Monty Hall Dilemma - Winning a GTI on a Game Show

[Dubya]

There is no need to test something that is mathematical FACT.

ANY choice where there are ONLY 2 OPTIONS is a 50/50, or 1 in 2 chance... It's not a theory.

On the contrary . . .

In theory, if you toss a coin 100 times you will get 50 heads and 50 tails.

In reality you might get something like 44 heads and 56 tails.

So, the "fact" that coin tosses result in a 50/50 outcome is only a theory, not a fact.

The theory is if you choose a door and stick with it you will win 33% of the time.

In reality, you might lose three times in a row and so have 0% success.

In theory, if you switch doors after the goat is revealed, you will win 67% of the time.

In reality, if you switch doors three times consecutively, you might win 3/3 times and so have 100% success.

The only FACTs are:

- There are two theories:

- One is a counter-intuitive theory that is founded in logic (the switching theory); and

- another theory that is based on flawed logic/intuition and says that it makes no difference whether you switch.

Now Coreying wrote elsewhere that what is logical depends on your "perspective".

I do not think logic is subjective.

However, if Rocket would get out three playing cards (2 x same, 1 x different) he would see in very short order that his theory is not supported by reality, whereas the 67% theory holds more or less true.

A shame he won't even test his own theory.

[Jarred]

by my train of thought, switching is an emotional response. The two doors have a 50/50 chance of having the prize, so really, it comes down to luck! A switch would only because by some insecurity/nerves/reverse physcology cause by something that's outside of the simplified problem.

Basically: Do you feel lucky, punk? Well. Do ya?

[Dubya]

Yea Sorry, I didn't bother to read all 14 pages of theories and such. I just thought I'd throw mine in there to mix it up further.

I approached the problem from an engineering perspective, rather than a mathematical one. Hence why I 'simplified' the problem, before 'solving' it.

Does that mean you have "switched"?

[team_v]

True...and it was yellowcake, too

That settles it, the cat is dead (In this universe anyway)

[Timbo]

Because there is no cake!

Schroedinger's cat ate it all.

True...and it was yellowcake, too

[Dubya]

You only ever had a 50% chance of winning.

Shouldn't that be 33%? Three doors, only one with a car.

I thought there was consensus that upon first choosing a door there was a 33% chance of picking a winner.

The thing the 50/50s have not explained is how the odds your first pick was correct rise from 33% to 50% upon learning which of the two other doors hides a goat.

You already knew one of them hid a goat. How does knowing which doors hides the goat make your first pick a 50% chance.

That information only has value if you switch to the other, unopened door, effectively allowing you to cover two doors instead of just one. So a 67% chance if you switch.

Cause regardless, the host will open a door, showing a goat. REGARDLESS. (cause he knows where they are)

IF the host RANDOMLY chose a door to open, (and picked a goat) then it would be be different.[/quote.]

You seem to be suggesting that if the host reveals a goat, the odds are different depending on whether he knew the goat was behind the door before he opened it.

But since the host will also show you a goat (regardless of whether you picked a goat or the GTI on the first choice) Which leaves you with two choices. You will always get to the point where you have only two choices. First part of the game is a there only for reverse physchology. (spelling)

That's right - two choices, but each has a different probability of yielding a favourable result.

Sticking with your first guess give you a 33% chance of winning the GTI - nothing can change this. Knowing which of the two other doors hides a goat makes no difference if you decide not to switch as you already knew one of them hid a goat. Whether it is door 2 or 3 makes no difference to your odds.

Switching to the two other doors, one of which just happens to be open, offers a probability of 67% of winning the GTI. Because they cover 67% of the field whereas your first choice covers 33% of the field.

You always had a 50% chance of winning.

Surely not 50% when you pick one from three?

(note this is a game of chance, not probability, no 50/50 game is probability ie coin toss)

While I am uncertain of the distinction, and game of chance it certainly is, there are probabilities associated with each possible outcome and there is a 67% chance one of the two doors you did not choose will hide a goat whether the host has opened one those to doors or not.

[Swallowtail]

This IS a mathematical issue. Nothing else. And it's a simple one once you look past your initial response. The simulator Timbo links to is spot on - 41 from 60 wins with switch, 22 from 60 wins with stick. Higher numbers and it would approach the true probability of 66.7% even closer.

Regardless of whether one door WILL be opened, you are choosing from three. If the math is beyond you or you can't be bothered reading it and trying to understand it then at least write out the possibilities so you will see the options...

[team_v]

Try the simulator. Stick with your original choice and keep repeating. You'll only win c.33.3% of the time. But if you switch each time, after the host reveals the goat, you'll win c.66.7%! So, why is that?

Because there is no cake!

Schroedinger's cat ate it all.

[Timbo]

You only ever had a 50% chance of winning.)

Try the simulator. Stick with your original choice and keep repeating. You'll only win c.33.3% of the time. But if you switch each time, after the host reveals the goat, you'll win c.66.7%! So, why is that?

[Jarred]

Yea Sorry, I didn't bother to read all 14 pages of theories and such. I just thought I'd throw mine in there to mix it up further.

