chap1210

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course Mth 152

11/11/2012 5:04PM

010.   Query 10 

 

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question: 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?

 

 

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: 

There is 1/6 chance of getting a 6, 5, or a 4, so 1/6 * $3 + 1/6 * $2 + 1/6 * $1 = $.50 + $.33 + .17 = $1.00 is a fair price to pay.

 

 

confidence rating #$&*: 3

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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 OK

 

 

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Self-critique Rating: OK

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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: 

There are 18 winning compartments, and 19 losing compartments, and if you win you get $1 and if you lose you get -$1. So 18/37 * $1 + 19/37 * (-1) = .49 + (-.51) = -$.02 is the expected net value of a bet on red.

 

 

confidence rating #$&*: 3

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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 OK

 

 

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Self-critique Rating: OK

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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: 

There arent anyways to get 1 or 2. To get 3 I can either draw 1 and 2, or 2 and 1. To get 4 I can either draw 1 and 3, or 3 and 1. To get 5 I can either draw 1 and 4, 4 and 1, 2 and 3, or 3 and 2. To get 6 I can either draw 1 and 5, 5 and 1, 2 and 4, or 4 and 2. To get 7 I can either draw 2 and 5, 5 and 2, 3 and 4, or 4 and 3. To get 8 I can either draw 3 and 5, or 5 and 3. To get 9 I can either draw 4 and 5, or 5 and 4. So there are 20 possibilities, so the probability of getting a 3, 4, 8, or 9 are 2/20, and the probability of get a 5, 6, or 7 is 4/20. So the expected sum is 2/20 * 3 + 2/20 * 4 + 4/20 * 5 + 4/20 * 6 + 4/20 * 7 + 2/20 * 8 + 2/20 * 9 = 120/20 = 6.

 

 

confidence rating #$&*: 3

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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 OK

 

 

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Self-critique Rating: Ok

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question: Query   Add comments on any surprises or insights you experienced as a result of this assignment.

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question: Query   Add comments on any surprises or insights you experienced as a result of this assignment.

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&#Your work looks very good. Let me know if you have any questions. &#