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]

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

Please cast your switch/not switch vote!!

http://www.vwwatercooled.org.au/newf...ad.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.