PHILOSOPHY - Probability: The Monty Hall Problem [HD]

The Monty Hall problem is a strange result arising from a very simple situation. In this video, Bryce Gessell (Duke University) explains why it seems so counterintuitive and why the solution isn't counterintuitive at all.

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[SEAL]

After watching a ton of these and still not understanding I thought about a scenario that makes it make sense.

We all agree at the start your door has a 1/3 chance and "his" doors have a 2/3 combined chance. Select your door. Now instead of him opening one door lets pretend he says "if you chose to switch, you can open BOTH of my doors and get the prize if its within either." You would take this and accept you now have a 2 in 3 chance of winning. After you switch he says he will reveal that one of his doors is empty. This is meaningless, it's obvious that one of his doors was empty, that doesn't mean his two combined don't equal a 2/3 chance. Now he says he will tell you which of his is the empty one. But again, this offer is completely useless to you. Opening the empty door (Knowing which one of his was empty) doesn't change the 2/3 odds because it revealed no new information. You already knew that one of his two was empty.

Think of his doors as a group with 2/3 odds and yours as a group (only containing one) with 1/3 odd. Him opening the inevitably empty door does nothing to change the group odds but does confirm that the 2/3 odds his group has rests purely inside the single door. The opened door is a complete trick, you already knew it was there when you determined the 2/3 odds, having him admit it's existence doesn't change anything.

[fisyr]

If you have trouble believing Monty Hall problem works like I used to, The best way to convince yourself is to imagine the same situation with not three, but one hundred million doors. You pick one, the host then opens 99 999 998 doors which have no car: then you are asked to switch... clearly the chance of having picked the right door in the first place is close to nothing and it's the best idea to switch.

[lwpack]

An essential piece that is always left out of the explanation for this problem is that the door that's chosen to be opened is chosen from the ones with nothing. It's not a random door that happens to have nothing. The host knew what was behind it and would never open the one with the prize.

[paul]

Ugh. The host KNOWS which door contains the prize. Inexplicable that this vital information is ignored in this video!


So... when you initially pick the wrong door, as you will 66% of the time, the host will be forced to show you the only remaining empty door. Therefore the 2 of 3 times you randomly pick the wrong door, you win each of those times by switching.

[XaxtonRevolution]

that is the best explanation of the monty hall problem ever

[tippyc]

easiest way to prove this to yourself is to draw a probability tree of the problem. theres a pretty small number of branches, so its pretty easy to do.

[ZoggFromBetelgeuse]

Another way to understand this is to consider a variant of this game : You chose one out of 12 doors, let's say it's door 1. The host opens doors 2, 3, 4, 5, 6, 8, 9, 10, 11 and 12 - they are all empty.
Should you switch to door 7 ?

[JoeyBlogs]

More simply, every door has 1/3 chance of the prize. One door/goat is eliminated by the host, who is not going to reveal the car until the final selection. If you unknowingly picked the car ( which occurs 1/3 of the time ) and switch, you lose ( as you switch to a goat ). However if you unknowingly picked a goat (which occurs 2/3 of the time) and one door/goat is eliminated, then when you switch you win ( as you switch to the car ). Thus 2/3 of the time switching wins and 1/13 of the time switching loses.

[nhwnhw]

Conditional probability is always elegant but can prove confusing. The words used generally new to most people. What if you consider this problem as there are 2 groups. For this example which number or letter is on the door isn't that important. The 2 groups are "first pick" and "not picked". The first "pick group" is composed of which door you pick first. The "not picked" group is the other 2 doors. A group of 1 door (FP) and a group of 2 doors (NP). There is only 1 winner and 2 losers spread across the 2 groups in a fashion unknown except to Monty. Monty shows you a loser from the "not picked" group every time. You know in this example any group of 2 has to contain at least 1 loser. So switch or stay? You should always switch for a better chance to win. Switching in a way gives you two chances. The "not picked" group always has 2 members. Not understanding this is why casinos are so nice...

