To advance to the next round of the competition. Unified state exam in mathematics. Solutions

Bets on the passage of the team in the line of bookmakers are very common. Perhaps, now all bookmakers offer bets on the passage in the following sports:

• Football. Basically, these are major world-class competitions: World Championship, European Championship, Confederations Cup, Club World Championship, Champions League, Europa League, Cup competitions of different football countries, etc.
• Basketball. A bet on the passage of a basketball team means the victory of one of the basketball teams over its opponent, taking into account overtime. It could also mean winning by the points difference the club needs to advance to the next round of the cup competition.
• Hockey. Similarly to basketball betting, the team wins in overtime in case of a draw in regular time. If we are talking about the playoffs, then the passage of the team to the next round is the object of the so-called bet on the passage (team to qualify).

Let's consider in more detail the bets on the pass in football. Bookmakers offer this type of bet only on matches that are played according to the Olympic system, i.e. right through. Such bets are not accepted for matches of regular championships, and there are no such bets in the betting lines. Cup competitions can consist of one match - for example, the FA Cup, the Italian Cup or two games - the Spanish Cup, etc. Accordingly, the bet on the team's passage to the next round will be made taking into account one or two matches, including a penalty shootout.

On large international tournaments the group tournament is short-lived and the player can place a bet in the office not only on the stage of knockout games (1/8, 1/4), but also on the exit from the group of the selected team. By and large, this category of bets can also be attributed to bets on the passage.

Another feature of bets on the passage of the team to the next stage in football are the odds that bookmakers set in their own. The odds for winning two matches in football can be an order of magnitude higher than in hockey or basketball. Let's say if one of the teams won the first match, then the coefficient for the passage of the second club to next stage competition will be overstated, this gives the player the opportunity to earn more on a successful bet.

Pass betting in basketball or hockey is different from football due to the rules of the game. In basketball and hockey matches, a draw can only be in regular time, and the winner is determined in overtime (or in a shootout in hockey).

In basketball and hockey, you can bet on winning a series of games that start in the playoffs. According to the regulations of the league, cup or championship, a series can go up to 3 or 4 victories of one of the teams, respectively, and the bet will cover all these games.

In hockey or basketball, bets on the run are a kind of insurance for a player who is not sure that the team will win in regular time. The odds of bookmakers will be lower than for the main outcome, but the chances that the bet will play will increase.

TB(4)

What does a sports bet on total over 4 mean? What is TB(4) in bookmaker bets? How to understand what is total...

Mission B10 Prototype (#320188) To go to next round competition, the football team needs to score at least 4 points in two games. If a team wins it gets 3 points, in case of a draw - 1 point, if it loses - 0 points. Find the probability that the team will be able to advance to the next round of the competition. Consider that in each game the probabilities of winning and losing are the same and equal to 0.4.

Task B10 (No. 321491) There are 33 students in the class, two of them are friends - Mikhail and Vadim. The class is randomly divided into 3 equal groups. Find the probability that Mikhail and Vadim will be in the same group.

Solution. According to the question of the problem, we are interested in the distribution of two guys into three groups (for convenience, we number these groups: group 1, group 2 and group 3). Therefore, the possible outcomes of the experiment under consideration are:

U 1 \u003d (Mikhail in the first group, Vadim in the second group) \u003d (M1, B2),

U 2 \u003d (Mikhail in the first group, Vadim in the third group) \u003d (M1, B3),

U 3 \u003d (Mikhail in the first group, Vadim in the first group) \u003d (M1, B1),

U 4 \u003d (Mikhail in the second group, Vadim in the first group) \u003d (M2, B1),

U 5 \u003d (Mikhail in the second group, Vadim in the second group) \u003d (M2, B2),

U 6 \u003d (Mikhail in the second group, Vadim in the third group) \u003d (M2, B3),

U 7 \u003d (Mikhail in the third group, Vadim in the first group) \u003d (M3, B1),

U 8 \u003d (Mikhail in the third group, Vadim in the second group) \u003d (M3, B2),

U 9 ​​\u003d (Mikhail in the third group, Vadim in the third group) \u003d (M3, B3),

Thus, the set U of all outcomes of the experiment under consideration consists of nine elements U= (U 1 , U 2 , U 3 ,… U 7 , U 9 ), and the event A - "Mikhail and Vadim were in the same group" - is favored by only three outcomes - U 3 , U 5 and U 9 . Let's find the probability of each of these outcomes. Since, according to the condition of the problem, a class of 33 people is randomly divided into three equal groups, then in each such group there will be 11 students of this class. Solely for the sake of convenience in solving the problem, imagine 33 chairs arranged in one row, on the seats of which numbers are written: the number 1 is written on the first 11 chairs, the number 2 is written on the next 11 chairs, and the number 3 is written on the last eleven chairs. The probability that Mikhail will get a chair with the number 1, equal to (11 chairs with the number 1 out of the total number of chairs). After Mikhail sat down on the chair with the number 1, there are only 32 chairs left, among which there are only 10 chairs with the number 1, therefore, the probability that Vadim will get the chair with the same number 1 is . Therefore, the probability of the outcome U 3 =(Mikhail in the first group, Vadim in the first group)=(M1, B1) is equal to the product and is equal to . Arguing in a similar way, we find the probabilities of the outcomes U 5 and U 9 . We have, P(U 5)=P(U 9)=P(U 3)=.

