The worlds first riddle!

Would have thought the probability that joe wins is 40% (0.4)as he goes first, and is not affected by the other two scores.

For butch to win, probability would be that joe loses (0.6) multiplied by proby that butch wins (0.4) = 0.24.

Then for Matilda to win, would be chance joe loses (0.6) * chance butch loses (0.76) * chance she wins (0.7) = 0.3192.

Sadly this doesn't equal 1 when added up, so I know that it's wrong!!

But hey! At least I tried!

Final thought: the 0.0408 left, could be broken down further making Joes proby of winning just over 40%, Butch's proby just over 24% and Matilda's just over 31.92%!?!?!

I'm looking forward to seeing the answer to this one!

COWS
I thought the cook received her cows after each division, reducing the number to be divided the next time.

FREE THROWS
Joe: .4310
Butch: .3879
Matilda: .1810

"I'm looking forward to seeing the answer to this one!"

That's Bizarre. :wink:


• Joe wins 50 times out of 116.
• Butch wins 45 times out of 116.
• Matilda wins 21 times out of 116

COWS: I think the question needs to be clearer.



As spring approaches, many people put up birdhouses for wild birds
thereby initiating the annual round of ``Squirrel Wars.''

Seasoned veterans of these wars have learned to thwart the cursed rodents by suspending the nest from wires. The question to answer is: What configuration will minimize the total length of wire needed

The details are as follows. The wire is to be strung between two trees standing a distance D apart, and are to be attached at a common height above the ground -- high enough so a person can walk under the wire close to the trees. T

he birdhouse is to be suspended midway between the trees, but at a distance d below the height at which the wires are attached to the trees, so that a person can reach the birdhouse easily.

• Any two-club members play a sport with one and only one other club member.

The same sport?

I'm not sure I understand the first condition. Here's an answer based on a potentially erroneous assumption.

3 members
3 sports

Mark:
"I'm not sure I understand the first condition. Here's an answer based on a potentially erroneous assumption."

3 members
3 sports


There are seven sports club members and seven sports. There are also a couple of solutions in which there is only one member (not much of a club) or three players all playing the same sport.





A gambler has in his pocket a fair coin and a two-headed coin.
He pulls one of the coins at random from his pocket and, when he flips it, it shows heads.

(a) What is the probability that it's the fair coin
Suppose he flips the same coin a second time and again it shows heads.

(b) Now what is the probability that it's the fair coin
Suppose that he flips the same coin a third time and it shows tails.

(c) And now what is the probability that it's the fair coin

When you say "two-headed coin", is this a coin with two "heads" or a coin with two faces the same? This effects the answer to (c) I think!

If the coin has two "heads" then after getting a tail you know for certain that the coin you flipped MUST be the fair coin.

As for the others, I haven't thought about them as I'm in the middle of completing a Su Doku! Ha ha!!

GAMBLER
a) 1/3
b) 1/5
c) 1

Bizarre
I'm in the middle of completing a Su Doku! Ha ha!!

They are mind bogglingly difficult. If you can complete them, you stand on the shoulders of giants.

Mark
GAMBLER
a) 1/3
b) 1/5
c) 1


: 1/3, 1/5, 1
a) There are two events: The selection of a coin -- resulting in the Unfair Coin, "U", or the Fair Coin, "F" -- and the flip -- resulting in heads, "H", or tails, "T", for the Fair Coin, and in one of the two heads, "H1" or "H2", for the Unfair Coin. Each of these events is considered a 50/50 proposition.
Event1 Event2
F H
F T
U H1
U H2
So the conditional probability P(F|H) = P(F and H)/P(H) = (1/4)/(3/4) = 1/3.
b) Similarly,
Event1 Event2 Event3
F H H
F H T
F T H
F T T
U H1 H1
U H1 H2
U H2 H1
U H2 H2
Now P(F|H,H) = P(F and H,H)/P(H,H) = (1/8)/(5/8) = 1/5.
c) It must be the fair coin. P(F|H,H,T) = 1.



I have before me three numeric palindromes (numbers which read the same backwards and forwards, like 838). The first is two digits long; the second is three digits, and when we add those two numbers together, we get the third number, which is four digits long. What are the numbers



For anyone with a free week could try to:

Find the only solution to ABCD x E=DCBA.
Each letter represents a different digit.
ABCD is a 4-digit number (implying that A is not zero).



With thieves I consort,
With the vilest, in short,
I'm quite at ease in depravity;
Yet all divines use me,
And savants can't lose me,
For I am the center of gravity.

What am I


I build up castles.
I tear down mountains.
I make some men blind,
I help others to see.

What am I



Voiceless it cries,
Wingless flutters,
Toothless bites,
Mouthless mutters.