I approached the problem from an engineering perspective, rather than a mathematical one. Hence why I 'simplified' the problem, before 'solving' it.

[team_v]

You only ever had a 50% chance of winning.

Cause regardless, the host will open a door, showing a goat. REGARDLESS. (cause he knows where they are)

IF the host RANDOMLY chose a door to open, (and picked a goat) then it would be be different. But since the host will also show you a goat (regardless of whether you picked a goat or the GTI on the first choice) Which leaves you with two choices. You will always get to the point where you have only two choices. First part of the game is a there only for reverse physchology. (spelling)

You always had a 50% chance of winning. (note this is a game of chance, not probability, no 50/50 game is probability ie coin toss)

That is exactly what i have been trying to argue.
The door was always going to be opened to reveal a goat so the choice will be one of 2 doors which is a 50/50 shot.

[Jarred]

You only ever had a 50% chance of winning.

Cause regardless, the host will open a door, showing a goat. REGARDLESS. (cause he knows where they are)

IF the host RANDOMLY chose a door to open, (and picked a goat) then it would be be different. But since the host will also show you a goat (regardless of whether you picked a goat or the GTI on the first choice) Which leaves you with two choices. You will always get to the point where you have only two choices. First part of the game is a there only for reverse physchology. (spelling)

You always had a 50% chance of winning. (note this is a game of chance, not probability, no 50/50 game is probability ie coin toss)

[Timbo]

Of course the odds in any binary choice scenario are 50:50, such as a coin toss, or the choice between two identical doors.

But that's not the Monty Hall game scenario.

The maths is well-described in the Wikepedia article, can be demonstrated with a simulator and (if you think that's rigged), using the three card method described by so many above.

The answer to the dilemma is that you should switch from your original choice because the probability of success is 2 in 3. It is 2 in 3 because of the conditional probability created by the host's intervention, and the rule of the game that says the host will always reveal a goat.

To the 50:50 guys, I commend you for your passion and determination, and I suggest that you read the literature available on this problem (there was a great discussion in the New York Times, where you'll find yourselves in excellent company). Some of those really, really brainy mathematicians -- you know the guys -- really prominent foreheads, a sliderule and 3 RPN calculators in their pockets -- also went down this path -- equally passionately, I might add. In the end, even after arguing the rule about the role of the game host not being clear, they've all conceded that the 1 in 2 probability is wrong, and that the probability of success if you switch is 2 in 3.

[Dubya]

An application is too easy to modify to say what you want it to say.

Maybe so, but in this case it is quite safe to rely on the integrity of the university lecturers, professors and students who have built the online simulators. After all, switching does not guarantee a win on any of the simulators I have seen and the results are not entirely predictable.

The only way you could test it is to do the card trick which i don't have handy.

Substitute 3 x identical objects (eg business cards) with a tiny mark on one of them.

You simply cannot keep the probabilities from a 1/3 situation when applying it to only 2 doors, it just doesn't work like that.

Get back to me after you have the test results.

The situation has changed and so should the probabilities to reflect the change in variables.

They don't.

So if you flip a coin, do you have a 50/50 chance of getting heads or tails or a 33/67 chance?

I am beginning to wonder . . .

[Dubya]

There is no need to test something that is mathematical FACT.

ANY choice where there are ONLY 2 OPTIONS is a 50/50, or 1 in 2 chance... It's not a theory.

No, it's not a mathematical fact, it is simply a theory until its is proved by testing it.

(And you are not choosing between two options, you choose between 3, giving your first choice a 33% chance of being right from the outset. These odds do not rise to 50% just because the other two cards are not revealed at exactly the same time.

So whether the facilitator reveals the two cards 10ms apart or 10 seconds apart, the chances that one of the other two cards conceals the winning card is 67%.

So whether you dessert your first choice in favour of the other two cards before or after the dud is revealed still leaves you with a 67% chance of winning if you switch.)

(Anyway) on the other hand, the 67ers theory has been shown to be true many times by, I imagine, millions of people. The 50/50 theory, conversely has been disproved as many times.

When a facilitator lays out three playing cards face down, you will pick the sole Ace first time 33% of the time.

If, after the "house" reveals one of the two jokers and you switch to the other card, you will win 67% of the time.

Switching to the other card after one of the Jokers is revealed gives you the same probability of picking the winning card as switching to the two other cards before the Joker is revealed.

However the revelation of the Joker guides your choice to the better of the two other cards.

Many of us have tested this theory and shown the 67% win rate using a switching strategy is a mathematical fact.

The 50/50 theory has been disproved in practice.

Why don't you try?

In the face of the considerable amount of logic, reason and argument submitted on this thread against your strongly-held belief in your theory, a continuing refusal to put both theories to a simple, 10-minute test and put the question beyond any doubt is bemusing.

That the 67% theorem is acknowledged to be "counter-intuitive" alone should give the 50/50s pause.

Fortunately, a test to prove the theory is within everyone's reach.

So if you believe it to be true, why not prove the 50/50 theory seeing it can so easily done?