The worlds first riddle!

OK, who is this Marilyn chick? How come we have not been introduced? You quote, "Marilyn correctly noted that in the modified version, it makes no difference - the contestant will win with probability 1/2 in either case."

Now, you have gotta remember that in the original problem I wrongly argued that there was 1/2 chance instead of 2/3. So I am probably not the best person to answer. However, luck favors the bold, so:

If Monty knows no more than you, and there are only two doors left, it would be tempting to agree. However, with n doors, and (n-2) empty doors opened by him, the probability of winning by changing from the original selection (1/1000) is (n-1)/n. So I would switch.

Mark:

CROSS COUNTRY
1: 6 total = 7, N > T, 6 > 3
2: 3 total = 5, N > T
3: 1 total = 4
4: 5 total = 9, N > T
5: 4 total = 9
6: 7 total = 13, N > T
7: 2 total = 9

You sure ran away with that answer.




How many sequences are possible in a seven-game World Series?

In other words, if two teams play the full seven games before one of them wins the required four games, how many different sequences of seven games are possible

[size=8]WORLD SERIES
The first six games are split, and there are C(6,3) ways for one team to win three of those games. There are two possible outcomes for the seventh game. That comes to 2 * C(6,3) = 40.[/size]

MONTY HALL
Not quite. You're applying the logic for the original problem to the modified problem. Try again, Tryagain.

Quote, "Try again, Tryagain."

Comedian eh!

Listen! I did not get to the bottom of the pile by getting it right first time…or second etc. However, nothing dulls my enthusiasm to keep trying.



The damn thousand-door problem

My chance of being right is 0.1%, and the chance of being wrong is 99.9%. Now, with open 998 doors open:

My first choice does not just magically became a 50/50? If so, I'm the luckiest person in the world to go picking the right door 50% of the time when faced with 1000 choices. I don't think you can use Bayes Theorem in this case because Monty does NOT know any more than me.

Switch.



Mark:

WORLD SERIES
The first six games are split, and there are C(6,3) ways for one team to win three of those games. There are two possible outcomes for the seventh game. That comes to 2 * C(6,3) = 40.





Mark after baffling poor Try, went to the king for his reward. The king explains the following:

1) The king holds two pieces of paper in his hands. One with a large number written on it, the other with a number that is half the first number written on it.

2) One of the two numbers is 1000, but the king is not telling whether 1000 is the bigger or the smaller number.

Mark was then asked by the king to pick one of the two pieces of paper. The king then revealed that the piece Mark chose has 1000 written on it, and offers Mark the chance to choose the other piece. Whatever is written on the piece Mark finally choose would be the amount of gold pieces Mark will get.

Should Mark change his choice?


Try, in an exclusive interview with A2Kay News in a down town bar said, "What freekin difference can it make!"

MONTY HALL
Monty doesn't know more than you, but he did get lucky. That eliminates lots of cases where he didn't get lucky.

For N doors, you've got a 1/N chance of selecting the prize.
If you didn't select the prize (probability = (N-1)/N)), Monty essentially must "select" the prize, and open all the other doors. His chances of doing that are 1/(N-1). (N-1)/N * 1/(N-1) = 1/N. So, you only find yourself in this situation with probability 1/N. Therefore, it doesn't matter if you switch or stick.

I think the best way to look at it is that Monty is playing the same game you are. You have each selected one of a thousand doors. The other 998 are opened and none contain the prize (the chances of that are small). Neither player has an advantage over the other, so the probability of winning is 1/2.

[size=8]KING
This is a paradox (50% chance of gaining $1000, 50% chance of losing $500). Mark should flip a coin.[/size]

im not sure
maybe the question is in the first part and the answer is in the 2nd part.
the riddle is also made before jesus was born so it probably has nothing to do with him unless the two men are pshycic, but other then that I don't know the answer.

Re: im not sure



Welcome, I have waited 4eternity for someone to say that. I do hope we will see more of you. A2K has a lot to offer.

Perhaps, as your original problem contained the given answer of 1/2 you will not be surprised to see I do not agree. For simplicity let us take the same scenario with 100 doors to equal 100%.

On 99 out of 100 occasions the other door will contain the prize, as 99 out of 100 times the player first picked a door with nothing.

To drive the point home, one only has to imagine the host having to start with the first door and go down the line, opening the doors but skipping over only the player's door and one other door. The player can then more easily appreciate the randomness of his first choice and the large amount of information he has gained since he made that choice and then see the wisdom in switching.

