Monty Hall Dilemma - Winning a GTI on a Game Show

[Dubya]

sorry dubya, i simply cant.

this is going to sound pretty claimish on my part, but no one has answered my question- if im really going to do this a thousand times with some cards, but knowing that the first one, as a rule, must be a failure, why would i bother even picking it up? i'll just deal with switching between the two remaining cards, where my chances are 1 in 2.

the 50/50ers were accused of not having the courage, bravado, nor the confidence blah blah to try this at home. i will. but i would like to know first, when im only going to be dealing with cards 2 and 3, not bothering to pick up card 1 because everyone knows it must fail- how will this prove in real life that the chances are 2/3?

Try it. But you need two people. Or try one of the online simulators.

Either way you will see a trend within 20 - 30 attempts:

If you never switch you will see that you win around a third of the time, as you would expect.

If you always switch, you will see that you win around two-thirds of the time, as you might not expect.

[Spilledprawn]

Dubya, perfect description. I'm sorry to say that if the 50/50s haven't got it yet, it is a lost cause.

[gareth_oau]

try the Poll

Ive just created a poll to see what the population thinks.

Please cast your switch/not switch vote!!

http://www.vwwatercooled.org.au/newforum/upload/showthread.php?t=42494

[Buller_Scott]

Dubya, perfect description. I'm sorry to say that if the 50/50s haven't got it yet, it is a lost cause.

the funny thing is, and not to pick any bones at all with anyone (i dont take any disagreement etc as an affront- this is actually a pretty worthy discussion imo), but i have posted two scenarios, one theoretical, and one practical.

and yet i dont see anyone disproving my conclusion RE the dub fest, the gti and the twenty raffle tickets reduced to ten- i dont see anyone proving that, should person A be allowed to switch until half the raffle tickets have been eliminated, that person A's chances of winning that GTI are now 55% or 11/20.

i feel i was pretty comprehensive in my logic, and i would have welcomed rebuttal, but no one challenged my numbers in that scenario at all.

then, i alluded to actually sitting at home and trying this 1000 times, but encountering the problem of KNOWING for a fact that the first card must fail, and thus my partner (the host) not even bothering to waste the energy to pick that first card up anymore- probability dealing with chance, i find it ironic that this first card is even relevant anymore as.... well..... there is no "chance" left- the rules dictate that it MUST NOT be the card i chose in the first place. it's no longer a "probable", and yet why do people insist that it must be counted?

so, spilledprawn- whilst some of the 2/3 gang are quick to dismiss the 50/50 guys as ''a lost cause'', i'd like to invite YOU to address my question- im going to do this 1000 times. im not even gonna bother picking up 3 cards- just the TWO THAT MATTER, before i get offered the switch. disregarding the third card as a waste of time seeing as it must, by rule, fail- how are my chances 2/3?

and please dont do what some of the others have done: "aww you dont get it, you're crap at math, you dont understand how it works, lol@your explanation that i will not acknowledge with any decent counter argument".

p.s. not to come across as uppity or a sourpuss at all, i just reckon that the 50/50 argument is well established, but i cannot see too many people challenging our explanations directly (namely, my two explanations as mentioned above).

i really really really really want someone to take my two scenarios and tear them apart- at this point of the thread, it'd be somewhat of a breath of fresh air.

cheers,

scotty

[The_Hawk]

I think Dubya said it best here: http://www.bigpond.com/internet/plan...d=bph-access-3

In short it's the culmative effect that make the probability of winning higher if you switch.

I'm not trying to offend, just want to try and lay out the logic in this scenario, to illustrate mote clearly, lets start with 100 doors, 99 goats and one GTI.

Your chances of choosing the GTI are 1/100 (and of getting a goat are 99/100).

The host then opens the doors one at a time, finally getting down to the final two doors then gives you the choice to switch.

When we started you had a 1% chance of being right and a 99% chance of being wrong. Now that the host has dragged things out to this point, slowly showing you all the goats you original odds haven't changed, there is still a 1% chance your first choice was right and a 99% chance you chose wrong.

So now, even though you have only two doors left, there is a 99% chance your door is still the wrong one, it's in your best interests to switch.

Lets say at this point they drag in a bystander off the street and say he gets what veers behind the other door. He has a 50/50 chance of getting a GTI since by the time he comes in there is only two options left.

So you see, it can be two things at the same time.

[team_v]

I'm not trying to offend, just want to try and lay out the logic in this scenario, to illustrate mote clearly, lets start with 100 doors, 99 goats and one GTI.

Your chances of choosing the GTI are 1/100 (and of getting a goat are 99/100).

The host then opens the doors one at a time, finally getting down to the final two doors then gives you the choice to switch.

When we started you had a 1% chance of being right and a 99% chance of being wrong. Now that the host has dragged things out to this point, slowly showing you all the goats you original odds haven't changed, there is still a 1% chance your first choice was right and a 99% chance you chose wrong.

