A very good riddle.....

[tauzreborn]

You get to choose one of three doors, behind which is a prize. Behindone door there is $1,000,000. Behind the other two are smelly goats.

So choose one of the three.

Once you choose, I show you one door containing a smelly goat. Now you have the option to switch to the other door or keep your original choice.

Do you switch or keep the original door?

[HOMEYCAT]

stick with original choice. Because, had I missed with the first choice, you would have awarded me the goat!

[tauzreborn]

No, the scenario is the same each and every time. You always get a chance to switch.

Look at it as if you did this 100 times, what decision would get you the cash most often.

[tauzreborn]

You should switch - every time.

At first you only have a 33% chance of getting it right.

Then when one goat is knocked out you have a fifty percent chance of getting it right.

By switching you get 17% more chance of getting it right.

[HOMEYCAT]

Oh, I didn't know that the choice was the same every time....but with a goat gone, how could it be the same every time?

[BOURBON TOWN CAT FAN]

tauzreborn wrote: You should switch - every time.

At first you only have a 33% chance of getting it right.

Then when one goat is knocked out you have a fifty percent chance of getting it right.

By switching you get 17% more chance of getting it right.
Once one door has been taken out of the equation each time you will always have a 50/50 chance afterward, which then I would probably stick with my original selection.

[HOMEYCAT]

I guess I'm cynical. If I choose a goat door, game over. I get a goat. (a smelly one at that) But, if I choose the million dollar door, the host offers me another choice, (the other goat) , so I stick to my original choice.

[tauzreborn]

But by sticking to your initial choice, youstill only getthe 33% chance that you had initially.

You made the decision with only 0.33 probability, if you stay withthe original choice you do not get that 50/50 shot.

It seems like you would havea 50% chance, but since you havent changed your pick the odds don't change either.

[Littlemeyer]

tauzreborn wrote: But by sticking to your initial choice, youstill only getthe 33% chance that you had initially.

You made the decision with only 0.33 probability, if you stay withthe original choice you do not get that 50/50 shot.

It seems like you would havea 50% chance, but since you havent changed your pick the odds don't change either.



I would stick with the original choice as well. The first time I picked, I overcame odds of 66%, and did not pick a goat. I'm hot, and papa needs a new pair of shoes...

[gerntz]

Do whatever your mood strikes you to do as it doesn't matter. no value in stressing over it - like most things in life.

[HOMEYCAT]

Well dammit! I want my frickin goat!

[Malachi]

This is the old 'The Price is Right' riddle. Let's just run through all the possibilites, there aren't that many. Let's say door #1 is the winner.

Scenario 1

I pick door #1,
Bob reveals door #3 (a goat)
I switch to door #2, I lose (switching loses)
I don't switch, I win (not switching wins)

Scenario 2

I pick door #2
Bob reveals door #3 (a goat)
I swtich to door #1, I win (switching wins)
I don't switch, I lose (not switching loses)

Scencario 3

I pick door #3
Bob reveals door #2 (a goat)
I switch to door #1, I win (switching wins)
I don't switch, I lose (not switching loses)

Adding up the results we see that switching wins 2 out of 3 times.

[Mr. Peanut]

malachi's math looks right to me... but, it sure is counter intuitive.

Seems thatthe final choice is a "new" choice, and the results should be 50/50. Its as if the first choice didn't even happen.No matter what happens in your first pick, step two is to bring your choice down to picking between 2 options. You know that whatever you do, you get a new choice of the two options left.

So, why wouldn't that just be a "pick em"? How is it different than Bob just giving you 2 choices in the first place?

[Malachi]

You're being tricked by Bob showing you a door.

On your first choice you had 2/3rd chance of being wrong. Bob opening a door for you doesn't change your chance of having picked wrong, he's just showing you the other wrong door.

So the door you picked is probably wrong, the door he shows is wrong, that leaves the last door as probably right.

Let's take it to an extreme. Let's say there are 100 doors, all but one with a goats behind them. You pick a door. You have 99 in 100 chance of having picked a wrong door. Bob then reveals all the other 'goat' doors (because he's NOT going to reveals the winning door). You know you almost certainly picked wrong to begin with, switching now is an almost certain win.

Does that help?

[Mr. Peanut]

Malachi wrote: You're being tricked by Bob showing you a door.

On your first choice you had 2/3rd chance of being wrong. Bob opening a door for you doesn't change your chance of having picked wrong, he's just showing you the other wrong door.

So the door you picked is probably wrong, the door he shows is wrong, that leaves the last door as probably right.

Let's take it to an extreme. Let's say there are 100 doors, all but one with a goats behind them. You pick a door. You have 99 in 100 chance of having picked a wrong door. Bob then reveals all the other 'goat' doors (because he's NOT going to reveals the winning door). You know you almost certainly picked wrong to begin with, switching now is an almost certain win.

Does that help?

Obviously you are right, and I understand why... but, I still stumble on the idea that that final choice is always between one door with and one without. You are free to choose. Just goes against my gut rxn to the question. Nice riddle and nice explanation Malachi.

[Littlemeyer]

Malachi wrote: You're being tricked by Bob showing you a door.

On your first choice you had 2/3rd chance of being wrong. Bob opening a door for you doesn't change your chance of having picked wrong, he's just showing you the other wrong door.

So the door you picked is probably wrong, the door he shows is wrong, that leaves the last door as probably right.

Let's take it to an extreme. Let's say there are 100 doors, all but one with a goats behind them. You pick a door. You have 99 in 100 chance of having picked a wrong door. Bob then reveals all the other 'goat' doors (because he's NOT going to reveals the winning door). You know you almost certainly picked wrong to begin with, switching now is an almost certain win.

Does that help?



F the odds! I'm on a roll....NO DEAL!

j/k The math makes perfect sense. Especially when you go with 100 doors.

[EricBigNally]

It all depends... on how big the dude opening the doors is. If he is little the quesiton is irrelevant because I'm just going to whack him if I'm wrong and run with the million. If he's big, then I have to think about it. I'd stick with my original answer and if I won, I get a million, if I was wrong, I push my odds and see if I can take him.



Thats right, if I lose, there is a very good chance I won a trip to the hospital...

[tauzreborn]

Since I just heard it yesterday, I couldnt find a way to explain it thoroughly without someone bringing up valid points. Thanks Malachi for truly putting it in perspective so even I can understand it better.