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1mrgrooism
Mark Evanier over at News From Me has a couple of posts about the Monty Hall Problem, literally, a winning strategy to picking What's Behind Door Number 2.
First, check this out.
Okay, that's part one. Makes no sense, right?
Now check this out.
Okay, it's more clear, but still, it's clearly wrong, switching can only ever possibly give you a 50-50 shot, since it's a choice between 2 doors, right!?
So, just to prove how wrong Mark was, I did this, and I ask you all to do the same. Take 3 index cards, write GOAT on two of them and CAR on the other, and mix them up REAL good. Choose a door completely at random, then flip them all over. Regardless of whether you picked a goat or a car, you remove an unpicked goat, then switch to the card you didn't originally pick.
Doing this, I got the car 8 out of 10 times!
So although it is counterintuitive, Mark is correct, you SHOULD ALWAYS SWITCH!
First, check this out.
Okay, that's part one. Makes no sense, right?
Now check this out.
Okay, it's more clear, but still, it's clearly wrong, switching can only ever possibly give you a 50-50 shot, since it's a choice between 2 doors, right!?
So, just to prove how wrong Mark was, I did this, and I ask you all to do the same. Take 3 index cards, write GOAT on two of them and CAR on the other, and mix them up REAL good. Choose a door completely at random, then flip them all over. Regardless of whether you picked a goat or a car, you remove an unpicked goat, then switch to the card you didn't originally pick.
Doing this, I got the car 8 out of 10 times!
So although it is counterintuitive, Mark is correct, you SHOULD ALWAYS SWITCH!
2belinthesun
Groo, I want you to know that while you were mixing the cards up really good, I snuck in and secretly switched one of the goat cards to a car card and therefore your experiment is null.
4Delirium9
Ok, I'm watching the video in your first link and I'm totally distracted by the man's voice. Is that Richard Ayoade speaking?
5belinthesun
Nope, Ron something. I thought the video was hilarious, couldn't help laughing.
6Delirium9
Heh, yeah, I now saw the end credits. Ron Clarke.
For me it was super funny because I was imagining Ayoade speaking. :P
For me it was super funny because I was imagining Ayoade speaking. :P
7MrsLee
I don't know, the first video made perfect sense to me (an extremely non math person), and I'm glad it did, because I couldn't get the second link to do anything. Still, I don't want a goat or a car, I want what's in the BOX!
8mrgrooism
#7 - Well I couldn't wrap my head around the logic. It seemed pretty clear to me that if you had two doors left then logically it was a fifty-fifty chance that you'd get it right so switching doors was merely a coin toss.
Having tried it now it seems switching doors is the right idea.
Having tried it now it seems switching doors is the right idea.
9billiejean
This is why I was a liberal arts major!
--BJ
--BJ
10JPB
A very simple way to think of the whole thing is this:
"1/3rd of the time, when Monty asks, I am sitting on a car. So best not to move. But 2/3rds of the time when Monty asks, I am sitting on a goat. So - best to swap. The best to swap case comes up 2/3rds of the time; the best to sit case comes up 1/3rd of the time."
If that is still confusing, consider this:
The reason it is hard to understand is that we often think that at *every* choice opportunity, we are playing a NEW GAME. This is usually true. Play slots, play roulette, and every single spin of the wheel is a NEW GAME. It doesn't matter what happened before.
But that's not the Monty Hall game. At the first choice, you are neither a winner nor a loser - you aren't even in the game. But, your first choice turns you into either a WINNER or a LOSER. At the time of the "swap choice" - you are either a current winner or a current loser. Since you are more likely a current loser at that swap choice, you have to make the choice that gives the best chances for a current loser, and ignore the choice that gives the best options for a current loser.
---------------------------------------------------
MORE THAN TWO DOORS:
What happens if there are 4 doors at the start?
1/4 of the time, you pick the right door, and swapping (picking another door) later always loses.
3/4 of the time, you pick the wrong door, and swapping later gives you a 50% chance of winning and a 50% chance of losing, whereas not swapping gives you a 100% chance of losing.
So swapping works .75*(.5) 37.5% of the time. Not swapping works 25% of the time. So it's still better to swap (but not much better)
Climb it to 5 doors and you get swapping working .8*(.33) = 26.4% of the time, but not swapping works only 20% of the time.
Going further... if there are N doors, not swapping works (1/N)% of the time. Swapping works
(1 - (1/N))*(1/(N - 2))% of the time.
