Query 10

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course Mth 152

6/30

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?

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Your solution:

1/6 chance for $3 = 1/6 * $3 = $.50

1/6 chance for $2 = 1/6 * $2 = $.33 1/3

1/6 chance for $1 = 1/6 * 1 = $.16 2/3

You could expect $1 * 1/6 + $2 * 1/6 + $3 * 1/6 = 6/6 = $1 fair price

confidence rating #$&*::2

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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?

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Your solution:

Expected value formula says

E = $1(18/37) + (-$1)(19/37)

E = -$.02702

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?

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Your solution:

With two picks, there is no sum of 1

With two picks, there is no sum of 2

With two picks, there are 2 ways to get sum of 3

With two picks, there are 2 ways to get a sum of 4

With two picks, there are 4 ways to get a sum of 5

With two picks, there are 4 ways to get a sum of 6

With two picks, there are 4 ways to get a sum of 7

With two picks, there are 2 ways to get a sum of 8

With two picks, there are 2 ways to get a sum of 9

9 is the largest combination sum.

There are 20 possibilities all together.

(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

Self-critique Rating:

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

This section of the chapter allowed for some practical knowledge in the use of probability.

Self-critique (if necessary):

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Self-critique rating: