Monty Hall Dilemma - Winning a GTI on a Game Show

[Rocket36]

slightly OT but I have to say it's quite cool that this can be so hotly discussed by so many without tempers/attitudes flaring - keep going guys! LOL!

[Dubya]

No. I think you seemed to have misread my point in it's entirety.

Step back once step.

The game show host will ALWAYS show you a goat - regardless wether you picked a goat or the GTI.

It simplifies down: You never had three options, only two. you can only choose a goat or GTI (two options not three). hence why uou have a 50/50 chance, and hence why it comes down to chance/luck. (regardless of wether you change your choice after the first illusion.)

(to eleborate: regardless of whether you pick a goat or GTI from the first 3 doors, the host can always pick a door with a goat. which eliminates one door and leaves you with two doors> one with a goat, one with a GTI. thats the part where it boils down to 50/50. The host's choice of door isn't dictated at all by your choice, He always chooses a goat, it's merely an illusion to create the insecurity in your choice, more for TV ratings than anything else.)

just my take on it all anyways.

Number of options = number of doors (not number of different outcomes).

Three doors means 3 options (not 2) and 33% chance of choosing correctly the first time and 67% chance of choosing correctly if you switch after all but two doors are opened.

100 doors means 100 options (not 2) and 1% chance of choosing correctly the first time and 99% chance of choosing correctly if you switch after all but two doors are opened.

In neither case does one have a 50% chance.

The only person who has a 50% chance is the casual bystander who enters the stage when only two doors remain shut with no idea which door the contestant original chose (hereinafter referred as the "new entrant").

When all doors are shut, the odds of picking the correct door = 1/n, where n = the number of doors.

But if you come in after the dud doors have been opened and do not know which door the contestant picked, the probability of your guessing the correct door is 50%.

However, if the new entrant finds out which door the contestant picked they should choose the other doors as the probability that the other door is the right one is inversely proportionate to the probability the contestant chose right, ie:

Contestant's probability of being right first time is:

1/n

ie 1/3 or 1/100 where number of doors (n) is 3 or 100, respectively.

Knowing this, the new entrant picks the opposite to the contestant as the odds of that door being right are:

n-1/n

ie (3-1)/3 (67%) or (100-1)/100 (99%) where n = 3 or 100, respectively.

If they are alive to this, the contestant switches too and shares the same probability as the new entrant whose probability is higher by a factor of n-1

ie 3-2 (2 times higher) or (100-1) (99 times higher) where n = 3 or 100, respectively.

Opening dud doors does not alter the probability that the initial selection will be correct. And so the door that remains must have probability 1-(1/n), or 67% or 99% where n = 3 and 100, respectively.

[Dubya]

Sorry I should have been clearer when defining exactly what I was referring to when I was referring to fact. The CHANCE you will win is 50/50 (or 1 out of 2) when there is two choices. THAT is mathematical FACT and cannot be disputed. Since the second choice is only ever from 2 options, you have a 50% CHANCE of winning and a 50% CHANCE of losing.

It is disputed.

Try the simulators or test the theory and and you will see why.

Switching increases your chances of winning by a factor of n-1, where n = number of doors.

[Swallowtail]

Sorry I should have been clearer when defining exactly what I was referring to when I was referring to fact. The CHANCE you will win is 50/50 (or 1 out of 2) when there is two choices. THAT is mathematical FACT and cannot be disputed. Since the second choice is only ever from 2 options, you have a 50% CHANCE of winning and a 50% CHANCE of losing.

Kris - you would be correct IF it was a two-choice scenario. No-one is disputing that in a 2 option situation it is 50/50. However, when you make a first choice, there are THREE options. Regardless of when the first wrong door is opened, your choice was made from three alternates. What happens after that can categorically be proven - mathematically.

Have you actually read the referenced pages on the mathematical probabilities that apply? This is like someone saying that 1+1=3 - you can believe it all you want, but it isn't true.

The mathematical equation for these probabilities is correct at 67% switch 33% stick. I have twice provided the formula in this thread, but the 50/50s seem to refuse to look at it, or to run the simulations - either online or themselves if they don't trust online ones.

There are numerous pages on the web showing the proof. This isn't theory it's proven math. Just because it seems counter-intuitive at first doesn't make it false.

For example:

http://mathworld.wolfram.com/MontyHallProblem.html shows in a four door situation sticking until the last choice gives you 75% chance, etc.

http://montyhallproblem.com/

This presents an interesting outlook on why people struggle to grasp it: http://www.montyhallproblem.com/philo.html

Monty Hall himself said: "Oh and incidentally, after one [box] is seen to be empty, his chances are no longer 50/50 but remain what they were in the first place, one out of three. It just seems to the contestant that one box having been eliminated, he stands a better chance. Not so." (http://www.montyhallproblem.com/as.html)

http://mathforum.org/dr.math/faq/faq.monty.hall.html

Could go on for pages and pages but hey, if you don't want to or can't be bothered to understand then we are just wasting keystrokes...

[Jarred]

Number of options = number of doors (not number of different outcomes).

Three doors means 3 options (not 2) and 33% chance of choosing correctly the first time and 67% chance of choosing correctly if you switch after all but two doors are opened.

100 doors means 100 options (not 2) and 1% chance of choosing correctly the first time and 99% chance of choosing correctly if you switch after all but two doors are opened.

In neither case does one have a 50% chance.

....

I'm sorry, but that has nothing to do with the original problem.

