words, the positive US coefficient reflects the potential net income with a bet of $100. The negative American coefficient reflects the amount of bet that must be made in order to receive a net winning of $100. For example, the coefficient (- 120) tells us that by betting $120 we will win $100.
How to calculate the probability of an event using American odds?
The probability of an event according to the American odds is calculated according to the following formulas:
(-(M)) / ((-(M)) + 100), where M is a negative American coefficient;
100/(P+100), where P is a positive American coefficient;
For example, we have a coefficient (-120), then the probability is calculated as follows:
(-(M)) / ((-(M)) + 100); we substitute the value (-120) instead of "M";
(-(-120)) / ((-(-120)) + 100 = 120 / (120 + 100) = 120 / 220 = 0,545 = 54,5%;
Thus, the probability of an event with an American coefficient (-120) is 54.5%.
For example, we have a coefficient (+150), then the probability is calculated as follows:
100/(P+100); we substitute the value (+150) instead of "P";
100 / (150 + 100) = 100 / 250 = 0,4 = 40%;
Thus, the probability of an event with an American coefficient (+150) is 40%.
How, knowing the percentage of probability, translate it into a decimal coefficient?
In order to calculate the decimal coefficient for a known percentage of probability, you need to divide 100 by the probability of an event in percent. For example, if the probability of an event is 55%, then the decimal coefficient of this probability will be equal to 1.81.
100 / 55% = 1,81
How, knowing the percentage of probability, translate it into a fractional coefficient?
In order to calculate a fractional coefficient from a known percentage of probability, you need to subtract one from dividing 100 by the probability of an event in percent. For example, we have a probability percentage of 40%, then the fractional coefficient of this probability will be equal to 3/2.
(100 / 40%) - 1 = 2,5 - 1 = 1,5;
The fractional coefficient is 1.5/1 or 3/2.
How, knowing the percentage of probability, translate it into an American coefficient?
If the probability of an event is more than 50%, then the calculation is made according to the formula:
- ((V) / (100 - V)) * 100, where V is the probability;
For example, we have an 80% probability of an event, then the American coefficient of this probability will be equal to (-400).
- (80 / (100 - 80)) * 100 = - (80 / 20) * 100 = - 4 * 100 = (-400);
If the probability of an event is less than 50%, then the calculation is made according to the formula:
((100 - V) / V) * 100, where V is the probability;
For example, if we have a probability percentage of an event of 20%, then the American coefficient of this probability will be equal to (+400).
((100 - 20) / 20) * 100 = (80 / 20) * 100 = 4 * 100 = 400;
How to convert the coefficient to another format?
There are times when it is necessary to convert coefficients from one format to another. For example, we have a fractional coefficient 3/2 and we need to convert it to decimal. To convert a fractional to decimal odds, we first determine the probability of an event with a fractional odds, and then convert that probability to a decimal odds.
The probability of an event with a fractional coefficient of 3/2 is 40%.
2 / (3+2) = 2 / 5 = 0,4 = 40%;
Now we translate the probability of an event into a decimal coefficient, for this we divide 100 by the probability of an event as a percentage:
100 / 40% = 2.5;
Thus, the fractional coefficient 3/2 is equal to decimal coefficient 2.5. In a similar way, for example, American odds are converted to fractional, decimal to American, etc. The hardest part of all this is just the calculations.
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
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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
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|
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 Irushed to the defense laboratory work, having learned the first 8 controlling 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…”. Last event clearly wider and inclusiveTinto 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
|
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.
Thus, . LIt is curious that this probability (as follows from the obtainedexpression) does not depend on the number of red balls in the box.
