Some puzzles

[biwema]

Key: G: door with goat P: door with price + number of cases


round1: G4 G4 G4 P1 P1
round2: G2 G2 P1 G3 G3
round3: G1 P1 G1 G1 P1
-----------------------------
product: 8 8 8 3 3
result: win loss win win loss

wins: 19
losses: 11

63% win

oops, I made a calculation mistake the post before (i took one factor 2 more in the seond column)

[fetofs]

9-Consider the square ABCD. The AB and BC sides are,respectively, 13 and 14 cm long. Consider the triangle BME, of a way that M=intersection of both diagonals of the rectangle; ME=MB and BE=BA. Prove that EC is parallel to BD.

10-In a city, there is a circular square which only entrance/exit is one road. There are "n" identical houses with the front turned to the center of the square and they are neighbours(consecutive houses) of one another, except from the two ones separated by the road. You got "k" colors of paint (k<n), and you wish to paint the front part of the houses in a way that neighbours don't have repeated colors. In how many ways can we paint them, considering that:
1-The separated houses aren't considered neighbours.
2-The separated houses are considered neighbours.

[fetofs]

9-Consider the square ABCD. The AB and BC sides are,respectively, 13 and 14 cm long. Consider the triangle BME, of a way that M=intersection of both diagonals of the rectangle; ME=MB and BE=BA. Prove that EC is parallel to BD.

10-In a city, there is a circular square which only entrance/exit is one road. There are "n" identical houses with the front turned to the center of the square and they are neighbours(consecutive houses) of one another, except from the two ones separated by the road. You got "k" colors of paint (k<n), and you wish to paint the front part of the houses in a way that neighbours don't have repeated colors. In how many ways can we paint them, considering that:
1-The separated houses aren't considered neighbours.
2-The separated houses are considered neighbours.

[rogue]

The Monty Hall problem is proving to be much more difficult (to me) than it first appeared. I'm certain that probability(x) is a function of probability(x-1), but I don't know what it is. Here are my formulae:

pw(n) = probability of winning with n doors
psc(n) = probability the selected door is the car door = 1/n
pwsc(n) = probability of winning if the selected door (of n doors) is the car door
psg(n) = probability the selected door is a goat door = (n-1)/n
pwsg(n) = probability of winning if the selected door (of n doors) is a goat door

pw(n) = psc(n) * pwsg(n-1) +
psg(n) * (pwsg(n-1) * psg(n-2) + pwsc(n-1) * psc(n-2))

pw(2) = 1/2
pwsg(2) = 0
pwsc(2) = 1

pw(3) = 2/3
pwsg(3) = 1
pwsc(3) = 0

pw(4) = 1/4 * 1 + 3/4 * (1 * 1/2 + 0 * 1/2) = 1/4 + 3/8 = 5/8
pwsg(4) = 1/2
pwsc(4) = 1

pw(5) = 1/5 * 1/2 + 4/5 * (1/2 * 2/3 + 1 * 1/3) = 1/10 * 8/15 = 19/30
pwsg(5) = 2/3
pwsc(5) = 1/2

Assuming this is correct (or close), this appears to be a very interesting series. I suspect that something might be missing, but I need some sleep to clear my head. I look forward to other contributions.

BTW fetofs, do you have a solution for this problem? Where did you find it or did you make it up yourself?

[wblipp]

Problem 5 - Tweaked Monte Hall

Were you wondering why you were permitted to use
the number "e" in the limit?


Let G(n) = Probability you will win if, at the n-door stage
you have selected a goat door.

C(n) = Probability you will win if, at the n-door stage
you have selected a goat door.

A(n) = Probability you will win if you begin the game
with n doors.

G(1) = 0
C(1) = 1

G(n+1) = (n-1)/n * G(n) + 1/n * C(n)

C(n+1) = G(n)

A(n) = (n-1)/n * G(n) + 1/n * C(n)

You can check that
G(n) = [tex]\normalsize\textstyle \sum_{k=0}^n\frac{(-1)^k}{k!}[/tex]

C(n) = [tex]\normalsize\textstyle \sum_{k=0}^{n-1}\frac{(-1)^k}{k!}[/tex]

G, C, and A all converge rapidly to 1/e.

[fetofs]

Originally Posted by wblipp

Problem 5 - Tweaked Monte Hall

Were you wondering why you were permitted to use
the number "e" in the limit?

You can check that
G(n) = [tex]\normalsize\textstyle \sum_{k=0}^n\frac{(-1)^k}{k!}[/tex]

C(n) = [tex]\normalsize\textstyle \sum_{k=0}^{n-1}\frac{(-1)^k}{k!}[/tex]

G, C, and A all converge rapidly to 1/e.

Wblipp's formula works up to 4. We just need to make sure this is the least number of operations needed to solve the problem

P.S: No, I don't have the answer.

P.S 2: Looks like LaTex doesn't work well spoilerized... I had to de-spoilerize it to read

[mfgoode]

Hi Fetofs,
How about my answer for Q1? Do you agree with it or not? Can you derive it?
I keep off 'probablity' so have not joined in the fray but the posts are interesting.
Mally

[fetofs]

Originally Posted by mfgoode

Hi Fetofs,
How about my answer for Q1? Do you agree with it or not? Can you derive it?
I keep off 'probablity' so have not joined in the fray but the posts are interesting.
Mally

Your answer for Q1 is perfect. (although not spoilerized )

[wblipp]

Originally Posted by fetofs

9-Consider the square ABCD. The AB and BC sides are,respectively, 13 and 14 cm long.

I don't understand this. AB and BC sound like adjacent sides of a square, but a square is usually defined to have equal sides.

[fetofs]

Originally Posted by wblipp

I don't understand this. AB and BC sound like adjacent sides of a square, but a square is usually defined to have equal sides.

D'oh! Rectangle!

[Zeta-Flux]

wacky,

That's right. I should have said "point" of symmetry.