Thus, P(A)=P(U 3)+P(U 5)+P(U 9)=.

Answer. 0.3125.

Comment. Many students, having compiled a set U of possible outcomes of the experiment under consideration, find the required probability as a quotient of dividing the number of outcomes U 3 , U 5 and U 9 that favor the event A to the number of possible outcomes U 1 , U 2 , U 3 ,… U 7 , U 9 , i.e. P(A)=. The fallacy of such a decision lies in the fact that the outcomes of the experiment under consideration are not equally probable. Indeed, P(U 1)=, and P(U 3)=.

Solution. According to the condition of the problem, the team plays two games, and the result of each such game can be either a win, or a loss, or a draw. So, the possible outcomes of this experience are: U 1 \u003d (B; B), hereinafter B - the team won the game, P - the team lost the game, H - the team played a draw, U 2 \u003d (B; H), U 3 = (V; P), U 4 = (P; V), U 5 = (P; N), U 6 = (P; P), U 7 = (N; N), U 8 = (N; P), U 8 \u003d (N; V). Thus, the set of possible outcomes of the experiment under consideration consists of 9 elements, and the event C - “the football team went to the next round of competitions” is favored by the outcomes U 1 = (B; B), U 2 = (B; H) and U 8 = ( N; C), since the occurrence of each of these outcomes guarantees the required number of points to enter the next round of the competition. Let's find the probabilities of outcomes U 1 = (B; B), U 2 = (B; H) and U 8 = (H; B). According to the condition of the problem, the probabilities of winning and losing are equal to 0.4, since the result of one game can be either a win, or a loss, or a draw, then the probability of a draw is equal to the difference 1-(U 2 +U 8) and is equal to 0.2. So, according to the theorem on the probability of the product of independent events, P(U 1)=0.40.4=0.16 and P(U 2)=P(U 8)=0.40.2=0.08. So, the desired probability is: P (C) \u003d P (U 1) + P (U 2) + P (U 8) \u003d 0.16 + 0.08 + 0.08 \u003d 0.32.

To advance to the next round of the competition, the football team needs to score
at least 9 points in two games. If the team wins, they get 5 points,
in case of a draw - 4 points if loses - 0 points. Find the probability
that the team will be able to reach the next round of the competition. Consider
that in every game the probabilities of winning and losing are equal 0,4 .

Obviously, the team can not lose. Both draws won't suit her either. What's left?
1) Win both times. 2) Win only once and draw the second game.

The probability of winning is 0,4 . The probability of winning both times is 0.4 0.4 = 0.16.

The probability of a draw is 1 - 0,4 - 0,4 = 0,2 . What is the probability of once
draw and win once? 0.4 0.2? No, it is equal 0.4 0.2 + 0.2 0.4.
The fact is that you can win the first game, or you can win the second game, this is important.
We now consider the probability of reaching the next round: 0,16 + 0,08 + 0,08 = 0,32 .

Answer: 0,32

We illustrate the solution graphically using the table 10 x 10 from 100 cells:

Red indicates victory, marsh indicates loss, and blue indicates draw.

Gray cell: first game - loss, second game - loss.
Red cage: the first game is a loss, the second game is a victory.
Green cell: the first game is a win, the second game is a draw.
Blue cell: first game - draw, second game - draw.

In this diagram, we will color both victories in yellow,
in blue - one win and one draw.

And one more illustrative scheme. At the first moment, the team has
three scenarios: win, draw and lose.

In each case, there are three options for the outcome of the second game.

We leave only those branches that suit the team.

Calculate the probability of each branch and add them.

Quest Source: Task 4. To go to the next round of the competition, the football team needs to score

Task 4. To advance to the next round of the competition, a football team needs to score at least 4 points in two games. If a team wins it gets 3 points, in case of a draw - 1 point, if it loses - 0 points. Find the probability that the team will be able to advance to the next round of the competition. Consider that in each game the probabilities of winning and losing are the same and equal to 0.4.

Solution.

Since the probabilities of winning and losing are 0.4 each, the probability of a draw is 1-0.4-0.4=0.2. Thus, a football team can advance to the next round with the following non-joint outcomes:

Won the first game and won the second game;

Draw the first game and win the second game;

Won the first game and drew the second game.

