one more try. If the expected # of throws is 3.5 to roll a six then the expected number of tries BEFORE rolling a six is 2.5 the expected amount of each trow is 3.5 so:
2.5 x 3.5 is 8.75

ANSWER is 8.75

seems a little high but the extra half throw must be pusihng up the average

Q29:
The expected number of throws is 3 (the average from a large number of sets of 6 throws).

Expected total before throwing a 6 is 7 (that's what I expect, anyway).

F.

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All of you are wrong.... Try again....

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- Luke.

Before rolling a 6, 5 rolls expected.
The expected sum goes up by 3 with each additional roll. (Not proved; observed for 1 through 5 rolls, using for do loops on MAPLE.)
(3 + 6 + 9 + 12 + 15) / 5 = 45 / 5 = 9, the expected sum ... ?

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Capitalize on this good fortune, one word can bring you round ... changes.

My Mistake!
Congratulations to guido.man!!! for correctly solving Q29!
Q29 Answer: 15
Q29 Working:
Let x denote the answer on any given trial of this experiment.
Let E(x) denote the expectation of x, in other words the answer to this problem.
E(x)=(1+E(x))/6 + (2+E(x))/6 + (3+E(x))/6 + (4+E(x))/6 + (5+E(x))/6
E(x)=(15+5*E(x))/6
E(x)=15

Standings:
1. Mr. Kevvy - 7 Points
2. William Rothamel - 6 1/2 Points
3. WinterKnight - 5 Points
4. Sarge - 5 Points
5. Dominique - 4 1/2 Points
6. Fred W - 2 1/2 Points
7. John McLeod VII - 1 Point
8. Labbie - 1 Point
9. Guido.Man - 1 Point
10. Scary Capitalist - 1/2 Point

Q32, 33, 34 coming later tonight...

Best Regards,
Luke.

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- Luke.

Here we go again: Another round of Questions....

Q32 - There is a one acre field in the shape of a right triangle, with sides of length x and y. At the midpoint of each side there is a post. Tethered to the posts on each side is a sheep. Tethered to the post on the hypotenuse is a dog. Each animal has a rope just long enough to reach the two adjacent vertices of the triangle. How much area outside of the field do the sheep have to themselves?

Q33 - Similar to a previous problem I posted... but different still... -
Ten people land on a deserted island. There they find lots of coconuts and a monkey. During their first day they gather coconuts and put them all in a community pile. After working all day they decide to sleep and divide them into two equal piles the next morning. That night one castaway wakes up hungry and decides to take his share early. After dividing up the coconuts he finds he is one coconut short of ten equal piles. He also notices the monkey holding one more coconut. So he tries to take the monkey's coconut to have a total evenly divisible by 10. However when he tries to take it the monkey conks him on the head with it and kills him. Later another castaway wakes up hungry and decides to take his share early. On the way to the coconuts he finds the body of the first castaway, which pleases him because he will now be entitled to 1/9 of the total pile. After dividing them up into nine piles he is again one coconut short and tries to take the monkey's coconut. Again, the monkey conks the man on the head and kills him. One by one each of the remaining castaways goes through the same process, until the 10th person to wake up gets the entire pile for himself. What is the smallest number of possible coconuts in the pile, not counting the monkeys?

Q34 - On a game show there are three doors. Behind one door is a new car and behind the other two are goats. Every time the game is played the contestant first picks a door. Then the host will open one of the other two doors and always reveals a goat. Then the host gives the player the option to switch to the other unopened door. Should the player switch?

Have a go! Best Regards,
Luke.

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- Luke.

Q34. Believe it or not the player should switch. The original probability of any door was 1/3 but now the probability of his original pick is still 1/3 but given that that door was not picked the remaining door has probability of 1/2 so he should definitely switch.

This is hard to see and will cause arguments I am sure. But if you try it with 100 doors you can see it right away. All doors are opened except your pick and one other --just as in Luke's problem. In this case the chance of your original door would be 1/100 and the other would be 99/100

coconut problem:

Easier than the last one. What is the smallest number that 10, 9, 8, 7, etc all divide evenly.

Have to use a spread sheet or look for primes ! This gets too big maybe need a BASIC computer program

I see that there is a tool for this: Says that the answer is 2520

so the answer to the coconut problem is just 2520-1 or 2519

Dog and Goat. Doesn't appear to be enough info for a numerical solution. So it sort of looks like the dog would always have enough rope to cover the goats' semi circle regardless of the aspect ratio of the triangle. It's too late to try to prove this so I will just guess:

The goats have no area left to themselves.

Dog and Sheep (I think Bill was in too much of a hurry to separate his Sheep from his Goats ;)

The area the Sheep will have for peaceful grazing will be twice the product of the lengths of their ropes.

If the sides of the right triangle are 2.a, 2.b and 2.c where 2.c is the hypotenuse then the area the sheep will have outside the field plus the area of the field is given by:

pi*a^2/2 + pi*b^2/2 + 2*a*2*b/2

The area the dog can cover on the sheeps' side of the hypotenuse is given by:

pi*c^2/2

Since it is a right triangle (2*a)^2 + (2*b)^2 = (2*c)^2 OR a^2 + b^2 = c^2

So the dog's area can be written as

pi*a^2/2 + pi*b^2/2

Subtracting this last from the first leaves 2*a*b

F.

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Dog and Sheep (I think Bill was in too much of a hurry to separate his Sheep from his Goats ;)

The area the Sheep will have for peaceful grazing will be twice the product of the lengths of their ropes.