What am I

Try,

I always believe you a faithful man :wink:

PALINDROMES
22, 979, 1001

THIEVES, GRAVITY, ETC.
V

ABCDxE
2178 * 4 = 8712

Hi, Try,

Here is the Einstein's enigma :

Données de base:

i: Il y a 5 maisons de 5 couleurs différentes.

ii: Dans chaque maison habite quelqu'un d'une nationalité différente.

iii: Il n'y a pas deux habitants qui boivent la même boisson, regardent le même sujet, ni jouent au même jeu.

Les faits:

1: Don1 habite la maison rouge.
2: Nimh joue au scrabble.
3: Francis boit du vin.
4: La maison verte est directement à gauche de la blanche.
5: Celui qui habite la maison verte boit du café.
6: Celui qui regarde le sujet général joue aux dominos.
7: Celui qui occupe la maison jaune s'adonne aux devinettes.
8: Dans la maison du milieu on boit du lait.
9: Tryagain habite la première maison, à gauche.
10: Celui qui s'occupe des autres langues est le voisin du joueur de dames.
11: Le jouer de bridge est le voisin de celui qui s'adonne aux devinettes.
12: Le buveur de bière s'occupe de politique
13: Walter s'occupe des nouvelles internationales
14: Tryagain habite à coté de la maison bleue.
15: Celui qui s'occupe d'autres langues a pour voisin un buveur d'eau.

Alors, dites moi, en vous servant de la logique pure: qui joue aux échecs?

5-digit number.

(b) Rearrange the digits in that number, and write it above the first number if it is larger, but below it if smaller.

(c) Subtract one from the other (you can use a calculator if you wish). :wink:

(d) Choose any digit (except zero) you want me to identify.

(e) Then post all the other (4)

Try,

As I told you before, the chess player is Walter!

Only it hapens he is german!!!

And what if he really dont play chess?

5-DIGIT TRICK (small font)

If post equals ABCD, then response equals 9-[(A+B+C+D) mod 9].

Francis, my language ability ranges from poor to very poor all in pidgin English. I still think It waz the German wot done it!

Mark writes in diddly, diddly writing the answer to the question I never got the chance to ask. Lets hear it for Mark who can give an answer before the question. (Did you know this from the past?
Or, did you work it out when you saw it?)

I got a message from Fred today, he asks:


"What is the smallest positive integer with the property that if the digit on the extreme right (i.e. the one's digit) is moved to the extreme left, the resulting number is one-and-a-half times the original number".

Can anyone help the guy out. I gotta take the cat to the dentist!

I've seen problems based on digital roots in several forms before.

HELP FOR FRED
285714

Mark:
HELP FOR FRED
285714

Fred sends his thanks.
He calls his original number x and writes x = abc...uvw, where a through w are the digits of x. Note that x is divisible by 2, so its one's digit must be even. Suppose, for the moment that x is a two digit number.

The equation 3(10a + b) = 2(10b + a)
would then have a single digit integer solution; simplifying gives
28a = 17b. Since the left-hand side is divisible by 7, the right is, too; but b is even, which is impossible.

He next looks at a three digit solution. The relevant equations are
3(100a + 10b + c) = 2(100c + 10a + b), 280a + 28b = 197c.
The left hand side is divisible by 7 again, which is impossible with c even. Having worked out these cases he sees that a solution requires finding a number of the form 199...997 which is divisible by 7.

The smallest such is 199997. This gives the equation 280000a + 28000b + 2800c + 280d +28 e = 199997f.

Since the left-hand side is divisible by 4 he gets that f must be 4 or 8. With f = 4 the solution x = 285714 comes out, while f = 8 gives x = 571428. Both are solutions, but the first is the smallest.



Pick 1998 points at random in the plane. Suppose no three of them lie on a line (which would be true if they were chosen truly at random).

Is there always a line in the plane which divides the set of points into two equal parts

Yes, and you can do it with any number of points.

Mark:
"Yes, and you can do it with any number of points."

Let S denote the set of slopes of lines joining two of the 1998 points. This is a finite set so there is some real number m not in the set. For each of the 1998 points the line of slope m passing through that point intercepts the y-axis at some point, let y1,..., y1998 be those points, labeled in such a way as to have y1 > y2 > ... > y1998. Notice that all of these points are different. Pick any point in the interval strictly between y999 and y1000 and draw the line through this point with slope m. That'll do it!


This one had me reaching for the pills.

A mutiny takes place on a pirate ship. Wanting to punish the mutineers and regain control of the ship, the captain puts the disloyal crew members off on different small islands.

The mutineers soon discover that communication between themselves is possible by sending messages attached to carrier pigeons. However, each mutineer can only communicate with one other plotter on any given day.
Suppose each of them has some unique piece of information he would like all the others to know and that on any given day he can pass along all the information he has to his chosen recipient.

There are eleven mutineers. How many days will it take before each of them has gathered all the information

One of the faithful crew member is found to be an accomplice and marooned like the others, making twelve mutineers in all. Now how many days will the communication take



Glittering points
That downward thrust,
Sparkling spears
That never rust.

What am I


Alive without breath,
As cold as death;
Never thirsty, ever drinking,
All in mail never clinking


What am I