The worlds first riddle!

[size=8]CEREAL GIFT
8 1/3 (I used the nifty state machine algebra you demonstrated several pages back)[/size]

I put out the questions; I give out the means to find the answer.
I must be as mad as a sack of polecats.


The Greek philosopher Pythagoras (c. 582-c. 507 B.C.) believed all relationships could be explained with numbers, so it's no wonder that a geometric theorem was named after him. The Pythagorean theorem (a^² + b^² = c^²) explains that in a right triangle, the sum of the squares of two sides equals the square of the hypotenuse (the longest side).

Two Olympic athletes want to race each other on foot. Mark runs due west for 6 kilometers and stops, and Rap runs due north for 9 kilometers and collapses.

Using the Pythagorean theorem why don't you, what is the shortest distance between them


How much further would each have to run to meet in the exact middle


You may assume Rap has by this time got his second wind.

[size=8]PYTHAGORAS
sqrt(117) km
sqrt(117)/2 km[/size]

Pennies

Count out 128 and flip them over

The number of heads in the selected group of 128 will equal the number of heads in the remaining group.

Rap

Cereal Boxes

[size=7]Using trial and error and an Excel spreadsheet, I wager it is somewhere near 8 boxes. I'm not sure how to come up with an analytical solution[/size]

Pythagorean race

[size=7]6^2+9^2=117
distance between the two is Sqrt(117) km half that is sqrt(117)/2[/size]

Rap

Footrace addendum

[size=7]Middle of the race or middle of the stopping distance?

If it's the middle of the race Rap needs to run an additonal [sqrt(117)/2-0.5] km and mark needs to run an additional [sqrt(117)/2+2.5] km

If it's the middle os the stopping distance see my previous post.[/size]

Rap

Mark:

PENNIES
Separate 128 pennies from the large group and turn them all over.

Rap:

Count out 128 and flip them over

The number of heads in the selected group of 128 will equal the number of heads in the remaining group.


(They will, I promise you. Try it).


Mark:

CEREAL GIFT
8 1/3 (I used the nifty state machine algebra you demonstrated several pages back) :wink:

Rap:

I wager it is somewhere near 8 boxes.

(If you're going to guess, then guess way out. An accurate guess is frightening)



First, if the probability of an event happening is p then the mean number of trials to obtain a success is 1/p.




Since there must eventually be a success the sum of probabilities is:
p + pq + pq^2 + pq^3 + ... = 1.

The mean number of trials (m) is: m = p + 2pq + 3pq^2 + 4pq^3 + ...
mq = pq + 2pq^2 + 3pq^3 + ...
m-qm= p + pq + pq^^3 + ...
m(1-q) = 1.
m = 1/(1-q) = 1/p.

Second, the answer to the problem can be express as the sum of the following:
• Number of trials to get a first toy
• Number of trials to get a second toy once you have one toy
• Number of trials to get a third toy once you have two toys
• Number of trials to get the final toy once you have three toys

The number of trials to get one toy is obviously one.
Once you have one the probability of getting a different toy in the next box is 3/4, thus the expected number of trials is 1/(3/4) = 4/3 to get the second toy.

By the same logic the number of trials to get the third is 1/(1/2)=2 and 1/(1/4)=4 for the final toy. Summing these yields 1 + 4/3 + 2 + 4 = 25/3.


Mark:

PYTHAGORAS
sqrt(117) km
sqrt(117)/2 km

Rap:
6^2+9^2=117
distance between the two is Sqrt(117) km half that is sqrt(117)/2

(Wait, there is more)

Footrace addendum (I don't know what this means, but he add's):

Middle of the race or middle of the stopping distance?

If it's the middle of the race Rap needs to run an additonal [sqrt(117)/2-0.5] km and mark needs to run an additional [sqrt(117)/2+2.5] km

(Who could disagree with that impeccable reasoning?) *






* Except for the fact I came up with the wrong answer!


Suppose there are n people in an office. In the New Year, they have a random gift exchange in which every name is written on scraps of paper, mixed around in a hat, and then everyone draws a name at random to determine whom they are to get a gift for... (Apologies for end of sentence preposition)

What is the probability nobody draws his or her own name

(What are yo'all waiting for? That's it, no numbers, no other information)
Question adapted from: Number combinations for 12 year olds.