So now, even though you have only two doors left, there is a 99% chance your door is still the wrong one, it's in your best interests to switch.

The thing is though, you have changed the probablitlities as the other doors are open and thus, their probablility shouldn't be included since they are no longer a probablility but a certainty.

As said earlier, it is a dynamic situation and the probabilities change to reflect that.
Each time a choice has to be made, the probabilities are reset and adjusted to suit the remaining choices.
It is simply not reasonable to say out of 2 choices, you have a 1/3 chance of getting a goat as there are only 2 options remaining (due to 1 door being opened and revealing the goat)

Just my view point though to back the 50/50.
Computer applications will never be truly random.

So you see, it can be two things at the same time.

*Insert schroedinger's cat*

[Dubya]

the funny thing is, and not to pick any bones at all with anyone (i dont take any disagreement etc as an affront- this is actually a pretty worthy discussion imo), but i have posted two scenarios, one theoretical, and one practical.

and yet i dont see anyone disproving my conclusion RE the dub fest, the gti and the twenty raffle tickets reduced to ten- i dont see anyone proving that, should person A be allowed to switch until half the raffle tickets have been eliminated, that person A's chances of winning that GTI are now 55% or 11/20.

i feel i was pretty comprehensive in my logic, and i would have welcomed rebuttal, but no one challenged my numbers in that scenario at all.

then, i alluded to actually sitting at home and trying this 1000 times, but encountering the problem of KNOWING for a fact that the first card must fail, and thus my partner (the host) not even bothering to waste the energy to pick that first card up anymore- probability dealing with chance, i find it ironic that this first card is even relevant anymore as.... well..... there is no "chance" left- the rules dictate that it MUST NOT be the card i chose in the first place. it's no longer a "probable", and yet why do people insist that it must be counted?

so, spilledprawn- whilst some of the 2/3 gang are quick to dismiss the 50/50 guys as ''a lost cause'', i'd like to invite YOU to address my question- im going to do this 1000 times. im not even gonna bother picking up 3 cards- just the TWO THAT MATTER, before i get offered the switch. disregarding the third card as a waste of time seeing as it must, by rule, fail- how are my chances 2/3?

and please dont do what some of the others have done: "aww you dont get it, you're crap at math, you dont understand how it works, lol@your explanation that i will not acknowledge with any decent counter argument".

p.s. not to come across as uppity or a sourpuss at all, i just reckon that the 50/50 argument is well established, but i cannot see too many people challenging our explanations directly (namely, my two explanations as mentioned above).

i really really really really want someone to take my two scenarios and tear them apart- at this point of the thread, it'd be somewhat of a breath of fresh air.

cheers,

scotty

I for one have not turned my mind to your purported analogy as we have a perfect analogy in the three playing cards and with the scenarios where there are more than three doors and only one GTI.

Accordingly, if it is perfectly analogous it is otiose and if it is not perfectly analogous it is irrelevant. In any event, none of us wants a second debate about whether it is perfectly analogous or not. There is simply no need to consider it while there is no overwhelming consensus on the main issue.

A number of people, most lately Hawk (a sceptic intitially, if I may say) with gleaming logic - but with incorrect link to my post, have explained how the higher odds from switching are as certain as a 50/50 long-run average from flipping a coin.

The 67ers are asking the 50/50s to simply apply the highly-testable theory. We're not talking about life on Mars here!

If the doubters do, they will see that if they never switch they will "win" ~33% of the time and if they always switch they will win ~67% of the time.

So instead of arguing until you're blue in the face, test the theory.

All of the 67ers have satisfied themselves in one or more ways such as:

- the ever more compelling logic as the number of doors is increased above 3 (with still only one GTI);

- used online simulators - the randomness of which is irrelevant as the only randomness is the order in which the cards are laid - you determine which card you will pick initially after the computer has determined what each card hides (or at least a proper simulator will);

- using three playing cards (one Ace and two Jokers, say) for 10 minutes with another person to test the theory.

By now it should only be a question of which of the 50/50s is going to take 10 minutes to test the theory and be first to share their enlightenment.

As opposed to arguing ad nauseum against a proposition that is so easily verifiable as being true in theory and reality!

So enough with theory guys, it's time to get real!

[team_v]

So enough with theory guys, it's time to get real!

And the reality is that it's 50:50.

[Dubya]

And the reality is that it's 50:50.

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).

They certainly wouldn't keep arguing tit for tat if the theory could so easily be disproved.

So as it stands, the 67ers are asking (by now pleading with) the 50/50s to submit the theory to the test and the 50/50s, team v for one at least, appear to be reluctant to do so.

Why the reluctance to prove the theory one way or the other by spending 10 minutes with some playing cards or an online simulator?

PS - just so long as we do not get any "you didn't explain" / "I didn't realise" excuses when the penny does drop...

[Rocket36]

And the reality is that it's 50:50.

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).

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.

[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.