Statistical Puzzle with Poll

Which door do you choose?


  • Total voters
    20
#1
Here's one:
You're on Let's Make a Deal.
Monty Hall offers you a choice of three doors. Behind one door is the Big Prize and behind the other two doors are Booby Prizes (goats and chickens)
You pick one.
Monty has Carol Merril reveal the goat behind one of the two doors that you didn't pick. Note that there will always be one door with a Booby Prize no matter which door you pick.
Now, Monty gives you the choice to stay with your first choice or switch.
What do you do?

Stick with your first choice.
Switch to the other door.
It doesn't matter, it's a toss-up.
Icy, you're making me crazy.
 
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Jim Wynne

Staff member
Admin
#2
Icy Mountain said:
Here's one:
You're on Let's Make a Deal.
Monty Hall offers you a choice of three doors. Behind one door is the Big Prize and behind the other two doors are Booby Prizes (goats and chickens)
You pick one.
Monty has Carol Merril reveal the goat behind one of the two doors that you didn't pick. Note that there will always be one door with a Booby Prize no matter which door you pick.
Now, Monty gives you the choice to stay with your first choice or switch.
What do you do?

Stick with your first choice.
Switch to the other door.
It doesn't matter, it's a toss-up.
Icy, you're making me crazy.
For a hands-on experience:

http://math.ucsd.edu/~crypto/Monty/monty.html
 

Tim Folkerts

Super Moderator
#3
A classic puzzle:agree1:. I won't vote because I've heard this one before.

Also, just to be crysal clear, the rules require that Monte will always show one of the Booby Prizes after you pick a door. He can't just do it if he feels like it.

Tim F
 

Jim Wynne

Staff member
Admin
#4
Tim Folkerts said:
A classic puzzle:agree1:. I won't vote because I've heard this one before.

Also, just to be crysal clear, the rules require that Monte will always show one of the Booby Prizes after you pick a door. He can't just do it if he feels like it.

Tim F
I didn't vote either, for the same reason. This one always starts "discussions" though.
 
D

davis007

#5
Hmmmm

OK I understand, sort of, the logic. But it still seems that at the point of the question "Do you want to switch doors?" The only information is that the car is behind one of two doors. Which to my feeble brain says 50:50 chance.

If the game was played without the preamble:

Monty: Here is the game. Three doors behind one is a car and behind two is a goat. You get to pick one.

Contestent: I pick..

Monty: Wait. Just to make it interesting lets open one door with the goat.

Now what are the contestents odds? 50:50

So I guess I do not understand how Monty opening one door after the contestent announced his choice changed the odds of which door the car was behind after all.

The only difference is the number of choices Monty has. If he opens one door before the contestent makes any choice he always has two choices. But if he waits until after the contestent chooses then he only has one choice 2/3 of the time. HMMMM I guess I get it after all.
 

Jim Wynne

Staff member
Admin
#6
davis007 said:
OK I understand, sort of, the logic. But it still seems that at the point of the question "Do you want to switch doors?" The only information is that the car is behind one of two doors. Which to my feeble brain says 50:50 chance.

If the game was played without the preamble:

Monty: Here is the game. Three doors behind one is a car and behind two is a goat. You get to pick one.

Contestent: I pick..

Monty: Wait. Just to make it interesting lets open one door with the goat.

Now what are the contestents odds? 50:50

So I guess I do not understand how Monty opening one door after the contestent announced his choice changed the odds of which door the car was behind after all.

The only difference is the number of choices Monty has. If he opens one door before the contestent makes any choice he always has two choices. But if he waits until after the contestent chooses then he only has one choice 2/3 of the time. HMMMM I guess I get it after all.
Here's a good explanation:
http://math.ucsd.edu/~crypto/Monty/montybg.html
 
B

Baldrick

#7
I love the Monty Hall problem - a search on the internet will reveal many different ways of explaining the solution. If you dig a little deeper you will find that this problem has a fascinating history, and that the correct solution, when it was first proposed, was rubbished by many "professional" statisticians, maths teachers etc.

How about this explanation:

Assuming there are no influences on the contestant’s initial choice, he/she has a 1 in 3 chance of picking the door with the car. This means that 2 times out of 3, the contestant will pick a booby prize.

Monty then has to reveal a booby prize (he can't reveal the car).

On the 2 occasions out of 3 where the contestant has picked a booby prize, Monty HAS NO CHOICE WHICH OF THE OTHER TWO DOORS TO OPEN. He MUST open the door with the remaining booby prize behind it. This means that the car has to be behind the door that he DOESN’T open. Remember, this will happen 2 times out of every 3.

