Fuzzy Math…

[Jaws]

It's our usual everyday reasoning that fails us in this situation, as common reasoning almost always does when dealing with long-run probabilities.

I understand the theory you have expressed but I think what their trying to say is:

When your on Montys' game show, your not going to have 10, 100 or 1000 chances to play the game so the long term probabilities are none existent. Just as flipping a coin 1000 times should give you very close to 500 heads and 500 tails. But in the short run say 10 flips you can just as easily get 10 heads or 10 tails.

And its the reason Las Vegas can exist and flourish. If everyone was well educated about probability, only the truly compulsive gambler would knowingly throw away money on any player vs house casino game (black jack if you can't count cards, craps, roullette, video poker, slots, etc.)

Yes, for the compulsive gambler, but not for the expert gambler who well educated about probability (only plays craps, blackjack and bacarat and only bets on the lowest house advantage), money management and playing for only a short time (the longer you play , the more likely you will lose).

That reminds me of when I was young and foolish. There was an older guy that would come around and shoot craps with us. He had a big wad of money with him and we had chump change. Well he use to let us shoot the dice and fade all our bets and eventually take all our money. One day I wised up to I told him I would shoot the dice but only one time win or lose and quit. So it turned out I won on the next 3 separate occasions and he would never shoot with me again.

Thanks

[Svennemir]

Joe 998:

Further assume ( because its not clear at the start) that Monty is given an instruction. He knows where the car is and is instructed to ALWAYS choose a goat

It is phrased like this:

However, the way the game works is that the door you pick does not get opened immediately. Instead, the host (Monty Hall) will open one of the other doors to reveal a goat.

Thus, the point is that a goat is revealed. No mistake there.

The probability is 2/3 if you change and 1/3 if you do not change. This is carefully explained on the site from which the riddle is pasted (follow the link in the original post). I would like to point out the error in the (any) calculation leading to 1/2 chance, but I cannot follow the numbers written in the post.

[Svennemir]

Einstein Riddle

Facts:

1. There are 5 houses in 5 different colors.
2. In each house lives a person with a different nationality.
3. These 5 owners drink a certain beverage, smoke a certain brand of cigar and keep a certain pet.
4. No owners have the same pet, smoke the same brand of cigar or drink the same drink.

Hints:

1. The Brit lives in a red house.
2. The Swede keeps dogs as pets.
3. The Dane drinks tea.
4. The green house is on the left of the white house.
5. The green house owner drinks coffee.
6. The person who smokes Pall Mall rears birds.
7. The owner of the yellow house smoke Dunhill.
8. The man living in the house right in the center drinks milk.
9. The Norwegian lives in the first house.
10. The man who smokes Blend lives next to the one who keeps cats.
11. The man who keeps horses lives next to the man who smokes Dunhill.
12. The owner who smokes Blue Master drinks beer.
13. The German smokes Prince.
14. The Norwegian lives next to the blue house.
15. The man who smokes Blend Has a neighbor who drinks water.

The question is: WHO KEEPS FISH?



(disclaimer: this will probably take some time)
Link

[mavraam]

The question is: WHO KEEPS FISH?

SPOILER FOLLOWS:












The Brit in the Red house who drinks beer, smokes Blue Master, also raises fish.

Here's how I have them laid out

Reading left to right:
A) Yellow, Norwegian, Water, Dunhill, Cats
B) Blue, Dane, Tea, Blend, Horse
C) Green, ?, Cofee, ?, ?
D) White, ?, Milk, ?, ?
E) Red, Brit, Beer, Blue Master, Fish

[Jaws]

Norwegians - yellow - water - dunhill - cats
Dane - blue - tea - blend - horses
Brit - red - milk - pall mall - birds
German - green - coffee - prince - fish
Swede - white - beer - blue master - dogs

[Mangudai]

Total: .9999#


# (effectively 1.000 because of rounding error)

No rounding error. 0.9999# is exactly 1!

Proof: Let
x = 0.9999#
10x = 9.9999#
10x - x = 9.9999#-0.9999#
9x = 9
x = 1

There is no trick. This is exact. In some proofs of Real analysis (ex. Cantor's diagonal argument that the Reals are uncountable) it is necessary to rule out all numbers that are ...9999# from some point onwards, because they are not unique. If you want to get a mathematician angry start talking about infintesimals.

[Mr.Frag]

If you want to get a mathematician angry start talking about infintesimals.

[:D]

[mavraam]

AAARGH.

I read number 8 wrong.

8. The man living in the house right in the center drinks milk

I thought it said right OF center not right IN center!!!!
[:@][:@][:@]

[Fred98]

Back to the goats and the car.

Some would say that my last car was a goat

ORIGINAL: Svennemir

However, the way the game works is that the door you pick does not get opened immediately. Instead, the host (Monty Hall) will open one of the other doors to reveal a goat.

Thank you for confirming that.