[PaulRubino]

Kind of unnecessarily lengthy. Quick analysis: Pick one of the doors. The chances you picked the correct door is 1/3. The chances that the prize is behind one of the other two doors is 2/3. Forget getting a peek behind one of the other two doors, would you swap your one door to have whatever is behind both the remaining two doors? That's it.

[JancenRevo]

so what does this teach us about our brains or about human logic?

[MrCman]

Have there ever been statistical tests to prove this phenomenon as true? From a logical perspective I understand the implications, but would it carry over to reality? If this test was done 1000 times would around 666 times the participant get the winning prize?

[bskeptic]

This is the shortest and best explanation of this problem that I've ever heard.

[shelleywinters]

good explanation. usually these problems make my head hurt LOL

[sebasforest]

So if you had to choose between 2 doors, A and B (so 1/2 chance), picked A, imagined a non existing door C, mentally discarded it and then switched your choice to door B, would your odds also improve to 2/3?

[cliffordhodge]

Consider a raffle in which each of 100 players buys one ticket from Monty, who then picks the winning ticket and puts it in an envelope, unannounced.  Unbeknownst to the players, the winning ticket is #50.  Imagine further that the holder of #50 stubbornly keeps his ticket to the bitter end,   while the other players go up to Monty, one by one, to find out what ticket number wins.  So one player goes up and asks Monty the winning number, then announces to the group of 100, "My ticket is not the winner."  At that point, all the players except the holder of #50 trade tickets with one another to increase their chances of winning.  After many iterations of this process - many rounds of trading - the traders should have increased chances of winning, if the principles put forth in the video apply.  But probability is a measure of ignorance, not of predispositional properties of the material world.  The probability that the holder of #50 wins remains exactly 1.0 from the drawing to the final announcement.  And the probability of any other player winning was, after the drawing, a perfect 0, and remains so.  The fact that they believe they increase their chances is irrelevant.

[cliffordhodge]

Final raffle consideration.  For any player to gain probability of winning, must not another player have diminished probability?  At no point does #50 become less than THE WINNER.  So how does trading enable anyone to gain?  The statement of the Monty Hall problem seems to suggest that winning probability is a property of the player and not a property of the winning non-winning door.  But in the raffle it seems clear that the holder of #50 is and remains the winner, while all others remain non-winners.  The enthusiastic circulation of tickets does not change that.  But it seems each ticket is like a door, and at each iteration of the announcement/trading, each player except the holder of #50 chooses a new door.

[dawyer]

Each sheep is also an independent and distinct individual. This game cannot mislead players by treating every sheep as the same thing.
There are three options, and two of the sheep also exist independently. If the player can distinguish them, there will be no chance of winning like 2/3.

[marksesl]

This is a conditional probability problem, so the ONLY proper explanation is to explain how the condition affects the outcome: Since Monty is restricted from revealing the car and gets two doors of the three, there is a 2/3 probability it will be behind the door he doesn't open. That's all. FINISHED.

[cliffordhodge]

Further raffle considerations.  Suppose that after the first round of trading in the raffle of 100 players, all the holders of the non-winning tickets trade tickets.  (The holder of #50, which is in fact the winning ticket, again keeps his ticket to the very end).  The video would dictate that each of the non-winners has, by trading, increased his chances of winning.  So imagine that at each round, when another non-winning ticket number is announced, one or more of the traders engages in multiple trades.  Does a 2nd or 3rd (or 60th) trade in each round vastly increase the winning probability with every single trade within a round?  Or does it increase only once with each round?  That is, does the increase in probability depend on the actual trade, or on the announcement of another non-winning number?  Further, if you claim this is not a parallel problem to the Monty Hall problem because the holder of #50 refuses to trade, how is that relevant from the epistemological standpoint of each trading player?  Hasn't each player gained the same kind of knowledge with each announcement of a non-winner, albeit his 'knowledge' is of the members of a larger group than just 2 or 3?