Two Donkeys and a Million Bucks

[Anaxogoras]

There are three doors.  Behind one of those doors is a million bucks.  Each of the other two doors hides a donkey.  Here's how the game works:  You pick a door at random, and then I open a different door and show you a donkey.  You now have the option of keeping your original pick, or switching.  What should you do?

A)  Keep my original choice
B)  Switch
C)  It doesn't matter, the two options are equivalent

[Ghosth]

Switch

Odds are much better.

[APDrone]

How angry are the bucks and how much ammo do I have?

Seriously, though.. Fresh off a statistical analysis test?

Your first choice implies a 33.33% chance of success.. with your second choice being 50% chance of success.  Theoretically, you should rechose, since your odds are 'better'.

However, since your final choice is 50/50, then it's hit or miss.

I suggest eeny-meeny algorithm started by coin toss.

 

[ODBAL]

I would guess C.  The remaining 2 options should both yield a 50% chance of success.

[druski85]

I know the correct answer is to switch, yet I have never been able to really understand why.  I've always been good at math, (85-90th percentile) but this simple probabilty question always threw me for a loop.  

Edit: So my answer is B.

[AKHog]

I'm in the same boat as Druski. Normally I have no problem with math but statistics are another story.

I understand the chances, your first pick you have a 1/3 chance and your second pick you have a 1/2 chance. Where I get lost is since one donkey is taken away, why does your first pick not automatically go to a 1/2 chance?

In other words once there are only 2 doors left, why do you have to re-pick to have your chances go up? If there are only 2 doors left and you have picked one, your chances are now 1/2 weather you re-pick or not.

I say option C. Even though the chances have changed, they changed when one door was eliminated, not when you re-pick.

[Getback]

So you are saying there is a 66.66666666666666666667% chance of getting those rabbits or deer.

[hammer]

I pick the door you opened. I always wanted a donkey! 

[ODBAL]

I pick the door you opened. I always wanted a donkey! 

That's the spirit!  I declare a winner.

[68valu]

no benefit to rechoosing as the only difference that has happened is that you may have chosen one of the doors that didnt have a donkey behind it on your first attempt, thus eliminating the one opened with the donkey in it. switching now would not increase your odds of success nor would it diminish your odds of success. You currently have 50/50 odds with your choice of the original 3 doors, with only 2 remaining.

[druski85]

no benefit to rechoosing as the only difference that has happened is that you may have chosen one of the doors that didnt have a donkey behind it on your first attempt, thus eliminating the one opened with the donkey in it. switching now would not increase your odds of success nor would it diminish your odds of success. You currently have 50/50 odds with your choice of the original 3 doors, with only 2 remaining.

See now this makes logical sense to me, but I'm certain I've heard people say switching increases odds.  Can someone please give the definate answer here?

Edit: I'll answer my own question.  I guess this is the best explanantion I've found to date.  I still don't like it. 

At first you have 33.33% chance of choosing the right door and there is 66.67% chance of the prize being somewhere else. You know that Anax is going to open an empty door so when he does, this should not change a thing about your belief of your door being the right one. You still have 33.33% chance of having selected the right door. Thus there must be 66.67% chance of the prize being somewhere else (behind the last door).

[APDrone]

Since you don't seem to have a big enough headache..

Suppose Anoxagoras RANDOMLY chose the door and it had a donkey.  Now, what are the odds that you have bucks or a donkey behind your door?

If you keep the door you chose, then you be betting that the first 2 doors Anoxagoras chose would both be donkeys. Probably less likely than one of each.

[Anaxogoras]

Thanks for the replies.  I've always thought this was a fun problem.  I can remember arguing about it for almost an hour with a buddy of mine until we actually tried a version of the game with playing cards.  Once it becomes empirically obvious that switching is better (evident after only 20 trials or so), you become more open to the mathematical answer, counter-intuitive though it seems.

When you first choose a door, your odds are necessarily 1/3.  When I open a door and show you a donkey, the odds that your initial guess was correct are still 1/3, but the other (unseen) door must have a million bucks 1/2 of the time.  Therefore you switch.

[AKHog]

When you first choose a door, your odds are necessarily 1/3.  When I open a door and show you a donkey, the odds that your initial guess was correct are still 1/3,

This is the part I don't get.

If you show me one donkey, then the odds of my initial guess being correct automatically go up to 1/2.

There are 2 doors left, and I have 1 picked, how are my odds not 1/2? Weather I pick again or stay with my original, the numbers are the same. 1/2.

Another part that adds to the confusion is why is simply 'switching' better? Why is the opposite of what you have currently in the two remaining options is somehow better?

[Anaxogoras]

If you show me one donkey, then the odds of my initial guess being correct automatically go up to 1/2.

Nope!    Your odds are not 1/2 because you made your guess with 3 possible choices.  What I show you has nothing to do with the probability of your first guess being correct.

Try the game with playing cards.  Do 20 trials where you keep the same choice, and then do 20 more where you switch.