Presume the car is behind door A
Presume I don’t switch

I choose Door A 1/3 I stick with door A ½ total = 1/6 WIN

I choose Door B 1/3 I stick with door B ½ total = 1/6 Lose

I choose Door C 1/3 I stick with door C ½ total = 1/6 Lose


Presume the car is behind door B
Presume I don’t switch

I choose Door A 1/3 I stick with door A ½ total = 1/6 Lose

I choose Door B 1/3 I stick with door B ½ total = 1/6 Win

I choose Door C 1/3 I stick with door C ½ total = 1/6 Lose


Presume the car is behind door C
Presume I don’t switch

I choose Door A 1/3 I stick with door A ½ total = 1/6 Lose

I choose Door B 1/3 I stick with door B ½ total = 1/6 Lose

I choose Door C 1/3 I stick with door C ½ total = 1/6 Win


Presume the car is behind door A
Presume I switch
Monty has revealed the goat so I always choose the third door

Choose Door A 1/3 Monty opens B. I switch to door C ½ total = 1/6 Lose

Choose Door B 1/3 Monty opens C. I switch to door A ½ total = 1/6 Win

Choose Door C 1/3 Monty opens B. I switch to door A ½ total = 1/6 Win


Presume the car is behind door B
Presume I switch
Monty has revealed the goat so I always choose the third door

Choose Door A 1/3 Monty opens C. I switch to door B ½ total = 1/6 Win

Choose Door B 1/3 Monty opens A. I switch to door C ½ total = 1/6 Lose

Choose Door C 1/3 Monty opens A. I switch to door B ½ total = 1/6 Win


Presume the car is behind door C
Presume I switch
Monty has revealed the goat so I always choose the third door

Choose Door A 1/3 Monty opens B. I switch to door C ½ total = 1/6 Win

Choose Door B 1/3 Monty opens A. I switch to door C ½ total = 1/6 Win

Choose Door C 1/3 Monty opens A. I switch to door B ½ total = 1/6 Lose




Therefore originally my odds were 33.33 %

But due to the strange rules:

If I don’t switch my odds increase to 50%
If I do switch my odds increase further to 66.667%


We Agree!

Its time for me to Escape. Ford Escape

[Mangudai]

Therefore originally my odds were 33.33 %

But due to the strange rules:

If I don’t switch my odds increase to 50%
If I do switch my odds increase further to 66.667%

Wrong, if you don't switch your odds are still 33.33%

Presume the car is behind door A
Presume I don’t switch

I choose Door A 1/3 I stick with door A ½ total = 1/6 WIN

I choose Door B 1/3 I stick with door B ½ total = 1/6 Lose

I choose Door C 1/3 I stick with door C ½ total = 1/6 Lose

For each of these six cases you presume the total probability should add up to one. Also it is improper to assign a probability to your choices in this manner. Since you are listing all the probabilities with deterministic variables, each thing is simply true or false not a fraction.


Presume the car is behind door A. Presume I don’t switch
I choose Door A I stick with door A = WIN
I choose Door B I stick with door B = Lose
I choose Door C I stick with door C = Lose

Chance of winning = 1/3

Presume the car is behind door A Presume I switch
Monty has revealed the goat so I always choose the third door

Choose Door A Monty opens B. I switch to door C = Lose
Choose Door B Monty opens C. I switch to door A = Win
Choose Door C Monty opens B. I switch to door A = Win

Chance of winning = 2/3

[Svennemir]

Hello again, here's anohter one - pasted for your convenience, but you may want to check out this link for information on the principle of induction (of course, induction is not really difficult to understand so you don't have to). It's not exactly easy, especially for people who are not accustomed to mathematics. Try to find the error in the "proof"!
--------------------------------------------------------------------------
All People in Canada are the Same Age
This "proof" will attempt to show that all people in Canada are the same age, by showing by induction that the following statement (which we'll call "S(n)" for short) is true for all natural numbers n:

Statement S(n): In any group of n people, everyone in that group has the same age.

The conclusion follows from that statement by letting n be the the number of people in Canada.

(....)

The Fallacious Proof of Statement S(n):

* Step 1: In any group that consists of just one person, everybody in the group has the same age, because after all there is only one person!

* Step 2: Therefore, statement S(1) is true.

* Step 3: The next stage in the induction argument is to prove that, whenever S(n) is true for one number (say n=k), it is also true for the next number (that is, n = k+1).

* Step 4: We can do this by (1) assuming that, in every group of k people, everyone has the same age; then (2) deducing from it that, in every group of k+1 people, everyone has the same age.

* Step 5: Let G be an arbitrary group of k+1 people; we just need to show that every member of G has the same age.

* Step 6: To do this, we just need to show that, if P and Q are any members of G, then they have the same age.

* Step 7: Consider everybody in G except P. These people form a group of k people, so they must all have the same age (since we are assuming that, in any group of k people, everyone has the same age).

