Monty Hall Dilemma - Winning a GTI on a Game Show

[Coaster]

It's pretty easy to demonstrate. You have 3 options; A door, B door, and C door. When one losing door is eliminated, switching doors will result in winning 2 out of 3 times. When you don't switch you only win 1 out of 3 times. Try it!

But there are is a variable that must be eliminated first; you have to pick a door before the game show host picks a door.

[Dubya]

It's pretty easy to demonstrate. You have 3 options; A door, B door, and C door. When one losing door is eliminated, switching doors will result in winning 2 out of 3 times. When you don't switch you only win 1 out of 3 times. Try it!

But there are is a variable that must be eliminated first; you have to pick a door before the game show host picks a door.

That order is certainly a dependency for the contestant's odds to increase from 33% to 67% through switching.

A first choice made after the host has revealed a goat would provide equal odds for success and failure.

No word from Rocket, et al, of late...

[Rocket36]

Can someone please tell me where I've said I wouldn't switch and that I have a 50% chance of winning? ALL I HAVE SAID is that when you have a choice between two options, with ANYTHING the result is one or the other, or 1 in 2, or 50:50. Therefore, the second choice has a 50% chance of picking the car. and a 50% chance of picking the goat.

Give someone two choices, get them to pick one only and they hav e a 50:50 choice. THAT is basic maths and if you don't understand that I'm sorry, go back to primary school.

Oh, and I live in Melbourne at the moment.

[abreut]

Can someone please tell me where I've said...that I have a 50% chance of winning?

Isn't that what you are saying right here, 2 sentences later?

Therefore, the second choice has a 50% chance of picking the car. and a 50% chance of picking the goat.

If the second choice has a 50% chance of winning, then presumably the first door/choice has a 50% chance as well because there are only 2 doors to choose from and the car has to be behind one of them, 100% - 50% = 50%. So you ARE effectively saying that you have a 50% chance of winning.

Anyway you are still wrong, yes there are 2 doors, 2 "choices", 2 "options" but one of those is twice as likely to have a car behind it than the other. Just because there are 2 options doesn't mean that they have to have the same probability of winning, and in this case they don't.

[Timbo]

Give someone two choices, get them to pick one only and they hav e a 50:50 choice. THAT is basic maths and if you don't understand that I'm sorry, go back to primary school.

[NickZ]

I find the easiest way to think about it is to apply the situation to 1,000 doors. The chances of you picking the right door first up is 1 in 1000. If the host then, after you picking a door, eliminates 998 of the options, only leaving two doors left, and offers you to switch, then seeing as your initial chance of picking the right door initially was minute, then it is strongly in your favour to switch (i.e. 999 out of 1000 times you will win!)

Yes, we can talk about the dealers motives, that chance comes into play, and other variables which will ultimately have an impact. These things have a much more significant impact when the choice is only 3 doors as opposed to when there is more doors. But the Monty Hall is purely to explain that as a mathematical equation and excluding other variable factors, the odds are in your favour to switch.

It is a bit like poker, each situation there is a mathematical formula giving you the probability of a particular hand winning, but ultimately other variable factors such as luck, skill, bluffing, pressure etc impacts on the pure mathematics behind the game.

[Achtung Tiggy]

ALL I HAVE SAID is that when you have a choice between two options, with ANYTHING the result is one or the other, or 1 in 2, or 50:50.

So if Australia plays Brazil in the final of the World Cup, does that mean we have a 50% chance of winning? (Since there are only 2 option:win or lose, draw is not permitted).

I hope you don't work for Tabcorp!

[Dubya]

Can someone please tell me where I've said I wouldn't switch and that I have a 50% chance of winning? ALL I HAVE SAID is that when you have a choice between two options, with ANYTHING the result is one or the other, or 1 in 2, or 50:50. Therefore, the second choice has a 50% chance of picking the car. and a 50% chance of picking the goat.

Give someone two choices, get them to pick one only and they hav e a 50:50 choice. THAT is basic maths and if you don't understand that I'm sorry, go back to primary school.

Oh, and I live in Melbourne at the moment.

Rocket, whatever the case, you've not yet agreed with those of us who have said that you should switch after the first goat is revealed because doing so doubles your chances from 1/3 to 2/3.

And as far as I can tell from your post above, you still believe you have a 50% chance of winning if you stick with your original choice and this, as has been explained in many different ways, is not correct.

You now appear to be suggesting that you would be willing to switch even though you believe your odds would not change.

If you believe that, why would you switch?

As you claim there is no benefit to switching, I cannot understand why it was unreasonable for anyone to infer that you would be disinclined to go against your initial instinct.

After all, if you believe the odds will not change whether you switch or not, most people would probably be more disappointed if they switched and lost than if they stuck with their initial choice and lost.

Whatever the case, as you seem to be firmly of the belief that the odds are 50% after the first goat has been revealed, your position on switching would seem to be more influenced by whimsy than by reason.

[GTom]

I look at it this way.

You have a 2/3 chance that the door you picked is incorrect.

When offered the chance to change the first door is out of the eqn. as you can't pick that door again you're left with a 1/2 chance the prize is behind the door.

1/2 < 2/3 chances of loosing
1/2 > 1/3 chances of winning

So you have a greater chance of not getting the prize with 3 doors in play.

That said you might have been lucky enough to pick the right door to begin with

I personally wouldn't change though....

[Treza360]

I look at it this way.
You have a 2/3 chance that the door you picked is incorrect.
When offered the chance to change the first door is out of the eqn. as you can't pick that door again you're left with a 1/2 chance the prize is behind the door.
1/2 < 2/3 chances of loosing
1/2 > 1/3 chances of winning
So you have a greater chance of not getting the prize with 3 doors in play.
That said you might have been lucky enough to pick the right door to begin with
I personally wouldn't change though....

Did you even play the games linked to earlier in the thread?