Monty Hall Problem

[Rod Paradise]

Correct.. Here's my explanation:

Basically at the start of the show you know you can pick one door which has a 1 in 3 chance of being the car.

This leaves a 2 in 3 chance that the car is behind one of the other doors.

You know that he will always give you the option to stick with your original choice or switch.

The exposure of the door with a goat is a red herring, it's completely irrelevant other than to confuse you into thinking the problem is now 50/50.

He may as well have said at the start,choose a door, and then decide if you want to stick with your original choice or switch to the other 2 doors.

If you are playing cards and you are up against two opponents and you all had the same hand AK o/s.

You are a 33.3% favourite to win therefore making the chances of either of your opponents winning as 66.6% - if you get one of your opponents to fold your chances of winning the pot dont stay at 33.3% - they rise to 50% as you now have an equal chance against just the one opponent.

Ok then by the same principle that you explain - how come in the example i have given your odds of winning the hand dont stay at 33.3%?

You've missed a  factor here - Monty Hall KNOWS the winner & will expose 1 loser.

[matt674]

I know - there is always a 100% chance that one of the "losing" hands/doors will be in the group of 2 then removed from the equation.

I got the problem after reading about it but it seemed that the only people posting answers on the thread were those who had seen and read about this "problem" before - and therefore knew the answer and explanation.

By asking other problems and throwing different examples into the equation it is easy to see who knows more other than just what they read from the original problem in the first place.

[ACE2M]

I know - there is always a 100% chance that one of the "losing" hands/doors will be in the group of 2 then removed from the equation.

I got the problem after reading about it but it seemed that the only people posting answers on the thread were those who had seen and read about this "problem" before - and therefore knew the answer and explanation.

By asking other problems and throwing different examples into the equation it is easy to see who knows more other than just what they read from the original problem in the first place.

I avoided your question as it had too many incorrect parts that i assumed you knew were incorrect anyway.   

[Decider]

I know - there is always a 100% chance that one of the "losing" hands/doors will be in the group of 2 then removed from the equation.

I got the problem after reading about it but it seemed that the only people posting answers on the thread were those who had seen and read about this "problem" before - and therefore knew the answer and explanation.

By asking other problems and throwing different examples into the equation it is easy to see who knows more other than just what they read from the original problem in the first place.

I did the same when I first got asked about it. 50/50 seemed obvious. When this program was on TV, a lady wrote in and suggested that players should always switch as it doubles there chances of winning. She was then rebuffed by a few top mathematicians in the country and told she was daft. Eventually they worked out she was right. It's not an easy problem even after it being explained as its counter intuitive.

[Mbuna]

This has been doing my head in.

if we pick a door at random we will pick the Car .333333 of the time and hence .66666 not.

if we always swap
.3333333 (when we pick the car first) we lose 0.0
.66666666(when we pick a goat)  we swap and win 1.0

net +ev= .33333*0 + .6666666*1=.6666666


If we always swap
.3333333 we win 1.0
.6666666 we lose 0.0

net +ev=.3333333*1+.66666666666*0= .333333333

if we randomly swap or stick on say a coin toss

.3333333 we win half the time .5
.666666666 we win half the time .5

Net +ev .3333333*.5 + .66666666*.5= .5

ie if we always swap we win 66% of the time
if we always stick we win 33% of the time

and if we swap 50% of the time we win 50% of the time.

[matt674]

This has been doing my head in.

if we pick a door at random we will pick the Car .333333 of the time and hence .66666 not.

if we always swap
.3333333 (when we pick the car first) we lose 0.0
.66666666(when we pick a goat)  we swap and win 1.0

net +ev= .33333*0 + .6666666*1=.6666666


If we always swap <shouldn't this be stick?>
.3333333 we win 1.0
.6666666 we lose 0.0

net +ev=.3333333*1+.66666666666*0= .333333333

if we randomly swap or stick on say a coin toss

.3333333 we win half the time .5
.666666666 we win half the time .5

Net +ev .3333333*.5 + .66666666*.5= .5

ie if we always swap we win 66% of the time
if we always stick we win 33% of the time <correct for the choice of 3 doors>

and if we swap 50% of the time we win 50% of the time. <correct if the two remaining doors are then swapped again so you do not know which door you picked originally>

[Royal Flush]

I can't believe 'top mathematicians' could get this wrong! It's just plain obvious!

[Mbuna]

This has been doing my head in.

if we pick a door at random we will pick the Car .333333 of the time and hence .66666 not.

if we always swap
.3333333 (when we pick the car first) we lose 0.0
.66666666(when we pick a goat)  we swap and win 1.0

net +ev= .33333*0 + .6666666*1=.6666666


If we always swap <shouldn't this be stick?>
.3333333 we win 1.0
.6666666 we lose 0.0

net +ev=.3333333*1+.66666666666*0= .333333333

if we randomly swap or stick on say a coin toss

.3333333 we win half the time .5
.666666666 we win half the time .5

Net +ev .3333333*.5 + .66666666*.5= .5

ie if we always swap  we win 66% of the time
if we always stick we win 33% of the time <correct for the choice of 3 doors>

and if we swap 50% of the time we win 50% of the time. <correct if the two remaining doors are then swapped again so you do not know which door you picked originally>

we dont know if we picked a car so if we swap 50% of the time when we pick the car we dump it 16.66666% of the time and when we origonally picked a goat and still swap 50% of the time we actually win 33.33333 % of the time .

I think this is correct even if we never loose sight of the first door

[Decider]

I can't believe 'top mathematicians' could get this wrong! It's just plain obvious!

I have no verification of this, this was the original story I had been given.

Having read the wikipedia article, it appears that in fact, the game show host, Monty, didn't offer anyone a chance to switch.   But he did expose the goat to make it more interesting.

The person who wrote in to say always switch, was in fact writing into a magazine which had updated the problem to include the switching part and hence to start the ball rolling on the discussions thereafter.

This link explains who the top mathematicians were:

http://www.letsmakeadeal.com/problem.htm

[mex]

surley the garage door has the car behind it, the barn door the goat?

[Decider]

surley the garage door has the car behind it, the barn door the goat?

Where's Sark when you need im

[thetank]

I remeber Sword, Rod and myself trying to explain this to tigscoco one night, and giving up after an hour and a half. 

[mikkyT]

The odds of you choosing the correct door in the first place, is 33%. 1 in 3 doors will contain a car.

Now that there are two doors, the odds of you having the correct door is 50%, 1 in 2 doors will contain the car.

But we pick the car only 33% of the time, and we pick a goat 66% of the time (as there are two goats!) so by always swapping we win a car 66% of the time making swapping teh +EV decision.

[roverthtaeh]

What if the goat's driving the car?

[Mbuna]

What if the goat's driving the car?

Theres only a 33% chance of that!