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You removed a set from the equation. The Odds are now 50/50.

Thats like saying all 3 doors are open and you picked the car, your odds of winnng the car are still 1 in 3.

The orginal odds were 1 in 3 as there were 3 unknowns, but one of the unknowns is a given. You can't add or subtract data and have the equation remain the same.

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You are wrong.

I think this is the best brain teaser I've ever seen.

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You are on let's make a deal. There are three doors. Behind one door is a brand-new car behind the other two doors are goats. The host of the show knows exactly what's behind each of the doors. You pick your door. He then opens up one of the doors revealing a goat.
You are then asked whether you would like to stick with your door or switch to the other door that you did not choose.

Would it be to your advantage to switch doors?

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I know the two doors that Stan Crowch prefers.

We have this thread every fall when some freshman get his mind blown in whatever class teaches this.

If we don't open the door how will we know if the goat is dead or alive?

I don't really want to win a dead goat.

what's in the boxxxxxxxxxxxxxxxxxxxxx?!

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You removed a set from the equation. The Odds are now 50/50.

Thats like saying all 3 doors are open and you picked the car, your odds of winnng the car are still 1 in 3.

The orginal odds were 1 in 3 as there were 3 unknowns, but one of the unknowns is a given. You can't add or subtract data and have the equation remain the same.

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watch the youtube video. you'll see your wrong.

The goat doesn't take off.

87,

it was already explained perfectly above.

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The goat doesn't take off.

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The goat is moot

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The goat doesn't take off.

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The goat is moot

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The cow is moo

What about on Wheel of Fish? Do you keep the Red Snapper, or take what's in the box?

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What about on Wheel of Fish? Do you keep the Red Snapper, or take what's in the box?

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Take the box. Red Snapper is out of season

288

The magician has essentially taken 2 of the doors and processed them down to one.


The determination of which door contains the goat or car was not reset upon eliminating one of the incorrect doors. At the beginning of the game, the doors each contained a 1/3 chance of holding the car. Nothing is shifted once the first goat door is revealed. You effectively get to pick both of the other doors if you switch.

It is easier to grasp if you consider this with larger set. Imagine it instead as essentially the same scenario of trying to find one particular item but with a deck of cards. Imagine you want to find the ace of spades in a 52 card deck. by selecting one card at random you have a 1/52 probability. Imagine you are given the same opportunity to switch, but after the dealer has removed 50 incorrect cards from among the 51 cards you didn't initially pick. By choosing from that group of 51 cards, you have a 51/52 probability of picking the correct card. The dealer has processed out all of the incorrect answers out of that 51 card set. Basically, think of it as this: you are picking out of the second group and if the second group contains the correct answer, you win. The wrong answers have been processed out of that second group.

Yes you switch, but your odds stay the same.

You had a 2/3 chance of picking a goat first.

Which means you have a 2/3 chance of picking the car when you switch.

Say the doors are A, B and C.

You pick A.

They open C and show a goat.

Doors left are A & B.

At this point you can select A or B. What you picked before has no bearing on that. It's a fresh pick. When you say "keeping my pick" it's the same as saying A...the same as if a stranger walked up and took over the game at that point and was given a choice of A or B without knowledge of the previous pick.

Incorrect, Cotton. There were three doors when you picked. Two doors hid a goat. No matter which door you pick, a door hiding a goat can be opened. This does not retroactively make your first choice 1 in 2, but still 2 in 3 to pick a goat.

When you switch, you will switch to the car 2 out of 3 times.

Let's say A has the car. The options are:

Pick A & switch = goat
Pick A & stay = car
Pick B & switch = car
Pick B & stay = goat
Pick C & switch = car
Pick C & stay = goat

By staying, 1 out of 3 times you get a car.
By switching, 2 out of 3 times you get a car.

Watch the dang video and it will visually explain why you always should switch your choice

So let's say I do the initial picking.

C is opened and their is a goat.

I step aside and you walk up with the choice of A or B. You don't know which door I originally picked. What are your odds of getting the car by picking A? Or B?

That's not how this game is played, Cotton. If we knew what your initial choice was then the odds would not be 50/50. There has been several great explanations and even a video on this thread that show you why you are wrong.

That's a different problem.

The second person only has 2 doors to choose, not 3.

If they are given knowledge of your choice, they should always pick the other.

Let's do this...

Play 1,000,000 times. You pick the doors. Behind one door, you get $1 from me. Behind the other two, you have to give me $1.

...we'll be even, right?

Or better yet, give him a $1.50 from you so he thinks he will actually be making money from the proposition.

1.75 even.

What if behind one of the doors lies 7 tickets to Midnight Yell?

You switch doors. The door you switch to has 2/3 chance of winning. It gained the odds of the open door because you chose randomly but the host acted with inside knowledge and not randomly.

It feels a little more intuitive if you extrapolate the problem out to 1,000 doors. If you pick a door, and the host opens 998 losing doors, there is a 99.9% chance you win if you switch.

I modeled the 3 door scenario with a random number generator, iterated through 100,000 instances, and got the 1/3:2/3 split you would expect.

What if the goat walks away?

What kind of goat and what kind of car is it? Because I'll take this goat over some crappy Ford Fiesta.

Think of it like this. You pick one door. The host offers you the chance to switch your pick to both the other doors or stick with the one you have.

But before you make that choice, he opens one of the doors.

In reading more on this, it looks like even some renowned mathmetician's disagreed with this arguing it was 50/50. However, it looks like with all the computer modeling and trials, it really does come out to 1/3 and 2/3 -- and after the trials, the mathmeticians relented, but were still a little bit in disbelief.

Coppell, they shouldn't be in disbelief. If the opening of the door were done blindly, or if the objects behind the doors were shuffled after the first door was opened then the odds would reset back to 50/50. But because the house by rule has to open one of the two non-car doors, you as the player an use that information to double your odds.

Let's say you pick door #1. If the car is behind door 3, the host will have to open door #2, leaving the car behind #3. If on the other hand the car is behind door #2, then host will have to open door #3, leaving the car behind door number 2.

Using the switch method, if you pick door 1, the only way you lose is if the car is behind door 1 which was your initial pick, which was a 1 in 3 chance. As long as your initial pick is one of the donkey doors ( a 2/3 probability) you will get the car if you switch.

Good Gawd. I just want to porn blast threads like this,

It's like a bunch of nerds using a micrometer to measure their deeks.

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Good Gawd. I just want to porn blast threads like this,

It's like a bunch of nerds using a micrometer to measure their deeks.

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If you know a better way then let us know