#$&* course MTH 152 Time: 5:12 PMDate: 7/22 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: `aA 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 ** Self-critique: - So I take the total expectation and multiply by its respective chance of winning and use the sum of the options to show the total “fair” price. - My initial reaction was to show the relationship between probability of the given numbers with respect to the set {1,2,3,4,5,6} ------------------------------------------------ Self-critique Rating: Ok. ********************************************* question: 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - $1.00 * 18/37 + (-$1.00) * 19/37 = 2.7 cents confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aIf your net gain is $1 for a win and -$1 for a loss the expected value is 18/37 * (+1) + 19/37 * (-1) = -$.027. ** Self-critique: I referred to the book’s example, being Example 5, on p. 765 “One simple type of roulette is played with an ivory ball and a wheel set in motion. The wheel contains thirty-eight compartments. Eighteen of the compartments are black, eighteen are red, one is labeled ‘zero,’ and one is labeled ‘double zero.’ (These last two being neither black nor red.) In this case, assume the player places $1 on either red or black. If the player picks the correct color of the compartment in which the ball finally lands, the payoff is $2; otherwise the payoff is zero. Find the expected net winnings.” In which it goes on to use the Expect Value Formula (hereafter known as EVF) E(net winnings) = ($1) 18/38 + (- $1) 20/38 = $1/19 or 2.7 cents. But the given solution is for the expected net loss compared to the expected net value of a bet on red. Having used the same formula (and extrapolated the method of adding a negative monetary value multiplied by the probability of either black or red plus the two 0 and 00 compartments) I see that my solution was indeed correct. Since I used this example in comparison with 12.5.10, would the solutions work regardless of what is asked for (being the expected net value or the expected net loss)? &&& Given the nature of x in the EVF, is it flexible to describe whatever the problem requires? I.e. can it mean multiple things with regards to net winnings/loss/value of a particular color? &&& ------------------------------------------------ Self-critique Rating: Okay. See my comment above and/or refer to p.765, Example 5. Because I am unsure we have the same text, I provided the referenced example. ********************************************* question: Query 12.5.20 exp sum of 2 of 5 cards 1-5. What is the expected sum of the numbers on the two cards drawn? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - My initial reaction was to make a permutation of P(5,2) = 10, but I see now that there are twice as many possibilities = 20 - The lowest possible sum and the highest possible sum are (1,2) = 3 and (4,5) = 9, given the constraint of non-repeating combinations. So there are (3, 4, 5, 6, 7, 8, 9) = 7 possible sums given. - 2 * 2/20 + 3 * 2/20 + 4 * 2/20 + 5* 2/20 + 6* 2/20 + 7 * 2/20 + 8* 2/20 + 9 * 2/20 ****** I see my error. 2 isn’t part of the program since it wasn’t considered as one of the 7 possible sums. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aYou 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. ** Self-critique In your solution “2/20 * 3 + 2/20 * 4 + 4/20 * 5 + 4/20 * 6 + 4/20 * 7 + 2/20 * 8 + 2/20 * 9.” Why does it begin with 2/20 rather than 3 + 2/20?