* Step 8: Consider everybody in G except Q. Again, they form a group of k people, so they must all have the same age.

* Step 9: Let R be someone else in G other than P or Q.

* Step 10: Since Q and R each belong to the group considered in step 7, they are the same age.

* Step 11: Since P and R each belong to the group considered in step 8, they are the same age.

* Step 12: Since Q and R are the same age, and P and R are the same age, it follows that P and Q are the same age.

* Step 13: We have now seen that, if we consider any two people P and Q in G, they have the same age. It follows that everyone in G has the same age.

* Step 14: The proof is now complete: we have shown that the statement is true for n=1, and we have shown that whenever it is true for n=k it is also true for n=k+1, so by induction it is true for all n.

[mavraam]

All People in Canada are the Same Age

That's a good one. But it makes an unstated assumption that
k+1 > 2.

This line of reasoning does not apply if k+1 = 2;

In an induction proof, any analysis of k+1 elements must be applicable to any case
where k > 0.

The reason this does not apply is because if we take P and Q out of a group of 2, WE
DON'T HAVE ANYONE LEFT THAT HAS TO BE THE SAME AGE AS EACH OF THEM.

Here's a few steps for the case where k+1 = 2:

* Step 5: Let G be an arbitrary group of 2 people; we just need to show that every member of G has the same age. OK, so far.

* Step 6: To do this, we just need to show that, if P and Q are any members of G, then they have the same age.
OK.

* Step 7: Consider everybody in G except P. These people form a group of k people, so they must all have the same age (since we are assuming that, in any group of k people, everyone has the same age).
Ok.

* Step 8: Consider everybody in G except Q. Again, they form a group of k people, so they must all have the same age.
OK.

* Step 9: Let R be someone else in G other than P or Q.

Nope. There is no R. There is no one else in the group other than P or Q

In order to invalidate induction, you only need to show that the rule doesn't apply to one case and then it invalidates the whole argument.

IMHO

[Mangudai]

Here are some funny arguments from Lewis Caroll. They are easy to resolve, but at the time he wrote them they were challenging for the basic symbolic logic that was around.

Men over five feet tall are numerous.
Men over ten feet tall are not numerous.
Therefore men over ten feet tall are not over five feet tall.

The meat that I eat is the meat that I buy at the market.
The meat that I buy at the market is raw meat.
Therefore I eat raw meat.

[Rainbow7]

Of course, not one of the problems in the entire thread falls under fuzzy logic.

[mavraam]

ORIGINAL: Mangudai

Here are some funny arguments from Lewis Caroll. They are easy to resolve, but at the time he wrote them they were challenging for the basic symbolic logic that was around.

Men over five feet tall are numerous.
Men over ten feet tall are not numerous.
Therefore men over ten feet tall are not over five feet tall.

The meat that I eat is the meat that I buy at the market.
The meat that I buy at the market is raw meat.
Therefore I eat raw meat.

I'm sure you know this but others may not:

What is Boolean Algebra?

Boolean Algebra

(boo´leen) , an abstract mathematical system primarily used in computer science and in expressing the relationships between sets (groups of objects or concepts). The notational system was developed by the English mathematician George Boole c.1850 to permit an algebraic manipulation of logical statements. Such manipulation can demonstrate whether or not a statement is true and show how a complicated statement can be rephrased in a simpler, more convenient form without changing its meaning. In his 1881 treatise, Symbolic Logic, the English logician and mathematician John Venn interpreted Boole's work and introduced a new method of diagramming Boole's notation; this was later refined by the English mathematician Charles Dodgson (better known as Lewis Carroll -this method is now know as the Venn diagram. When used in set theory, Boolean notation can demonstrate the relationship between groups, indicating what is in each set alone, what is jointly contained in both, and what is contained in neither. Boolean algebra is of significance in the study of information theory, the theory of probability, and the geometry of sets. The expression of electrical networks in Boolean notation has aided the development of switching theory and the design of computers.

Lewis Carroll's fascination with logic problems helped create a mathematics that was years ahead of its time. Its only after the development of the digital computer that we can appreciate the power of his work.

As a programmer, I use Boolean algebra on a daily basis in the same way an accountant uses a calculator. Its a basic tool of software design on all levels. But Carroll couldn't possibly have foreseen its use in 1881.

[Mangudai]

Of course, not one of the problems in the entire thread falls under fuzzy logic.

O contraire. Math is a subset of logic.

[Rainbow7]

I was refering to the fuzzy part.

[Rainbow7]

To resurrect a dead horse, this exchange was recently highlighted over at The Straight Dope:

The Monty Hall problem.

I love how strongly everyone always feels about this problem.

[Zap]

ORIGINAL: Mangudai

Of course, not one of the problems in the entire thread falls under fuzzy logic.

O contraire. Math is a subset of logic.

Yea, well math is all greek to me![:D]