My comments were based on the original problem. I'm not arguing the maths behind it or anything, and yes I understand the 66% and everything, hence why my comments were 'my take on it all".

Your post was based on the maths behind the problem (which I'm not arguing) by post was based on the situation itself.

"
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a GTI; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to switch from door No. 1 to door No. 2?" Is it to your advantage to switch your choice?
"

Do you understand my point?

[Treza360]

Ah statistics, seemingly witchcraft and magic.

At the end of the day Dubya's original post is correct and the mathematics for this problem which is all over the internet (not just Wikipedia) and in mathematical text books is sound. You have better odds of winning the car/prize by switching.

Of course though the 2/3 answer is an ideal gameshow where they always open a second door with the dud prize and offer the switch.

In reality it is most likely that they will offer the switch everytime when you've already picked the good prize and then offer the switch a proportion of the time when you have picked the dud prize.

Because no one but the show producers will know what that proportion is, it now makes it impossible to decide whether to switch or not. If the proportion value is greater than 1/2 then make the switch but if it's less than or equal to 1/2, stick with what you originally had.

Oh and I'm sure many smarter people than any of us have tried to disprove this but the problem is is that in over 250 years, Bayes' theorem has remained sound and valid.

On a side note I'm so glad I became an Engineer and not a Mathematician, having to understand the formulas makes my head hurt let alone coming up with the formulas which would make it explode.

Cheers,
Trent

[Spilledprawn]

Enough postulating.

I've tested the switching theory and it works. Online and with cards. It works with dogs and cats (I didn't have a goat or GTi)

Any 50/50's care to share the results of their testing?

Stop arguing why you should bother doing the experiment, do it if only to show us we're wrong (make sure you use at least three options, and not two as suggested by some nay sayers, that is kind of the point!).

[Timbo]

Rocket, as you live in Canberra, as do I, do you fancy setting up a session at the Casino, as a custom 3 card game, following Monty's rules? I'll run the switch strategy, based on my understanding of the game, where i should win 67%, and you can stick to your original choice strategy, where you think you'll win 50%.

We'll run it over, say, 100 rounds each at $2 a round. The biggest loser pays for drinks at the end. Up for it??

[Jarred]

Rocket, as you live in Canberra, as do I, do you fancy setting up a session at the Casino, as a custom 3 card game, following Monty's rules? I'll run the switch strategy, based on my understanding of the game, where i should win 67%, and you can stick to your original choice strategy, where you think you'll win 50%.

We'll run it over, say, 100 rounds each at $2 a round. The biggest loser pays for drinks at the end. Up for it??

but who would be the one that removes ( and knows which ones are) the 'goat' cards? there's ya problem.

[Treza360]

Your post was based on the maths behind the problem (which I'm not arguing) by post was based on the situation itself.

Jarred, see my post above. I think this should cover your problem with the actual circumstances.
Cheers,
Trent

[Jarred]

Jarred, see my post above. I think this should cover your problem with the actual circumstances.
Cheers,
Trent

Too right Trent.

I'm just trying to get people to see the different was of problem solving & the differences between the pure maths & real situations.

I'm also glad I'm not becoming a mathematician either!

I was going to make a post about the assumptions involved in the real problem & mathematical problem, but then my head started to hurt, so I didn't bother.

Thats enough posting from me, it's a friday afternoon and my head hurts! haha

[Dubya]

but who would be the one that removes ( and knows which ones are) the 'goat' cards? there's ya problem.

The "house".

Timbo will be the house and give Rocket odds of, say, 2.5:1 on the basis Rocket is not allowed to switch (which he does not want to do as he thinks he has a 50/50 chance either way).

Rocket will accept these odds as he believe he has a 50% chance of winning if he does not switch and so believes at 2.5:1 he will come out in front.

Timbo will come out in front as Rocket will only win 33% of the time.

It would be amusing indeed if it were the prospect of losing real dollars that made the penny drop and Rocket realise his odds are stuck on 33% if he does not switch!

And if Rocket's chance is 33%, the house's chance (and the house sits on the card Rocket could have switched to) must be 67%.

QED.

[Timbo]

but who would be the one that removes ( and knows which ones are) the 'goat' cards? there's ya problem.

We get a neutral dealer/croupier, hence running it at the casino. Rocket will be sure to tip her/him with his winnings

[Timbo]

Hmmm....I like Dubya's idea MUCH better than mine

Mate, you come down and observe -- I'm sure Rocket will stand drinks for you, too

[Treza360]

The "house".
Timbo will be the house and give Rocket odds of, say, 2.5:1 on the basis Rocket is not allowed to switch (which he does not want to do as he thinks he has a 50/50 chance either way).
Rocket will accept these odds as he believe he has a 50% chance of winning if he does not switch and so believes at 2.5:1 he will come out in front.
Timbo will come out in front as Rocket will only win 33% of the time.
It would be amusing indeed if it were the prospect of losing real dollars that made the penny drop and Rocket realise his odds are stuck on 33% if he does not switch!
And if Rocket's chance is 33%, the house's chance (and the house sits on the card Rocket could have switched to) must be 67%.
QED.

That is if they play for long enough though. It is still possible for Rocket to come out ahead. That's the joy of statistics.

For example I played 100 rounds on that website linked to earlier and I came back with only 63%. Yes if you played forever the house would come out ahead but there is still the possibility that in only a "small" amount of games the player might come out ahead by not switching.

If they play for long enough though, the house will win.

Cheers,
Trent