Even if only one of the 100 doors is opened, switching still increases the player's chances of finding a car. The 99/100 chance that the car is not behind the door the player picked is spread evenly over 98 doors after the host reveals an empty space.

Each of those 98 doors (all doors other than the one the player picked and the one the host reveals) has a 99/9800 chance of having the car, so by switching the player improves his chances slightly from a .0100 chance to just over .0101. This is an improvement of 99/98 or just over 1%.

My first choice had a 99/100 chance. I now have 98/100 more bits of information. The other door does indeed have a 1/2 chance. However, a 99/100 and 1/2 does not average out at 1/2. I still say, switch no matter what your friend says.

Come back!

(Re: KING. I will be back, I want to re-check the numbers)

Mark:

KING
This is a paradox (50% chance of gaining $1000, 50% chance of losing $500). Mark should flip a coin.




Oh, goody goody we have a disputed call (again). Let's add some (sum) numbers to this problem:


Knowing that one of the numbers is 1000, the two numbers in the king's hands are either (500,1000) or (1000,2000). Assume it is equally likely (as implicit in the problem) for either of those cases to occur, the raw expected value for a random choice is:

(500+1000+1000+2000)/4 = 1125

However, knowing which piece has the 1000 written on it, means that the expected value of choosing the other piece would be:

(500+2000)/2 = 1250

This is a whole lot better than the 1000 which Mark currently holds. So switch! :wink:




Mr. Baker, Mr. Carpenter, Mr. Hunter, and Mr. Walker were a baker, a hunter, a carpenter, and a walker, but not necessarily in that order. ( Well, definitely not in that order). They were in the habit of wearing a brown, a charcoal, a heliotrope, and a white shirt.

No man's profession was the same as his name and the color of each man's shirt began with a letter which was different from the initial letter of both his name and his profession.

Mr. Hunter and the professional walker dined together regularly. The hunter never would wear a brown shirt. Mr. Carpenter was the baker.

What is the profession of each man and the color of his shirt

Mr. Carpenter is a baker and wears a white shirt.
Mr. Walker is a hunter and wears a charcoal shirt.
Mr. Hunter is a carpenter and wears a brown shirt.
Mr. Baker is a walker and wears a heliotrope shirt.

KING
Touche! In my haste, I mistook this for the paradox which is similar.

MONTY
First of all, leaving more than one door unopened is changing the problem; so let's ignore that.

"My first choice had a 99/100 chance. I now have 98/100 more bits of information. The other door does indeed have a 1/2 chance."

I think you meant to say, "My first choice had a 1/100 chance."
The sums of the probabilities for the two unopened doors must equal one. If you agree that the other door has a 1/2 chance, then so does yours.

"To drive the point home, one only has to imagine the host having to start with the first door and go down the line, opening the doors but skipping over only the player's door and one other door. The player can then more easily appreciate the randomness of his first choice and the large amount of information he has gained since he made that choice and then see the wisdom in switching."

Here's the large amount of information I would gain from the process:
Hmmm, with no a priori knowledge, Monty was able to correctly pick 98 doors that didn't contain the prize. Wow! Either he's clairevoyant, or there's a damn good chance that he was able to do it because my door has the prize.

I'll go back to my previous post, but state it differently. This is equivalent to you and Monty each picking a door at random. Then, all of the other doors are opened and they just happen (with low probability, but it is a key condition of the problem) to contain nothing. Now, did Monty do a better job of picking a door at random than you did?

I think you are overlooking the fact that to successfully open N-2 doors, many possibilities for the location of the prize had to be eliminated.

Let's look at a 4-door situation, and you pick the first door. Here are the possibilities (P=prize, E=empty).

PEEE (prob = 1/4)
With probability 1, Monty will be able to open 2 non-P doors.
With probability 1/4, you win if you stick.

EPEE (prob = 1/4)
With probability 1/3, Monty will be able to open 2 non-P doors.
With probability 1/12, you win if you switch.

EEPE (prob = 1/4)
With probability 1/3, Monty will be able to open 2 non-P doors.
With probability 1/12, you win if you switch.

EEEP (prob = 1/4)
With probability 1/3, Monty will be able to open 2 non-P doors.
With probability 1/12, you win if you switch.

Adding things up reveals:
Stick: 1/4
Switch: 3/12 = 1/4

The other half of the time, the game didn't make it to the end because Monty guessed wrong.

So, by sticking, your probability of winning is .25/.5 = 1/2, and by switching, your probability of winning is .25/.5 = 1/2.

