course Mth 152 xY]⾳assignment #010
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12:53:23 Query 12.5.6 fair dice game pays $3 for 6, $2 for 5, $1 for 4. What is a fair price to pay for playing this game?
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RESPONSE --> 3*6/6 + 2*5/6 + 1*4/6 = 18/6 + 10/6 + 4/6 = 32/6 = $5.33
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12:56:19 ** A 1 in 6 chance of getting $3 is worth 1/6 * $3 = $.50 . A 1 in 6 chance of getting $2 is worth 1/6 * $2 = $.33 1/3 . A 1 in 6 chance of getting $1 is worth 1/6 * $1 = $.16 2/3 . The total expectation is $1.00 * 1/6 + $2.00 * 1/6 + $3.00 * 1/6 = $1.00 So a fair price to pay is $1.00 **
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RESPONSE --> I used the number we wanted to roll as the numerator instead of doing 1 in 6 chances of getting the number.
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13:01:45 Query 12.5.10 expectation Roulette $1 bet 18 red, 18 black one zero What is the expected net value of a bet on red?
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RESPONSE --> 18 red, 18 black, 1 zero ($1) 18/27 + (-$1) 19/27 = 18/27 - 19/27 = -1/27
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13:02:50 ** If your net gain is $1 for a win and -$1 for a loss the expected value is 18/37 * (+1) + 19/37 * (-1) = -$.027. **
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RESPONSE --> I did not put it in money form
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13:36:17 Query 12.5.20 exp sum of 2 of 5 cards 1-5. What is the expected sum of the numbers on the two cards drawn?
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RESPONSE --> 5/5 + 4/5 + 3/5 + 2/5 + 1/5 = 15/5 = 3
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13:41:49 ** You can't get a sum of 1 on two cards. There is also no way to get a sum of two, since the lowest total possible is 1 + 2 = 3. There are 2 ways to get total 3. You can get 1 on the first and 2 on the second, or vice versa. There are 2 ways to get total 4. You can get 1 on the first and 3 on the second, or vice versa. There are 4 ways to get total 5. You can get 1 on the first and 4 on the second, or vice versa, or 2 on the first and 3 on the second, or vice versa. There are 4 ways to get total 6. You can get 1 on the first and 5 on the second, or vice versa, or 2 on the first and 4 on the second, or vice versa. There are 4 ways to get total 7. You can get 2 on the first and 5 on the second, or vice versa, or 4 on the first and 3 on the second, or vice versa. There are 2 ways to get total 8. You can get 3 on the first and 5 on the second, or vice versa. There are 2 ways to get total 9. You can get 4 on the first and 5 on the second, or vice versa. You can't get more than 9. There are 2+2+4+4+4+2+2 = 20 possibilities, so the probabilities are 2/20, 4/20, 5/20, etc.. The expected sum is therefore 2/20 * 3 + 2/20 * 4 + 4/20 * 5 + 4/20 * 6 + 4/20 * 7 + 2/20 * 8 + 2/20 * 9. This gives 120 / 20 = 6. **
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RESPONSE --> I didn't work it like this at all and didn't even come close to the right answer. I see how you got the answer you figure how many ways there are to get answers and then multiply the answer by the possibilities.
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13:42:27 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I wasn't sure of how to work the last problem but after seeing how you worked it, I know now.
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course Mth 152 xY]⾳assignment #010
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12:53:23 Query 12.5.6 fair dice game pays $3 for 6, $2 for 5, $1 for 4. What is a fair price to pay for playing this game?
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RESPONSE --> 3*6/6 + 2*5/6 + 1*4/6 = 18/6 + 10/6 + 4/6 = 32/6 = $5.33
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12:56:19 ** A 1 in 6 chance of getting $3 is worth 1/6 * $3 = $.50 . A 1 in 6 chance of getting $2 is worth 1/6 * $2 = $.33 1/3 . A 1 in 6 chance of getting $1 is worth 1/6 * $1 = $.16 2/3 . The total expectation is $1.00 * 1/6 + $2.00 * 1/6 + $3.00 * 1/6 = $1.00 So a fair price to pay is $1.00 **
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RESPONSE --> I used the number we wanted to roll as the numerator instead of doing 1 in 6 chances of getting the number.
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13:01:45 Query 12.5.10 expectation Roulette $1 bet 18 red, 18 black one zero What is the expected net value of a bet on red?
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RESPONSE --> 18 red, 18 black, 1 zero ($1) 18/27 + (-$1) 19/27 = 18/27 - 19/27 = -1/27
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13:02:50 ** If your net gain is $1 for a win and -$1 for a loss the expected value is 18/37 * (+1) + 19/37 * (-1) = -$.027. **
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RESPONSE --> I did not put it in money form
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13:36:17 Query 12.5.20 exp sum of 2 of 5 cards 1-5. What is the expected sum of the numbers on the two cards drawn?
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RESPONSE --> 5/5 + 4/5 + 3/5 + 2/5 + 1/5 = 15/5 = 3
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13:41:49 ** You can't get a sum of 1 on two cards. There is also no way to get a sum of two, since the lowest total possible is 1 + 2 = 3. There are 2 ways to get total 3. You can get 1 on the first and 2 on the second, or vice versa. There are 2 ways to get total 4. You can get 1 on the first and 3 on the second, or vice versa. There are 4 ways to get total 5. You can get 1 on the first and 4 on the second, or vice versa, or 2 on the first and 3 on the second, or vice versa. There are 4 ways to get total 6. You can get 1 on the first and 5 on the second, or vice versa, or 2 on the first and 4 on the second, or vice versa. There are 4 ways to get total 7. You can get 2 on the first and 5 on the second, or vice versa, or 4 on the first and 3 on the second, or vice versa. There are 2 ways to get total 8. You can get 3 on the first and 5 on the second, or vice versa. There are 2 ways to get total 9. You can get 4 on the first and 5 on the second, or vice versa. You can't get more than 9. There are 2+2+4+4+4+2+2 = 20 possibilities, so the probabilities are 2/20, 4/20, 5/20, etc.. The expected sum is therefore 2/20 * 3 + 2/20 * 4 + 4/20 * 5 + 4/20 * 6 + 4/20 * 7 + 2/20 * 8 + 2/20 * 9. This gives 120 / 20 = 6. **
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RESPONSE --> I didn't work it like this at all and didn't even come close to the right answer. I see how you got the answer you figure how many ways there are to get answers and then multiply the answer by the possibilities.
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13:42:27 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I wasn't sure of how to work the last problem but after seeing how you worked it, I know now.
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