Which means, if there are 100 doors, you get a 1% of winning by not swapping, but a 1.01% chance of winning if you swap. It's always better to swap.
Now that this is understood, it's no longer counter-intuitive why swapping works. It's only difficult to grasp because we humans have a really hard time keeping anything more than "the choice available to us right now" in our heads.
"1/3rd of the time, when Monty asks, I am sitting on a car. So best not to move. But 2/3rds of the time when Monty asks, I am sitting on a goat. So - best to swap. The best to swap case comes up 2/3rds of the time; the best to sit case comes up 1/3rd of the time."
If that is still confusing, consider this:
The reason it is hard to understand is that we often think that at *every* choice opportunity, we are playing a NEW GAME. This is usually true. Play slots, play roulette, and every single spin of the wheel is a NEW GAME. It doesn't matter what happened before.
But that's not the Monty Hall game. At the first choice, you are neither a winner nor a loser - you aren't even in the game. But, your first choice turns you into either a WINNER or a LOSER. At the time of the "swap choice" - you are either a current winner or a current loser. Since you are more likely a current loser at that swap choice, you have to make the choice that gives the best chances for a current loser, and ignore the choice that gives the best options for a current loser.
---------------------------------------------------
MORE THAN TWO DOORS:
What happens if there are 4 doors at the start?
1/4 of the time, you pick the right door, and swapping (picking another door) later always loses.
3/4 of the time, you pick the wrong door, and swapping later gives you a 50% chance of winning and a 50% chance of losing, whereas not swapping gives you a 100% chance of losing.
So swapping works .75*(.5) 37.5% of the time. Not swapping works 25% of the time. So it's still better to swap (but not much better)
Climb it to 5 doors and you get swapping working .8*(.33) = 26.4% of the time, but not swapping works only 20% of the time.
Going further... if there are N doors, not swapping works (1/N)% of the time. Swapping works
(1 - (1/N))*(1/(N - 2))% of the time.
Which means, if there are 100 doors, you get a 1% of winning by not swapping, but a 1.01% chance of winning if you swap. It's always better to swap.
Now that this is understood, it's no longer counter-intuitive why swapping works. It's only difficult to grasp because we humans have a really hard time keeping anything more than "the choice available to us right now" in our heads.
12Delirium9
Excellent explanation, JPB! :)
BTW, the video made perfect sense to me. But that was the funny explanation, yours was the scholarly one. :D
BTW, the video made perfect sense to me. But that was the funny explanation, yours was the scholarly one. :D
13readafew
Another way to look at it is,
take 100 doors, your chance of picking the right door is 1% now Monty removes ALL the doors except one removing 98 goats, there is the door you have already picked and one more door. One of the doors has the car one has the goat.
You most likely (%99) picked a goat the first time so the other door is a likely (%99) the car. It has to do with the fact that there is not a random initial condition after the first round because Monty took away a KNOWN goat.
I believe that studies have been done and around %80 of people will stick with their first choice.
take 100 doors, your chance of picking the right door is 1% now Monty removes ALL the doors except one removing 98 goats, there is the door you have already picked and one more door. One of the doors has the car one has the goat.
You most likely (%99) picked a goat the first time so the other door is a likely (%99) the car. It has to do with the fact that there is not a random initial condition after the first round because Monty took away a KNOWN goat.
I believe that studies have been done and around %80 of people will stick with their first choice.
14reading_fox
Topical toushstone the curious incident of the dog
Do TV shows vet the guests as to whether they know statistics or not? After all the profit margins are probably worked out on 80% of people sticking, reducing the prize payout.
/end cynic.
Do TV shows vet the guests as to whether they know statistics or not? After all the profit margins are probably worked out on 80% of people sticking, reducing the prize payout.
/end cynic.
15JPB
The truth is, Let's Make a Deal, Monty Hall's show, never offered the 'switch' choice as described here in this problem. They kinda knew it would be stupid to do that.
16JPB
#13 Yup.
If Monty, upon hearing your guess, removes all but one door from consideration, and only removes goat doors, then the math is strongly in one's favor the more doors there are, because the formula for swapping being wrong with N doors is still (1/N)%, but the formula for being RIGHT with swapping is ((N-1)/N)%.
As N increases, the latter formula approaches 100%.
If Monty, upon hearing your guess, removes all but one door from consideration, and only removes goat doors, then the math is strongly in one's favor the more doors there are, because the formula for swapping being wrong with N doors is still (1/N)%, but the formula for being RIGHT with swapping is ((N-1)/N)%.
As N increases, the latter formula approaches 100%.
