Fuzzy Math…

[mavraam]

ORIGINAL: Svennemir

I'm not sure that makes sense, but one thing I've found that will convince some is to get 3 playing cards, 1 red, 2 black and simulate the situation 20 times. The first ten, never switch. The second ten always switch. The second 10 should almost always win over the first barring some statistical anomoly.

It's funny, I also like to use playing cards to convince people. When they've tried to not switch 10 times, it becomes clear that they could as well select the door/card right away without the host revealing anything, and then the 1/3 follows.

Lo and behold! The power of google enabled me to find the one about the waiter you mentioned. (and in the first try!)

Here it is: Three men go to stay at a motel and the clerk
charges them $30.00 for the room. They split the cost ten
dollars each. Later the manager tells the clerk that he over-
charged the men and that the actual cost should have been
$25.00. He gives the clerk $5.00 and tells him to give it to the
men. But he decides to cheat them and pockets $2.00. He then
gives each man a dollar. Now each man has paid $9.00 to stay in
the room and 3 X $9.00 = $27.00. The clerk pocketed $2.00.
$27.00 + $2.00 = $29.00. So where is the other $1.00?

Link

Thanks for finding that one!!!! [:D]

[Fallschirmjager]

You guys need to get laid...and soon

[dinsdale]

Hang on, isn't this the gamblers fallacy?

As soon as there are 2 doors left then probability resets. It is a 1:2 chance regardless of which choice. Probability changes with perspective, so if you need to make the switch decision before the first dooe opens then the probability will be different than a decision made after the switch.

This is no different from determining the probability of a coin flip if the previous 10 flips have all been heads: the probability is still 1:2 because there are still 2 choices.

[Rainbow7]

@ mavraam: The brute force method isn't very intellectually satisying, and it doesn't really explain anything. That's my last resort.

@dinsdale: The probabilities don't get reset, because nothing has altered the fact that you only had a 1 in 3 chance with your first pick of a door (or 1 in 1000 chance with my 1000 door setup). What's telling is that Monty won't open your door and he won't open the correct door. So you could stick with your 1/1000 chance door. But not being a gambling man, I'd choose the remaining door at 999:1 odds. And this is quite different from the coin flip example because each coin flip is independent of all previous ones, whereas here you don't get a second choice after Monty opens a door (you're stuck with your first choice).

[dinsdale]

ORIGINAL: Rainbow
@dinsdale: The probabilities don't get reset, because nothing has altered the fact that you only had a 1 in 3 chance with your first pick of a door (or 1 in 1000 chance with my 1000 door setup). What's telling is that Monty won't open your door and he won't open the correct door. So you could stick with your 1/1000 chance door. But not being a gambling man, I'd choose the remaining door at 999:1 odds. And this is quite different from the coin flip example because each coin flip is independent of all previous ones, whereas here you don't get a second choice after Monty opens a door (you're stuck with your first choice).

But at the moment when you are asked to chose a new door or keep the old one something has changed: you've been asked to choose. An entirely new probability equation has formed at that instant. It doesn't matter if you change or not, you have been asked to make a decision and that begins a new. I can see the "trick" because the claim is that you remain at 1:3 instead of switching and being at 1:2 but I disagree that no intervening event has occured.

[Blackhorse]

Re: the Monty Hall test.

You switch. You start off "owning" 1/3rd of the doors; Monty has 2/3rds.

The unstated key to the puzzle is advance knowledge. Monty *knows* which door contains the prize, and at least one of his two doors has to contain a goat. So even after he reveals a goat he still "owns" a 2/3rds chance of winning.

If neither you nor Monty know which door contains the prize, and he randomly reveals one, and it does not contain a car, *then* your odds are 50:50 to stay or switch.

"let's look behind Door Number 2, which Carrol Merrill is now showing us" . . . mmmm, Carrol Merril. [;)]

[Rainbow7]

@dinsdale: The confusion always rests at this second "choice", as though you were being asked to choose between two doors, ignoring everything that came before. But that's not the setup of the problem. You're being asked to remain with your first choice or swtich to the "remaining" door. And given how Monty will reveal information, what remains has the most probability of being the correct door. There isn't an equal probability distribution here at the second choice.

[mavraam]

@ mavraam: The brute force method isn't very intellectually satisying, and it doesn't really explain anything. That's my last resort.

I know. But some people who don't have a background in probability will never quite get it.

Another way of looking at it is this:

Two out of 3 times, you WON'T pick the right door on the first try. Then, after the other wrong door is open, you'll pick the right door after you switch.

One out of 3 times, you WILL pick the right door on the first try. Then, after the other wrong door is open, you'll lose the right door after the switch.

So by switching, you'll win 2 out of 3 times.