The probability of the first outcome is . Probability of the second outcome . Probability of the third outcome . The desired probability of entering the next round of the competition is equal to the sum of the probabilities of these three independent outcomes.

Football matches are different. It can be just a friendly match, a match of the regular season of the country, a match in a group stage, a two-legged play-off cup match, a single knockout cup match, as a result of which one team has to go through and the other is relegated. In some matches, such as championship games or group tournaments, the result is fixed by regular time. In knockout matches, there may be options, up to extra time and determining the final winner in a penalty shootout. So, for such matches, they accept a bet not only on the result itself, but also for the team to pass to the next round or the final victory if it's final. We will talk about such rates in more detail.

So, a football match of any regular season ends after 90 minutes and a few minutes added by the referee. The result of such a match can be a victory of one of the teams or a draw. The winner gets 3 points, the loser gets 0 points. If it's a tie, both teams get 1 point. The same situation with matches of group tournaments. In case of equality of points, no additional games and halves are assigned, but additional indicators are counted - face-to-face meetings, goals, etc. However, there are such match formats when a team may not win in regular time, but go further. Consider examples.

One-match confrontation. Matches of domestic cup competitions of some countries, final matches of European cups, play-off matches of the World Championships, Europe, etc., are held in the form of one match. The host of the match is determined by lot, or the game takes place on a neutral field. If in such a match one of the teams won, then everything is simple - it goes further, and the loser leaves the tournament. But, in regular time, a draw can be fixed. What then? In some cups, a replay is scheduled on the field of another team (such a format, in England, for example). In other situations, assigned additional time- two halves of 15 minutes. And if this is not enough to determine the winner, then a series of post-match penalties takes place.

We know that bookmakers accept bets on the main outcome of the match: the victory of one team, the victory of the second team and a draw. In the case of such games, a draw can be fixed in regular time and the bet is calculated based on this draw result. The rate on the final winner, the team that will go further or receive the cup is accepted separately. it team passing bet.

Pass bets can be found in the extra line by going inside a specific match, in which the main outcome may not be the same as the pass outcome.

In different bookmakers, such a block of bets is drawn up and called differently ...

… but the essence is the same.

Two-match confrontation. In some domestic cups, European cups, World Cup qualifiers, European Championship play-offs, etc., the play-off format, knockout games, implies a two-legged confrontation. One game at home, one away. There may be several options here.

A team can win one match and draw the other. And she passes. So, if you didn’t bet on the second game, but on the pass, then you will win. And the bet on winning will lose, because. there was a draw.

Moreover, a team can win one match and lose the second. And the team that won with the largest difference in the sum of two games passes. If the difference is zero (for example: 2:1, 0:1), then the team that scored more goals on a foreign field goes on. If the scores are identical (3:1, 1:3), then extra time is assigned in the second match, as in the situation with a one-match playoff.

Obviously, the team can win the second match and not pass. For example, a team loses an away match 2:0 and wins at home 1:0. As a result, the match is won, and the corresponding bet on the main outcome of the match plays. But, the bet on the passage of such a team, just loses.

Teams can play two games in a draw. If both matches, in regular time, ended with the same draw score (0:0, 0:0, or 2:2, 2:2), then extra time is assigned, and then a penalty kick. So, all bets on winning teams in such games are lost. But, all the same, some team goes further.

Different draws can be fixed, for example 0:0 and 1:1. Then the team that scored on the road passes like this. And, again, the bet on the passage of the corresponding team plays, and bets on wins are broken due to draws in regular time.

A vivid example of the results of a two-legged confrontation is the game of the ¼ finals of the current Champions League. Real Madrid lost 2-0 to Wolfsburg. And before the return game, Real Madrid's pass was no longer as ridiculous as it was originally. Still, a 2-goal loss and no away goals is serious.

So, in the relevant matches, it is necessary to distinguish between the result of the game itself and the result of the confrontation in the playoffs. We must not forget that a team can draw, even lose - but pass.

One more example. Sevilla - Atleti Bilbao. Meetings in the playoffs of the Europa League 2015-2016. Sevilla wins the away match 1:2. And so, what would you like to bet on the return home game? As a result, Sevilla lost at home with the same score 1:2, breaking a long home unbeaten streak. But, at the same time, she went further, beating her opponent in a penalty shootout.

conclusions. After a victorious result in the first match, it is extremely dangerous to bet on the team's victory in the second match. In such series, teams often play by result. They can play frankly for a draw, but in the end, they can lose. So, sometimes, you should give preference to bets on the passage, and not on the main result of the match. Or, to correlate the bet on the main outcome with the real motivation of a particular team for a particular fight.

If you are confident in the strength of the team and predict its final success, then it is better to bet on the pass. In a stubborn struggle, teams can even draw in regular time, and the victory, in the end, will go to the same team that is the strongest and most experienced.

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