If you still aren't buying this, then let's get together and play a round. You can be Monty. I'll always stick with my first choice. If I lose, I'll pay you $10,000, but if I win, you'll pay me $100,000. Surely, you'll find those odds in your favor. :wink:

"BTW (2) I played the 100 version with two packs of cards. You got to choose first and did not change. I am delighted to inform you that you lost 11 in a row and therefore owe me $110,000."

I don't think you had time.

"BTW (1) I fully agree with your four door setup!"

Does that mean you're convinced?

[size=8]A2K GIRLS
Why didn't they note the weight when only one girl was on the scale?
91, 92, 95, 99, 101[/size]

BEATING A KING'S POTENTIALLY DEAD HORSE

Let's look at the N-door situation, and you pick the first door. Here are the possibilities (P=prize, E=empty).

PEEE... (prob = 1/N)
With probability 1, Monty will be able to open 2 non-P doors.
With probability 1/4, you win if you stick.

...EPE... (prob = (N-1)/N) (This is N-1 cases rolled into one)
With probability 1/(N-1), Monty will be able to open N-2 non-P doors.
With probability (N-1)/N * 1/(N-1) = 1/N, you win if you switch.

That gives us:
Stick: 1/N
Switch: 1/N

The other (N-2)/N of the time, the game didn't make it to the end because Monty guessed wrong.

So, by sticking, your probability of winning is (1/N)/(2/N) = 1/2, and by switching, your probability of winning is (1/N)/(2/N) = 1/2.

This is mathematical art, and deserves it's place in history.

Mark:
"Does that mean you're convinced?"

Regretfully, your use of fact, logic and science, has extinguished the flickering flame of rhetoric. However, I have a goodly supply of kindling and may well relight the fire of debate when hell freezes over.

Mark:

A2K GIRLS
"Why didn't they note the weight when only one girl was on the scale?"

A2k girls don't wear sensible shoes, they are wild and free spirited. But maffs aint their strong suit, if you know what I mean! (Much like me actually)


91, 92, 95, 99, 101


Since 5 items taken two at a time is 5!/(2!3!) = 10 and since the ten readings are distinct:

(1) they represent all the possible combinations of the five weights;
and (2) no two weights can be the same.

We can, therefore, denote the weights by a, b, c, d, e where a < b < c < d < e. The two lightest girls must therefore be in the first weighing and the two heaviest in the last. So we know that

a+b = 183, d+e = 200

By adding the ten weighings, we obtain 4a + 4b + 4c + 4d + 4e = 1912
Therefore, a+b+c+d+e = 478. Hence, c = 478 - (200 + 183) = 95

Consider the final weighing. Assuming integer weights for the moment, the possible values of e are restricted to 104, 103, 102 and 101. Adding these to c would give 199, 198, 197 and 196--only the last of which was actually recorded. Hence it must be that

e = 101, d = 99

For the first weighing, possible values of b are 94, 93 and 92. If b = 94, then b+e = 195, which is not a recorded weighing. If b = 93, then b+c = 188, which again is not a recorded weighing. Hence,

b = 92, a = 91

This gives us 91 < 92 < 95 < 99 < 101 and






A map of downtown Collins, Colorado is shown below.
Some students at Collins High School, which is labeled A in the map below, wish to visit their friends at Poudre High School, labeled C.

Using only the main roads shown (notice that Overland Trail ends at Drake and that Laporte ends at College Ave.), and always traveling North and West, how many different routes are possible to get from Collins
High to Poudre High

How many of the paths will take them by the other high school in the city, Rocky Mountain High (labeled B)

[size=8]COLLINS
If I didn't screw up the addition,
120 and 40.[/size]

Mark:

COLLINS
120 and 40.


Using a table similar to Pascal's triangle, the numbers in the following table show how many different ways you can get to a particular corner. Remember, you can only travel north and west. In each block or square, you would add the top right hand corner number and the bottom left hand corner number to get the number for the top left hand corner of that square.







For all positive integers n, the symbol n! denotes the product of the first n positive integers. In other words, 4! = 4x3x2x1 = 24.

Find the value of n for which
(3!)(5!)(7!) = n!



Due to internet connection problems , may I leave you with:


The Christmas Tree / Hanukah Bush Problem / Add your own...

Hang each of the numbers from 1 to 10 as the ornaments so that every line adds up to the same number. There are five lines (there are three lines containing four ornaments, one line with three ornaments, and one line with just two ornaments).

(That's a triangle of 1,2,3,4.)



I hope to be back soon, but if not, may I wish you all a happy holiday and a healthy and peaceful 2007.