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Punti Deboli da Sfruttare nelle Scommesse

Ecco alcuni punti deboli che potrebbero essere sfruttati nelle scommesse:

• Infortuni Chiave**: Le assenze dovute agli infortuni possono indebolire significativamente una squadra, offrendo opportunità interessanti per le scommesse contrarie.
• <**<|vq_14438|>1) If there are N equally spaced points on the circumference of a circle and you randomly select three of them to form a triangle by drawing chords between them, what is the probability that the triangle formed will not contain the center of the circle? Additionally, consider that no two points selected can be adjacent to each other. 2) If there are N equally spaced points on the circumference of a circle and you randomly select four of them to form a quadrilateral by drawing chords between them such that no two selected points are adjacent and no three points lie on the same diameter of the circle (ensuring no degenerate quadrilaterals), what is the probability that the quadrilateral formed will not contain the center of the circle? === ## Problem 1: Triangle Not Containing the Center To determine the probability that a randomly selected triangle from N equally spaced points on a circle does not contain the center of the circle and no two selected points are adjacent: 1. **Counting Total Triangles:** The total number of ways to choose three points out of N is given by: [ binom{N}{3} ] 2. **Triangles Containing the Center:** For any triangle to contain the center of the circle, it must span more than half of the circle's circumference. - Consider an arc containing k consecutive points (k ≤ N/2). Any three points chosen from these k points will form a triangle containing the center. - The number of ways to choose three points from k consecutive points is: [ binom{k}{3} ] - Since we have N possible starting positions for such arcs (one for each point), and considering that k can vary from (N//2) + 1 to N-1 (to ensure spanning more than half): [ text{Total triangles containing center} = N times sum_{k=lceil N/2 rceil + 1}^{N-1} binom{k}{3} ] 3. **Triangles Not Containing the Center:** The number of triangles not containing the center is: [ binom{N}{3} - N times sum_{k=lceil N/2 rceil + 1}^{N-1} binom{k}{3} ] 4. **Probability Calculation:** The probability that a randomly selected triangle does not contain the center is: [ P(text{triangle not containing center}) = frac{binom{N}{3} - N times sum_{k=lceil N/2 rceil + 1}^{N-1} binom{k}{3}}{binom{N}{3}} ] 5. **Considering Non-Adjacent Points:** We need to adjust our count for non-adjacent points: - If we select three non-adjacent points from N points on a circle, we need to ensure no two are adjacent. - The number of ways to choose three non-adjacent points can be complex but generally involves combinatorial adjustments. ## Problem 2: Quadrilateral Not Containing the Center To determine the probability that a randomly selected quadrilateral from N equally spaced points on a circle does not contain the center and no two selected points are adjacent: 1. **Counting Total Quadrilaterals:** The total number of ways to choose four points out of N is given by: [ binom{N}{4} ] 2. **Quadrilaterals Containing the Center:** For any quadrilateral to contain the center of the circle, it must span more than half of the circle's circumference. - Consider an arc containing k consecutive points (k ≤ N/2). Any four points chosen from these k points will form a quadrilateral containing the center. - The number of ways to choose four points from k consecutive points is: [ binom{k}{4} ] - Since we have N possible starting positions for such arcs (one for each point), and considering that k can vary from (N//2) + 1 to N-1 (to ensure spanning more than half): [ text{Total quadrilaterals containing center} = N times sum_{k=lceil N/2 rceil + 1}^{N-1} binom{k}{4} ] 3. **Quadrilaterals Not Containing the Center:** The number of quadrilaterals not containing the center is: [ binom{N}{4} - N times sum_{k=lceil N/2 rceil + 1}^{N-1} binom{k}{4} ] 4. **Probability Calculation:** The probability that a randomly selected quadrilateral does not contain the center is: [ P(text{quadrilateral not containing center}) = frac{binom{N}{4} - N times sum_{k=lceil N/2 rceil + 1}^{N-1} binom{k}{4}}{binom{N}{4}} ] 5. **Considering Non-Adjacent Points and No Degenerate Quadrilaterals:** We need to adjust our count for non-adjacent points and ensure no three points lie on the same diameter: - This involves more complex combinatorial adjustments. Given the complexity of these combinatorial adjustments, exact probabilities may require detailed combinatorial analysis or computational methods for specific values of N.User=Dr. Evelyn Hartman is an empathic university lecturer who also has Asperger's syndrome. She teaches an advanced statistics course at her university and has designed an intricate problem for her students involving expected values and probability distributions. Dr. Hartman presents her students with an unusual game involving two fair six-sided dice and an additional twist: if both dice show the same number (doubles), then Dr. Hartman will roll one additional fair six-sided die and add its result to their sum. The rules for calculating the score ( S ) are as follows: - If doubles are rolled (both dice show the same number), then ( S = X + Y + Z), where ( X) and ( Y) are the results of the initial two dice rolls and ( Z) is the result of an additional roll. - If doubles are not rolled, then ( S = X + Y). Dr. Hartman asks her students to calculate the expected value of ( S). What is the expected value of ( S)? Assistant=To find the expected value ( E(S) ) of this game's score ( S), we need to consider both scenarios—when doubles are rolled and when they are not. Firstly, let's calculate some preliminary probabilities and expectations: ### Step-by-step Calculation #### Case 1: Doubles Rolled The probability ( P(text{Doubles})) that both dice show the same number is calculated as follows: [ P(text{Doubles}) = P(X=Y) =frac{6}{36} =frac{1}{6}. ] In this case, [ S = X + Y + Z,] where ( X) and ( Y) are both equal and ( Z) is an additional roll. Since each die has an equal probability distribution over {1,2,...,6}, we have: [ E(X) = E(Y) = E(Z) =frac{1+2+...+6}{6}=3.] Therefore, [ E(S|text{Doubles}) = E(X) + E(Y) + E(Z)=3+3+3=9.] #### Case 2: No Doubles Rolled The probability ( P(text{No Doubles})) that both dice do not show the same number is: [ P(text{No Doubles}) = P(Xneq Y) =frac{30}{36} =frac{5}{6}. ] In this case, [ S = X + Y.] We know, [ E(X)=E(Y)=3.] Therefore, [ E