The only time it will be successful to stay with your original choice is when you were lucky enough to pick the car at the first attempt, which will happen 1 in 3 times.

Therefore swapping will be successful 2 in 3 times.

Why isn’t it 50-50? The randomness disappears because Monty doesn’t have a free choice which door to open 2 times out of 3. And on these occasions, the door he DOESN’T open MUST have the car behind it.

If you’re still struggling, try applying the same logic to the situation where there are 100 doors, 1 car and 99 goats. The contestant again picks one at random (a 1 in 100 chance of success, yes?). Monty then opens 98 of the remaining 99 doors, each one being a goat. Do you stay with your original choice? Or switch? Do you still think it’s a 50-50?
 
#8
This thing is really counter-intuitive. My original figuring was a toss up. Your odds of being right changed from 1 in 3 to 1 in 2 after Monty opens a door. Considering Steve Prevette chose this one, I feel much better. I found this puzzle and an interesting article with some interesting answers @:
.:: Dean's World: The Brainteaser That Changed My World ::.

Here’s a sample explanation:
Before the other door is opened, you have three alternate futures, all equally likely. Opening one door closes off one, but it equally consistent with the other two. That is why people thing, "hmmm, 50/50".
BUT, the door that stayed closed ONLY stays closed in half of the future where you have the right door. And the door that stayed closed stays closed in ALL of the futures where IT is the right door.
SO:
All the futures started out equally likely
ALL the futures with the prize behind the opened door are foreclosed
HALF the futures with the prize behind your door are foreclosed
ALL the futures with the prize behind the remaining closed door are open.

Try something simpler. Here are your possible combinations at the beginning with the (P)rize, Goat (B)ooby Prize, and Chickens (B)ooby Prize. I eliminated the possible permutations because they don’t really matter.
Door......1 2 3
Combo 1 P B B
Combo 2 B B P
Combo 3 B P B

If you choose door number 1 and don’t switch, your chances are 1 in 3 of getting this correct. The same goes for any other door choice. (Let's say you always choose #1.) Monty then ALWAYS reveals A SINGLE DOOR (x) with a booby prize leaving you:
Door......1 2 3
Combo 1 P B x
Combo 2 B x P
Combo 3 B P x

2 times out of 3, Monty has eliminated the remaining Booby Prize leaving only the Prize behind the door you did not pick. Remember, Monty does not know which door the prize is behind. However, HE DOES NOT OPEN A DOOR AT RANDOM. The lovely Carol Merill only opens a door that has a Goat or Chicken behind it. The online version of the game that Jim posted above bears out this conclusion: You should always switch, you'll be right more often.

Would the two Steve’s, Statistical and Prevette like to chime in? Cari, how about you?
 
Last edited:
R

Rob Nix

#9
Good one Icy, now I'll make it slightly graphical:

Here is another simple way to explain it:

LEGEND:

C = correct choice
Ia = incorrect choice (a)
Ib = incorrect choice (b)
# = stay the same
> = switch doors

You have nine original choices, after which Monty takes away three choices:

C # C
Ia > C
Ib > C
Ia > Ib Monty takes away this choice
Ia # Ia
Ib > Ia Monty takes away this choice
Ib # Ib
C > Ia Monty takes away this choice (or the “b” - no matter)
C > Ib

Leaving a 2/3 chance, when switching, to get the correct door, and
leaving a 1/3 chance, when staying to get the correct door.

In other words, with your original choice you have a 1/3 chance of being right, and since you only have one choice left at the end, it must be a 2/3 chance of being right!
 
Last edited by a moderator:

Jim Wynne

Staff member
Admin
#10
Rob Nix said:
Good one Icy, now I'll make it slightly graphical:

Here is another simple way to explain it:

LEGEND:

C = correct choice
Ia = incorrect choice (a)
Ib = incorrect choice (b)
# = stay the same
> = switch doors

You have nine original choices, after which Monty takes away three choices:

C # C
Ia > C
Ib > C
Ia > Ib Monty takes away this choice
Ia # Ia
Ib > Ia Monty takes away this choice
Ib # Ib
C > Ia Monty takes away this choice (or the “b” - no matter)
C > Ib

Leaving a 2/3 chance, when switching, to get the correct door, and
leaving a 1/3 chance, when staying to get the correct door.

In other words, with your original choice you have a 1/3 chance of being right, and since you only have one choice left at the end, it must be a 2/3 chance of being right!
That's a good explanation, but for those who still want to cling to 50-50, you have the right answer, but the wrong question. At the point when you must choose between two doors, the odds are indeed even, but the question is, given 3 doors before Monte's intervention, should you stick with your original choice after Monte changes the sample space. Rob's illustration clearly shows that switching is the best choice.
 
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