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A friend offers you a game in which you roll two 8-sided die.You get $10 if the sum of the die is 12. 48, 84, 57, 75, 66You get $6 if the sum of the die is 10. 28, 82, 37, 73, 46, 64, 55You get $3 if the sum of the die is 8. 17, 71, 26, 62, 35, 53, 44You get $1 if the sum of the die is 4. 13, 31, 22Otherwise, you pay your friend $2.Is it a fair game?

Question

A friend offers you a game in which you roll two 8-sided die.You get 10ifthesumofthedieis12.48,84,57,75,66Youget10 if the sum of the die is 12. 48, 84, 57, 75, 66You get 6 if the sum of the die is 10. 28, 82, 37, 73, 46, 64, 55You get 3ifthesumofthedieis8.17,71,26,62,35,53,44Youget3 if the sum of the die is 8. 17, 71, 26, 62, 35, 53, 44You get 1 if the sum of the die is 4. 13, 31, 22Otherwise, you pay your friend $2.Is it a fair game?

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Solution

To determine if the game is fair, we need to calculate the expected value of the game. The expected value is the average amount of money you would expect to win (or lose) per game if you played the game many times.

First, let's calculate the probability of each outcome:

  • The sum of the dice is 12: There are 5 outcomes that result in a sum of 12 (48, 84, 57, 75, 66). Since there are 64 total outcomes (8 sides on the first die times 8 sides on the second die), the probability of this outcome is 5/64.

  • The sum of the dice is 10: There are 7 outcomes that result in a sum of 10 (28, 82, 37, 73, 46, 64, 55). So, the probability of this outcome is 7/64.

  • The sum of the dice is 8: There are 7 outcomes that result in a sum of 8 (17, 71, 26, 62, 35, 53, 44). So, the probability of this outcome is 7/64.

  • The sum of the dice is 4: There are 3 outcomes that result in a sum of 4 (13, 31, 22). So, the probability of this outcome is 3/64.

  • The sum of the dice is anything else: There are 42 other outcomes (64 total outcomes minus 5 outcomes for a sum of 12, minus 7 outcomes for a sum of 10, minus 7 outcomes for a sum of 8, minus 3 outcomes for a sum of 4). So, the probability of this outcome is 42/64.

Next, let's calculate the expected value:

  • If the sum is 12, you win 10.So,theexpectedvaluefromthisoutcomeis10. So, the expected value from this outcome is 10 times the probability of this outcome, or 10(5/64)=10 * (5/64) = 0.78125.

  • If the sum is 10, you win 6.So,theexpectedvaluefromthisoutcomeis6. So, the expected value from this outcome is 6 * (7/64) = $0.65625.

  • If the sum is 8, you win 3.So,theexpectedvaluefromthisoutcomeis3. So, the expected value from this outcome is 3 * (7/64) = $0.328125.

  • If the sum is 4, you win 1.So,theexpectedvaluefromthisoutcomeis1. So, the expected value from this outcome is 1 * (3/64) = $0.046875.

  • If the sum is anything else, you lose 2.So,theexpectedvaluefromthisoutcomeis2. So, the expected value from this outcome is -2 * (42/64) = -$1.3125.

Finally, add up all the expected values to get the total expected value of the game: 0.78125+0.78125 + 0.65625 + 0.328125+0.328125 + 0.046875 - 1.3125=1.3125 = 0.50.

Since the expected value is positive, you would expect to win money on average if you played this game many times. Therefore, the game is not fair to your friend, but it is favorable to you.

This problem has been solved

Similar Questions

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Two players, A and B.A goes first. They roll a dice that has random outputs between 1 and 30. The player who gets the higher number wins, and the loser pays the winner the amount that the winner gets on his dice.1. What is the expected winnings?2. Would you prefer being Player A or Player B? Why?

Alice and Bob are playing a board game.On their turn, a player must roll 𝑁N standard 66-sided dice, and their action will be determined by the sum of values on the top faces of the 𝑁N dice.On a certain turn of Alice, she rolls the dice and obtains the sequence of values 𝐴1,𝐴2,…,𝐴𝑁A 1​ ,A 2​ ,…,A N​ on the top faces.However, Bob isn't paying attention, allowing Alice to cheat a little!Alice can choose a die, and flip it - so the opposite face is upward.Note that these are standard 66-sided dice, so the sum of values on opposite faces is 77. That is:11 is opposite to 66.22 is opposite to 55.33 is opposite to 44.So as to not make Bob suspicious, Alice can perform this flipping operation at most 𝐾K times.What's the maximum score (i.e, sum of values of top faces of the dice) she can obtain?Input FormatThe first line of input will contain a single integer 𝑇T, denoting the number of test cases.Each test case consists of two lines of input.The first line of each test case contains two space-separated integers 𝑁N and 𝐾K — the number of dice and the number of times Alice can flip dice.The second line contains 𝑁N space-separated integers 𝐴1,𝐴2,…,𝐴𝑁A 1​ ,A 2​ ,…,A N​ — the initial values of the dice.Output FormatFor each test case, output on a new line the maximum score Alice can obtain if she flips a die at most 𝐾K times.Constraints1≤𝑇≤1051≤T≤10 5 1≤𝑁≤2⋅1051≤N≤2⋅10 5 1≤𝐾≤𝑁1≤K≤N1≤𝐴𝑖≤61≤A i​ ≤6The sum of 𝑁N over all test cases won't exceed 2⋅1052⋅10 5 .Sample 1:InputOutput42 13 44 23 4 4 55 31 2 3 2 16 26 5 4 3 2 1

Sure! The text you provided is actually a list of numbers. Each number represents a sum of two dice rolls. Let me explain it to you. When you roll a pair of dice, there are different combinations of numbers that can come up. For example, if you roll a 6-sided die, you can get numbers from 1 to 6 on each die. When you add the numbers on both dice together, you get a sum. In the text you provided, the first number is 8 and it is followed by 5/36. This means that when you roll two dice, the sum of the numbers on the dice will be 8, and this outcome has a probability of 5 out of 36. Similarly, the second number is 9 and it is followed by 4/36. This means that the sum of the numbers on the dice will be 9, and this outcome has a probability of 4 out of 36. The pattern continues for the next numbers. The third number is 10 with a probability of 3 out of 36, the fourth number is 11 with a probability of 2 out of 36, and the fifth number is 12 with a probability of 1 out of 36. To understand the probabilities better, let's imagine rolling two dice many times. If we count the number of times we get a sum of 8, we would expect it to happen about 5 out of every 36 rolls. Similarly, for a sum of 9, we would expect it to happen about 4 out of every 36 rolls, and so on. So, the numbers and fractions in the text represent the different sums you can get when rolling two dice, along with the probability of each sum occurring.

An ordinary (fair) die is a cube with the numbers 1 through 6 on the sides (represented by painted spots). Imagine that such a die is rolled twice in succession and that the face values of the two rolls are added together. This sum is recorded as the outcome of a single trial of a random experiment.Compute the probability of each of the following events.Event A: The sum is greater than 8.Event B: The sum is divisible by 4 or 6 (or both).Write your answers as fractions.(a) PA = (b) PB =

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