There’s a fun and notorious probability problem known as the Monty Hall problem. It’s notorious for tons of reasons. The setup is simple.
You’re on a game show and are presented with three closed doors; a prize behind 1 of the doors; non-prizes behind 2 of the doors. Usually it’s something like, a car behind one of the doors and a goat behind each of the other doors.
The rules of the situation are that once you choose a door, one of the goat-doors from the remaining doors will be revealed. After the reveal, you are allowed to switch your choice, or keep it the same.
Assuming you want the prize, what strategy should you adopt? Should you stay or should you switch?
Turns out, you should switch. You have a 2/3 chance of getting the prize over the long run by switching, while a 1/3 chance of getting the prize if you stay.
It’s a notorious problem because most people — especially somewhat math-inclined people — are generally tempted to reject the answer. They will insist that the odds are 50-50.
There’s a YouTube video that mentions the Monty Hall problem, and the correct solution. A person left a comment saying that the real odds are 50-50 — and confronted with the possibility that someone is wrong on the internet — I engaged with YouTube user: adolthitler.
I tried several ways to explain the reasoning behind the correct answer, and after a lot of back and forth, the last volley is (currently)
jbwhitmore (me) writes:
@adolthitler I don’t forget that the door is removed — and it isn’t just removed, it’s revealed — and that is crucial to why a person should switch.
(a) If you happened to have picked a door with a goat, when the only other goat in the situation is revealed, you are guaranteed to win the car IF YOU SWITCH.
(b) And since you agree that 2/3 of the time you will pick the door with a goat, it follows by necessity that switching will win you the car 2/3 of the time.
It’s not opinion, it’s math.
adolthitler (other guy) writes:
@jbwhitmore its opinioinated math. There are two doors left. NOT THREE. Its now 50% chnace the car is behind door 1, and 50% its behind door 3, thats simple math.
Who cares when you made the choice? It does not affect the outcome.
You have fallen prey to a fallacy. Count the doors before you switch. 2 doors, and there is nothing magical making one more likely. Its simple logic. Your maths works fine, your logic is shit.
2/3 of the time you chose a door with a goat, and 1 goat is revealed, so its now 1/2 times you chose a goat. Got it you removed one door, you did not alter your math. Redo the math based on the current situation, not the historical situation.
2 doors, not three the removed door boosts your choice just as much as the alternate door.
Right, so adolthitler is both wrong and impervious to being corrected — and publicly proclaiming that he’s correct at my expense. The way I generally deal with these situations in real life is to make a bet with the person. That way, even if he won’t change his mind, he’ll get punished monetarily for holding false and uncorrectable beliefs, and I’ll benefit. It’s win-win as far as I’m concerned.
The conundrum was — how do I set the terms of the bet in such a way that I can actually make the bet and win?
Here’s part of what I’ve come up with — since adolthitler resolutely thinks that the odds are 50-50 whether you switch or stay with your first choice, and I resolutely think that the odds are 66-33 for switching — we could pay each other depending on whose strategy wins.
Every time staying wins — I’d pay adolthitler $1.08.
Every time switching wins — adolthitler would pay me $1.00.
That’s an 8% payout difference on what adolthitler thinks is a 50-50 outcome, and if he’s got conviction he should really want to take this bet — and to play as many times as possible.
How the Bet is Set Up
Using a random number generator to determine which door holds the prize. (Here is a way to choose one number from [1,2,3] randomly a thousand times).
Step 1: We always choose the same door to start with — let’s just say we choose Door 1 every time.
Step 2: One goat is revealed (from behind Door 2 or 3).
I am forced to always switch to the other closed door.
adolthitler is forced to always stays with Door 1.
Step 3: Whoever ends up picking the right door gets paid by the other person. (The going rate is $1.08 that I would have to adolthitler; and $1.00 that adolthitler would pay me).