[dinsdale]

ORIGINAL: mavraam

I know. But some people who don't have a background in probability will never quite get it.

Not really true, it's a trick question designed to hide when the probability is calculated.

Two out of 3 times, you WON'T pick the right door on the first try. Then, after the other wrong door is open, you'll pick the right door after you switch.
One out of 3 times, you WILL pick the right door on the first try. Then, after the other wrong door is open, you'll lose the right door after the switch.

So by switching, you'll win 2 out of 3 times.

This ignores the fact that you might lose when the first door opens. It's altering the probability because it's always assumed that you never lose with the first pick this is in effect a loaded dice scenario.

[Svennemir]

Okay, let's say we ALWAYS choose to change door. So there are two scenarios depending on what is behind the door of our first guess:

1) We pick the door with the car. Probability: 1/3.
The host opens a door with a goat, and we pick the other door with a goat.

Result: goat
Probability: 1/3

2) We pick a door with a goat. Probability: 2/3.
The host opens the other door with a goat, and the remaining door, which we pick, contains the car.

Result: car
Probability: 2/3.

One might say that whenever there are two possible choices, the chances are equal. This is not true. Remember, our first guess DICTATES the behaviour of the host - he HAS to open the door with the other goat if we hit a goat. This is the subtle information. He is forced to do it that way, and we can take advantage of that.

[Belisarius]

ORIGINAL: Svennemir

By the way, it can be tricky! But I've figured it out (and it's not about rounding errors - come to think of it, maybe you thought about another one, Belisarius?)

Yeah sorry I did. It's not rounding errors, it's how you phrase it.

The $2 the clerk keep are the ones they pay on top of the $25. They already got the other $3 back.

[Svennemir]

[Fred98]

Regarding the cars and the goats:

ORIGINAL: mavraam

But in fact are choosing 2 of 3 if you switch instead of 1 of 3.

I'm not sure that makes sense.....

No, it makes no sense. You are saying that if I switch that the odds change. But if I switch the odds remain the same. The car is behind door A. Whether I switch or not the car is still behind door A. It makes no difference to the odds



Option 1

Assume I am the contestant and Monty is the game show host.

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


1. I choose
2. Monty chooses
3. Then I choose again


There are a set number of outcomes. Note that Monty will ALWAYS choose a goat.


goat = 2/3
goat = 1/2
goat = 1/2 total = 2/6

goat
goat
car = 2/6

car
goat
car = 1/6

car
goat
goat = 1/6



Check total = 6/6

And for success my chances are 3/6 or 50%

This is an improvement because to start with my chances were 1/3 or 33.3%



But there is another option

Monty has no idea where the car is. When it is his time to choose he must choose at random


goat = 2/3
goat = 1/2
goat = 1/2 total = 2/12

goat
goat
car = 2/12

goat
car
car = 2/12

goat
car
goat = 2/12

car
car
car = 1/12

car
car
goat = 1/12

car
goat
car = 1/12

car
goat
goat = 1/12


Check total: = 12/12

The positive results add up to 6/12 or 50%

Again, as I have the option to switch, my chances increase from 33.3% to 50%

But the act of switching or not switching does not change the odds

-

[mavraam]

ORIGINAL: Svennemir

That is one of the most clever geometry puzzles I have ever seen.

I have figured it out so don't read if you don't want a spoiler!!








OK, I warned you!


Here's a hint:

The two triangles do not have the same slope.

Do the math and you will see that the larger triangle
has a slightly lower slope than the smaller one.

That's all I'm saying.

[Rainbow7]

ORIGINAL: Joe 98


Again, as I have the option to switch, my chances increase from 33.3% to 50%

But the act of switching or not switching does not change the odds

-

I don't follow much of what you wrote, but the chances should add to 100%. If your first door only had a 33% chance of being correct, and Monty opens a goat door (which is the only door he'll open) which then takes on a value of 0%, the remaining door is left with ~66% chance of having a car.

Listen everyone. Please do a google search of 'monty hall probability' and take a look at the dozens of sites that discuss this problem. It's well established that switching is always recommended.

[Fred98]

But that’s’ the fallacy.

I choose Door A. It does not matter whether I am right or wrong – nobody is going to open the door. The 1/3 chance does not exist. That is your error

Monty opens a door with a goat behind and then asks me to choose again.

I have ½ chance of choosing correctly.

The real problem is that the story line is incomplete. Without the full story we cannot resolve the problem. Example

After my first choice, is the door opened? If it is the car do I automatically win?

Does Monty choose a door at random or does he always open a door with a goat behind?