If the sides of the right triangle are 2.a, 2.b and 2.c where 2.c is the hypotenuse then the area the sheep will have outside the field plus the area of the field is given by:

pi*a^2/2 + pi*b^2/2 + 2*a*2*b/2

The area the dog can cover on the sheeps' side of the hypotenuse is given by:

pi*c^2/2

Since it is a right triangle (2*a)^2 + (2*b)^2 = (2*c)^2 OR a^2 + b^2 = c^2

So the dog's area can be written as

pi*a^2/2 + pi*b^2/2




Subtracting this last from the first leaves 2*a*b

F.

Did you account for the displacement of the centers of the two circles (each sheep and the dog) ??

Dog and Sheep (I think Bill was in too much of a hurry to separate his Sheep from his Goats ;)

The area the Sheep will have for peaceful grazing will be twice the product of the lengths of their ropes.

If the sides of the right triangle are 2.a, 2.b and 2.c where 2.c is the hypotenuse then the area the sheep will have outside the field plus the area of the field is given by:

pi*a^2/2 + pi*b^2/2 + 2*a*2*b/2

The area the dog can cover on the sheeps' side of the hypotenuse is given by:

pi*c^2/2

Since it is a right triangle (2*a)^2 + (2*b)^2 = (2*c)^2 OR a^2 + b^2 = c^2

So the dog's area can be written as

pi*a^2/2 + pi*b^2/2




Subtracting this last from the first leaves 2*a*b

F.

Did you account for the displacement of the centers of the two circles (each sheep and the dog) ??

Yes. Reference this picture:


Direct Link

The area (outside the field) available to sheep (a) is the semi-circle on the left = pi*a^2/2

The area (outside the field) available to sheep (b) is the semi-circle below = pi*b^2/2.

Add these two areas to the area of the field (= (2*a)*(2*b)/2) and we have an area that encompasses the area that can be covered by the dog (the red area).

Subtract the red area (pi*c^2/2) and we have the area of succulent meadow available for peaceful grazing.

F.

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Did I inspire the inclusion of the pictures?
What program did you use to make that, Fred?

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Capitalize on this good fortune, one word can bring you round ... changes.

Did I inspire the inclusion of the pictures?
What program did you use to make that, Fred?

Would you believe MS Word copy/pasted into Paint where I added the numbers and the infill colours.

F.

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Did I inspire the inclusion of the pictures?
What program did you use to make that, Fred?

Would you believe MS Word copy/pasted into Paint where I added the numbers and the infill colours.

F.

Cool Fred !

How did you generate the BB code ?? Did you start with a JPEG

Did I inspire the inclusion of the pictures?
What program did you use to make that, Fred?

Would you believe MS Word copy/pasted into Paint where I added the numbers and the infill colours.

F.

Cool Fred !

How did you generate the BB code ?? Did you start with a JPEG

From Paint, I saved as a .png. Uploaded that to ImageShack (free hosting service). When you upload images to ImageShack, they provide the links for Message Boards, Direct Links, etc.

F.

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Bill wins yet again!!! :-)

Correct Answer to Q33: 2519
Correct Answer to Q34: Yes
2 more points...

Standings:
1. William Rothamel - 8 1/2 Points
2. Mr. Kevvy - 7 Points
3. WinterKnight - 5 Points
4. Sarge - 5 Points
5. Dominique - 4 1/2 Points
6. Fred W - 2 1/2 Points
7. John McLeod VII - 1 Point
8. Labbie - 1 Point
9. Guido.Man - 1 Point
10. Scary Capitalist - 1/2 Point
11. TBD...

Congratulations, Bill is now the 3rd Problem Solving Champion!
Also - Sorry, Bill, Fred, etc. you all have Q32 wrong. I'm looking for a answer in acres please...

Q32 - There is a one acre field in the shape of a right triangle, with sides of length x and y. At the midpoint of each side there is a post. Tethered to the posts on each side is a sheep. Tethered to the post on the hypotenuse is a dog. Each animal has a rope just long enough to reach the two adjacent vertices of the triangle. How much area outside of the field do the sheep have to themselves?

Best Regards,
Luke.

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- Luke.

Bill wins yet again!!! :-)

Correct Answer to Q33: 2519
Correct Answer to Q34: Yes
2 more points...

Standings:
1. William Rothamel - 8 1/2 Points
2. Mr. Kevvy - 7 Points
3. WinterKnight - 5 Points
4. Sarge - 5 Points
5. Dominique - 4 1/2 Points
6. Fred W - 2 1/2 Points
7. John McLeod VII - 1 Point
8. Labbie - 1 Point
9. Guido.Man - 1 Point
10. Scary Capitalist - 1/2 Point
11. TBD...

Congratulations, Bill is now the 3rd Problem Solving Champion!
Also - Sorry, Bill, Fred, etc. you all have Q32 wrong. I'm looking for a answer in acres please...

Q32 - There is a one acre field in the shape of a right triangle, with sides of length x and y. At the midpoint of each side there is a post. Tethered to the posts on each side is a sheep. Tethered to the post on the hypotenuse is a dog. Each animal has a rope just long enough to reach the two adjacent vertices of the triangle. How much area outside of the field do the sheep have to themselves?

Best Regards,
Luke.

Message to self: "Read the b******* question!!"

1 acre.

F.

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What do you mean by "outside of the field"? Is the right triangle not completely within the 1 acre field?

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Capitalize on this good fortune, one word can bring you round ... changes.

What do you mean by "outside of the field"? Is the right triangle not completely within the 1 acre field?

Yes, but the rope can twist all the way round the post (360 deg) and the hedge isn't very good. See my picture for how I understood the problem.

F.

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