Monty Hall Dilemma - Winning a GTI on a Game Show

[team_v]

But how do you know for sure something that is contentious if you have not tested the theory with a simple experiment, just as any researcher with a theory would do (if they possibly could).

An application is too easy to modify to say what you want it to say.
The only way you could test it is to do the card trick which i don't have handy.


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

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

[Swallowtail]

I am enjoying watching this....

I agree with Dubya - and the Wikipedia page with its excellent explanation. The situation is not a 50/50, as elements of the original 67/33 apply, and the pre-knowledge of the host of which door holds the car changes the scenario. It is not a "fresh" situation once the first door is removed.

Read through the Wikipedia page with its full explanation, not just the quick one.

This is the math of the probability of winning by switching:

[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?

[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 . . .

[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.

[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)

[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]

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.

[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?

[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.