Theorise:
I choose a door at random and the door is opened – its a goat
Monty opens the second door with a goat – because he knows where the goats are
The third door is a car so I can’t lose


Theorise
I choose a door at random. It remains closed
Monty opens a door with a goat – because he knows where a goat is
The third door has either a goat or a car – 50% chance of my choosing the car


Theorise
I choose a door at random. It remains closed.
From the remaining doors, Monty chooses one at random and opens the door
If its a car I lose – a 50% chance
If its’ a goat, then there are 2 doors remaining. I have 50% chance of getting it right.


Every possibility points to 50%
-

[dinsdale]

ORIGINAL: Joe 98

But that’s’ the fallacy.

I choose Door A. It does not matter whether I am right or wrong – nobody is going to open the door. The 1/3 chance does not exist. That is your error

Monty opens a door with a goat behind and then asks me to choose again.

I have ½ chance of choosing correctly.

The real problem is that the story line is incomplete. Without the full story we cannot resolve the problem. Example

After my first choice, is the door opened? If it is the car do I automatically win?

Thats the key, the way the question is phrased it assumes that you do not lose or win on your first selection so it was never 1:3 and always as you say 1:2

[Rainbow7]

ORIGINAL: Joe 98

But that’s’ the fallacy.

That is your error

The real problem is that the story line is incomplete.

Theorise:
I choose a door at random and the door is opened – its a goat

Every possibility points to 50%
-

Okay, I pretty much give up. Please do it the brute force way with a friend. But note that the problem IS complete as stated. Your one example above never happens because Monty (of course) never opens your door. That defeats the purpose. You guys are correctly trying to update probabilities with new information. The real problem here is that you're doing it without the benefit of Bayesian theory, which is all about updating probabilities. There is no fallacy. There is no trick. It's our usual everyday reasoning that fails us in this situation, as common reasoning almost always does when dealing with long-run probabilities.

[mavraam]

OK,

Let's try it this way:

We'll calculate the expected value of each of the strategies.

Here's how you calculate the expected value:

1) You enumerate all possible scenarios and the probability of them happening. The total of all the probabilities must == 1 or you have something wrong.

2) For each scenario, you multiply the value of the outcome by the probability of it happening.
3) You sum these together to get the expected value.

We will assume there are 3 doors labeled A,B,C. We always pick door A first but it would be the same no matter which door we pick. The car can be behind any of the 3 doors therefore:

Ending up with the car has a value of 1, ending up with the goat has a value of 0.

Strategy A: NEVER SWITCH DOORS.
This means we will always pick door A, then Monty will always open one of the other doors and we will always stick with door A.

1) Enumerate all possible scenarios:

S1: Car is behind door A. (1 in 3 chance = .3333)
S2: Car is behind door B. (1 in 3 chance = .3333)
S3: Car is behind door C. (1 in 3 chance = .3333)
Total: .9999#

2) Multiply the probability of each by the value of the outcome:

S1 = 1/3 * 1 = 1/3 (We pick A, keep A, car is behind A, we win)
S2 = 1/3 * 0 = 0 (We pick A, keep A, car is behind B, we lose)
S3 = 1/3 * 0 = 0 (We pick A, keep A, car is behind C, we lose)

3) Sum them up end we get 1/3 or .3333.

This means strategy A has an expected value of .3333 which is the equivalent of saying you have a 33.33 % chance of winning the car.

Strategy B: ALWAYS SWITCH DOORS.
This means we will always pick door A, then Monty will always open one of the other doors and we will always switch door A with the remainin door.

1) Enumerate all possible scenarios:

S1: Car is behind door A. (1 in 3 chance = .3333)
S2: Car is behind door B. (1 in 3 chance = .3333)
S3: Car is behind door C. (1 in 3 chance = .3333)
Total: .9999#

2) Multiply the probability of each by the value of the outcome:

S1 = 1/3 * 0 = 0 (We pick A, Monty flips B or C, we switch to other, we lose)
S2 = 1/3 * 1 = 1/3 (We pick A, Monty flips C, we switch to B , Car is behind B, we win)
S3 = 1/3 * 1 = 1/3 (We pick A, Monty flips B, we switch to C , Car is behind C, we win)

3) Sum them up end we get 2/3 or .6667.

# (effectively 1.000 because of rounding error)

------------

This means strategy B has an expected value of .6667 which is the equivalent of saying you have a 66.67 % chance of winning the car.

If you don't have a background in probability but know someone who does, have them verify or contradict my calculations.

[mavraam]

There is no fallacy. There is no trick. It's our usual everyday reasoning that fails us in this situation, as common reasoning almost always does when dealing with long-run probabilities.

Well said! 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.) Our common reasoning fools ourselves into thinking we 'have a shot' which allows normally rational people to throw away money on losing games.

Did I mention lotteries??? [:'(]