Here’s how the first 10 rounds would play out by picking the first 10 random prize placements:
Round 1
Prize is behind Door 2
Step 1: We both choose Door 1
Step 2: Door 3 is revealed to hold the other goat (this is forced)
Step 3: I am forced to switch to Door 2 (this forces me to win)
Result: I win $1.00 (My total: +$1.00)
Round 2
Prize is behind Door 3
Step 1: We both choose Door 1
Step 2: Door 2 is revealed to hold the other goat (this is forced)
Step 3: I am forced to switch to Door 3 (this forces me to win)
Result: I win $1.00 (My total: +$2.00)
Round 3
Prize is behind Door 1
Step 1: We both choose Door 1
Step 2: Door 2 or 3 is revealed to hold a different goat (choice)
Step 3: I am forced to switch to either 2 or 3 (this forces me to lose)
Result: I lose $1.08 (My total: +$0.92)
Round 4
Prize is behind Door 2
Step 1: We both choose Door 1
Step 2: Door 3 is revealed to hold the other goat (this is forced)
Step 3: I am forced to switch to Door 2 (this forces me to win)
Result: I win $1.00 (My total: +$1.92)
Round 5
Prize is behind Door 3
Step 1: We both choose Door 1
Step 2: Door 2 is revealed to hold the other goat (this is forced)
Step 3: I am forced to switch to Door 3 (this forces me to win)
Result: I win $1.00 (My total: +$2.92)
Round 6
Prize is behind Door 1
Step 1: We both choose Door 1
Step 2: Door 2 or 3 is revealed to hold a different goat (choice)
Step 3: I am forced to switch to either 2 or 3 (this forces me to lose)
Result: I lose $1.08 (My total: +$1.84)
Round 7
Prize is behind Door 3
Step 1: We both choose Door 1
Step 2: Door 2 is revealed to hold the other goat (this is forced)
Step 3: I am forced to switch to Door 3 (this forces me to win)
Result: I win $1.00 (My total: +$2.84)
Round 8
Prize is behind Door 1
Step 1: We both choose Door 1
Step 2: Door 2 or 3 is revealed to hold a different goat (choice)
Step 3: I am forced to switch to either 2 or 3 (this forces me to lose)
Result: I lose $1.08 (My total: +$1.76)
Round 9
Prize is behind Door 3
Step 1: We both choose Door 1
Step 2: Door 2 is revealed to hold the other goat (this is forced)
Step 3: I am forced to switch to Door 3 (this forces me to win)
Result: I win $1.00 (My total: +$2.76)
Round 10
Prize is behind Door 1
Step 1: We both choose Door 1
Step 2: Door 2 or 3 is revealed to hold a different goat (choice)
Step 3: I am forced to switch to either 2 or 3 (this forces me to lose)
Result: I lose $1.08 (My total: +$1.68)
How many rounds would you like to play? A thousand? A million? Ten million? I’ll play until someone goes bankrupt.
Side-bet to any interested readers: How long before adolthitler gives up?
Comments on this entry are closed.
I have never heard the solution to this riddle described so elegantly. Well done!
I wouldn't upset adolthitler too much. I just visited his YouTube channel and he sounds like a pro-wrestler…
Thanks Cedric!
Jonathan,
I want some of your action! Great job, our school system really does a poor job at teaching basic logical thinking and math concepts.
I hope all is well and you have a super Christmas
Unfortunately, the riddle as you proposed it in fact does have 50% 50% odds. You have left out the criterion that Monte Hall knows which door hides the prize. So I'm afraid your friend is correct for saying 50%, although his reasoning is possibly not accurate.
Say without loss of generality you always pick door A and are stubborn to stick with your original choice. Then 1/3 of the time you are correct, and Monte opens another door, and staying wins. The other 2/3 of the time you pick the wrong door, but since Monte Hall doesn't know the location of the door, then given this situation, 1/2 of the time he will open the wrong door (one of the two remaining doors holds the prize). However, we know that in the current instance he actually does open the correct door, so we must exclude these failed possibilities from the result. So he opens the correct door with probability 1/3+2/3*1/2=2/3. So then we must renormalize our probability with respect to the chances that Monte Hall opens the correct door, so the full probability is 1/3/(2/3)=1/2. If you don't believe that, then think: how can someone who doesn't know the location of the prize convey information about said prize by opening a door? He cannot. It may be useful to imagine it this way: when Monte doesn't know the location of the door, if you choose the prize originally, then he's guaranteed to open the correct door (all that's left are empty doors), whereas if you DO choose the correct door, he only opens the door with 1/2 probability (because only one of the two remaining doors has the prize). This means the events where you originally choose a non-prize door sometimes get "tossed out" because Monte opens the wrong door. That's great for you if you're stubborn on not switching because you want to delete all records of yourself picking the wrong door, and 1/2 of the time Monte opens the wrong door and thus throws out that test automatically. However, if you're switching, then Monte's carelessness is harmful to you; you only win when you originally pick the wrong door, but these are the only cases where Monte sometimes makes a mistake. When you would usually win, Monte now screws up 1/2 of the time, and thereby occasionally eliminates the record of your win happening in the first place. I hope this has helped you, and you might find it worthwhile to write another little script that goes through the same process but this time does not "know" which door is the right one. It is a very important thing to mention that Monte knows where the prize is in the problem statement.
"You have left out the criterion that Monte Hall knows which door hides the prize."
You are absolutely correct that it has to be known beforehand which door has a goat before it is reveal, but I don't think that I left this out of my setup as far as I can tell. Here is my setup:
"The rules of the situation are that once you choose a door, one of the goat-doors from the remaining doors will be revealed. After the reveal, you are allowed to switch your choice, or keep it the same."
It isn't just "you choose a door, a goat happens to be revealed" (which makes it 50-50); it's a rule of the game (as I write it) that "one of the goat-doors will be revealed".
how you arrived at your odds are interesting [your $1.08 to my $1.oo] will you take the same bet if the odds are your $3 to my $1 ? If not , how about your $2 to my $1… One bet one game.
No, I would not take the $3:$1 nor the $2:$1 bets.
thanks for your reply. chibayer@gmail.com