The Monty Hall Problem: Help!

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A hypothetical . (I love hypothetical)

I made an assumption with all the previous discussions, that Monty changes the door between games.

But let us suppose that the car is always behind the same door; say door A.

There are 100 contestants away in a room somewhere and they are called out one by one to play. Then they leave without being able to contact any other contestants.

So, now the only variable avaliable is the 1 in 3 choice the contestant makes. As both other doors have goats every time.

Monty knows this, but obvioulsy the contestants don't.

So, will the effect be the same.

Should anyone like to write a program to run this I'm sure we would be duly appreciative.
 
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Anyway message is, if you do do a test again take out the 50:50 bit.
Having seen the exchange since I started writing this, this may not help your understanding. But it helps mine.

O
The two ball choice on the second run represents my random choice of stick or swap.
 

oxo66

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What is interesting to me is that apparently in real life Game Show scenarios, so few Contestants switch when given the choice.
Yes you are right. The psychology is the really interesting bit. People generally value more highly something they already have over something they haven't got yet. I think psychologists have a term for it.

By swapping
Your new decision is effectively betting that you were initially wrong.
and people find it hard to do that.

in simple terms - people feel that having the correct answer and throwing it away is worse than correcting an initial wrong answer is 'good'
 
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Think it was Int21 who, many posts ago asked what if a new player is given the choice of the remaining two doors, after the goat door has been revealed.

If the new player is not aware of the original player's choice, the odds would be 50/50 wouldn't they?
 

Ringo

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Think it was Int21 who, many posts ago asked what if a new player is given the choice of the remaining two doors, after the goat door has been revealed.

If the new player is not aware of the original player's choice, the odds would be 50/50 wouldn't they?
Yes they would. If the new player was just asked to pick one of the two remaining doors, without knowing he original choice, then their odds would be 50/50. Likewise, the original players odds would be reduced to 50/50 if the host had the option of shuffling the things around behind the two remaining closed doors.

But as it's the same person playing, and they have previous experience, they have an advantage.
 
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The Wikipedia page on the problem has a conditional probablity table, to show the odds of switching versus sticking in 300 cases.

It is quite the most lucid way to present the problem. In the two-out-of-three cases where you are initially wrong, the host has no choice which door he can open. If you switch in these cases, you must now win. In the one-out-of-three cases where you are were right, any switch will now cause you to lose. So the odds are 2/3 in favour of the switch.

I would still hate to have to explain this in a maths class! :frust:
This post cleared this up for me.

If anyone's still struggling to get their head around this, maybe try thinking of it from Monty's perspective as if he's trying to win (by not giving the contestant the car).

In the scenario where the contestant switches, Monty is relying on the contestant picking the car door first time, as picking a goat door will mean that Monty has to reveal the other remaining goat door, leaving the car door as the one the contestant would be switching to.

Essentially, if you switch choices, you're always going to get what wasn't behind your original door and, since there are twice as many goat doors as car doors, you're twice as likely to end up on the car door by switching.
 
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Essentially, if you switch choices, you're always going to get what wasn't behind your original door and, since there are twice as many goat doors as car doors, you're twice as likely to end up on the car door by switching.
Well, yes..

I do believe that I have worked out what the problem was. And in a way it was like the three meals problem. And it is that not all the facts are taken into consideration.

Firstly, I agree that at the beginning you have one out of three chances of being correct. Or to put it another way, 33% chance of choosing correctly, 66% chance of picking the wrong one.

But what threw me, and is the reason I still would not swap, is that once Monty removes one bad door, The odd are NOW 50/50 that if I swap I will win.
I emphasis this because, in my practical test, I picked the correct door 14(I think) times. And thus by swapping I would have lost.

This factor does not appear to be taken into consideration.

Now, If I had more than one try the odds of moving to the correct door improve. And If I had 100 tries then, indeed, I would probably win the majority of the time. But not every time.

And as I have only one chance, I may as well stick with my first choice as, for me, the odds are simply win or lose.
 
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And as I have only one chance, I may as well stick with my first choice as, for me, the odds are simply win or lose.
Any punter is faced with the same 50/50 decision. The larger stick/switch question compares their choice to that of others in their situation.

Two out of three punters will benefit from the switch. Success is not guaranteed. :)
 
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