Job source: Task 4. To advance to the next round of competition, the football team needs to score

Task 4. To advance to the next round of competition, a football team needs to score at least 4 points in two games. If a team wins, it receives 3 points, in case of a draw - 1 point, if it loses - 0 points. Find the probability that the team will 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 equal to 0.4, 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 incompatible outcomes:

Won the first game and won the second game;

Drew the first game and won 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 required probability of reaching the next round of competition is equal to the sum of the probabilities of these three independent outcomes.

Football matches are different. It could be just a friendly match, a match of the regular national championship, a match in a group tournament, a two-legged cup playoff match, a single knockout cup match, as a result of which one team must advance and the other must be eliminated. In some matches, such as championship games or group tournaments, the result is recorded in regular time. In knockout matches, there may be options up to extra time and a penalty shootout to determine the final winner. So, on such matches they take a bet not only on the result itself, but also for the team to advance to the next round or the final victory if this is the final. Let's talk about such rates in more detail.

So, a football match of any regular season ends after 90 minutes and several minutes added by the referee. The result of such a match can be a victory for one of the teams or a draw. The winner gets 3 points, the loser gets 0 points. If there is a draw, then both teams receive 1 point. The situation is the same with matches in group tournaments. If the points are equal, no additional games or halves are assigned, but additional indicators are calculated - head-to-head matches, goals, etc. However, there are match formats where a team may not win in regulation time, but will advance further. Let's look at examples.

One-match confrontation. Matches of domestic cup competitions of some countries, final matches of European cups, playoff matches of the World and European Championships, 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 one of the teams wins in such a match, then everything is simple - it moves on, and the loser leaves the tournament. But, in regulation time a draw can be recorded. What then? In some cups, a replay is scheduled on the field of the other team (this is the format in England, for example). In other situations, extra time is assigned - two halves of 15 minutes each. 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 may be recorded in regular time and the bet is calculated based on this draw result. Bets on the final winner, the team that will advance further or receive the cup are accepted separately. This bet on team pass.

Pass bets can be found in the additional line by going inside a specific match, in which the main outcome may not coincide with the pass outcome.

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

...but the essence is the same.

Two-match confrontation. In some domestic cups, in European cups, in the playoffs of selection for the World and European Championships, etc., the playoff format, knockout games, implies a two-game confrontation. One game is at home, the second is away. There may be several options here.

A team can win one match and draw the second. And she passes. This means that if you bet not on the second game, but on the pass, you will win. And the bet on victory will lose, because... there was a draw.

Moreover, a team can win one match and lose the second. And the team that won with a larger difference on 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 advances. If the scores are identical (3:1, 1:3), then extra time is assigned in the second match, as in a situation with a one-game playoff.

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

Teams can play two games in a draw. If both matches in regular time end with the same draw score (0:0, 0:0, or 2:2, 2:2), then extra time is awarded, and then a penalty. So, all bets on team victories in such games are off. But still, some team moves on.

Various draws can be recorded, for example 0:0 and 1:1. Then the team that scored on the road goes like this. And, again, the bet on the passage of the corresponding team plays, and the bets on victories are lost due to draws in regulation time.

A striking example of the results of a two-game confrontation is the quarter-final game of the current Champions League. Real Madrid lost away to Wolfsburg 0:2. And before the return game for Real's pass, the odds were no longer as ridiculous as originally. Still, a defeat by 2 goals and a lack of 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, or even lose, but still pass.

One more example. Seville – Atleti Bilbao. Meetings in the playoffs of the Europa League 2015-2016. Sevilla wins the away match 1:2. So what do you want to bet on the return home game? As a result, Sevilla lost at home with the same score 1:2, ending their long unbeaten home streak. But, at the same time, she advanced 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 based on the result. They can openly play for a draw, but in the end they can lose. So, sometimes, you should give preference to betting on the pass rather than on the main outcome of the match. Or, the bet on the main outcome should be correlated with the real motivation of a particular team for a particular match.

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 bitter struggle, teams can play in a draw in regulation time, and the victory, in the end, will go to the same team, which is the strongest and most experienced.

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Prototype of task B10 (No. 320188) To advance to the next round of competition, a football team needs to score at least 4 points in two games. If a team wins, it receives 3 points, in case of a draw - 1 point, if it loses - 0 points. Find the probability that the team will 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, among them two 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 will number these groups: group 1, group 2 and group 3). Therefore, the possible outcomes of the experiment under consideration are:

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

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

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

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

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

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

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

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

U 9 ​​= (Mikhail in the third group, Vadim in the third group) = (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 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 conditions of the problem, a class of 33 people is randomly divided into three equal groups, each such group will contain 11 students from this class. Purely for the sake of convenience in solving the problem, let’s imagine 33 chairs arranged in one row, with numbers written on the seats: 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 you 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 sits 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 a chair with the same number 1 is equal to . 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 . Reasoning in a similar way, we find the probabilities of 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 desired probability as the quotient of dividing the number of outcomes U 3 , U 5 and U 9 favoring event A by the number of all possible outcomes U 1 , U 2, U 3 ,… U 7, U 9, that is, P(A)=. The fallacy of such a decision lies in the fact that the outcomes of the experiment in question are not equally probable. Indeed, P(U 1)=, and P(U 3)=.

Solution. According to the problem, the team plays two games, and the result of each such game can be either a win, a loss, or a draw. This means that the possible outcomes of this experiment are: U 1 = (B; B), here and further B - the team won the game, P - the team lost the game, H - the team played a draw, U 2 = (B; H), U 3 = (B; P), U 4 = (P; B), U 5 = (P; N), U 6 = (P; P), U 7 = (N; N), U 8 = (N; P), U 8 = (N; V). Thus, the set of all possible outcomes of the experiment under consideration consists of 9 elements, and the event C - “the football team advanced to the next round of competition” 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 advance to the next round of competition. Let's find the probabilities of the outcomes U 1 = (B; B), U 2 = (B; H) and U 8 = (H; B). According to the conditions 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, 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. This means, 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 equal to: P(C)= P(U 1)+ P(U 2)+P(U 8)=0.16+0.08+0.08=0.32.

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

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

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

The probability of a draw is 1 - 0,4 - 0,4 = 0,2 . What is the probability of one time
draw and win once? 0.4 · 0.2? No, it's equal 0.4 0.2 + 0.2 0.4.
The point is that you can win the first game, and you can win the second, this is important.
Now we calculate the probability of reaching the next round: 0,16 + 0,08 + 0,08 = 0,32 .

Answer: 0,32

Let's illustrate the solution graphically using a table 10 x 10 from 100 cells:

Red color indicates victory, swamp color indicates loss, blue color indicates a draw.

Gray cell: the first game is a loss, the second game is a loss.
Red cell: 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 square: the first game is a draw, the second game is a draw.

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

And another one visual diagram. At the first moment the team has
three options for the development of events: victory, draw and loss.

In each case, there are three possible outcomes for the second game.

Let's leave only those branches that suit the team.

Let's calculate the probability of each branch and add them up.