[size=8]GIFT EXCHANGE
Take your pick:

a(0) = 1, a(n) = [ n!/e + 1/2 ] for n > 0

a(n) = n!*Sum((-1)^k/k!, k=0..n)

a(n) = (n-1)*(a(n-1)+a(n-2)), n>0

a(n) = n*a(n-1)+(-1)^n[/size]

Aaaaaaaaagh! A pre 9am answer! Why do you torment me so? And with four answers, surprisingly none of which is the one I have. However, all perfectly valid I am sure.


One, I confidently predict you will not be able to answer (before 9am).

You choose one of two identical looking bags at random. One bag has three black marbles and one white marble. The other has three white marbles and one black marble. After choosing a bag you draw one marble out at random. You notice it is black. You then put it back and draw another marble out of the same bag at random.

What is the probability that the second marble drawn is black

Xmas pool



Rap

Try

If you take markr's 1st and 2nd solutions and let n go to infinity--you'd come up with n!/e. This conclusion is almost trivial for solution 2.

markr your solutions seem to be based upon counting non-matching permutations. To convert to probabilities wouldn't you divide solutions 1 and 2 into probability by dividing by the total number of permutations (n!).

And, I'm also curious about 3. The one seems to be based upon a fibonacci series. Could you embellish?

Rap

Conditional probability

[size=7]There are two bags each selected at random, the probability of black balls in one bag is 3 out of 4 and the other is 1 out of 4.

The probability of choosing the black ball rich bag is ½ and the black bag ball poor bag is ½.

So the first probability needed is the odds of picking that first black ball.
So
P(B1)=1/2*3/4+1/2*1/4=3/8+1/8=1/2

So the probability of drawing 1 black ball is ½

So now what is the probability of drawing two black balls from a single bag chosen at random-with replacement?

P(B)∩P(B1)=1/2*3/4*3/4+1/2*1/4*1/4=9/32+1/32=10/32=5/16

Now comes conditional probability

What is the probability of drawing a second black ball given that one has already been drawn?

From definition

P(B|B2)=[P(B)∩P(B1)]/P(B1)=(5/16)/(1/2)=5/8[/size]

Rap

Yes, I gave the number of derangements. I forgot to divide by n!

I didn't actually compute these; so I can't embellish. I wrote a program to compute the number of derangements, then looked up the formula(s) for the sequence.

[size=8]MARBLES
3/4 * 3/4 + 1/4 * 1/4 = 10/16 = 5/8[/size]

Nice cartwheel!

You've got spheres (10 inches in diameter) that are placed in a pile that is a pyramid with a triangular base.
1. How many spheres are in a pyramid with X levels?
2. How high is a pyramid with X levels?

Inside a unit square, you are to inscribe a regular hexagon. How big can you make it (maximum side length)?

You have a hollow right circular cone (base radius is 1 meter, height is 3 meters). A cube is placed inside the cone, such that the bottom face is resting on the base of the cone and the four top vertices are contacting the side of the cone. What size is the cube?

For which real numbers X (greater than zero) does the following expression evaluate to an integer?
cbrt(3 + sqrt(X)) + cbrt(3 - sqrt(X))
cbrt = cube root
sqrt = square root

You encounter two members of a tribe who tell the truth 25% of the time. One makes a statement, and the second says it's true. What is the probability that the first statement is true?

Quick solutions

Spheres

[size=7]1) the triangular number S=X(X+1)/2
2) Side length is 10X and height is 5X+10[/size]

Inscribed hexagon

[size=7]Hexagon side =.5 Area= 6√3/4[/size]

What X>0?

[size=7](3 + √(X))^1/3 + (3 - √(X))^1/3=I
if X=∞
(3 + √(∞))^1/3 + (3 - √(∞))^1/3=(3 + (∞))^1/3 + (3 - (∞))^1/3=
(∞)^1/3 + (-∞)^1/3=∞-∞=0
so I=0[/size]

Truth telling

[size=7]Truth Table says that either both natives are telling the truth or both are lying
So the probability that the first statement is true is
(1/4)(1/4)/(1/4)=(1/4)[/size]

Rap

P.S.
Exact answers required - no rounding - use radicals, etc.

Sphere stack Correction
Height h=[10√3(x-1)]/3+10
Rap