gambler's fallacy

[dr_codex]

This is usually framed in a sneaky way, which is what causes a lot of confusion. Monte Hall does not open a door at random. If he did, the odds would be 50:50.
Instead, he opens one of the two doors with the booby prize. That non-random act adds information.

The other way to think of this is the "Deal or no Deal" version. If you get to the end, 23 of 25 cases have been opened. Howie then offers you a choice: stick with your initial selection, or swap with the remaining one.

This truly is a 50/50 decision. There's no information that you can have about which is higher, and nobody has had a thumb on the scale anywhere during the elimination rounds. (Well, I assume that the game isn't rigged.)

Note that the last cases in Deal or no Deal don't necessarily contain the million dollars -- that's almost never the case (so to speak). So, there was a 1/25 for each of the 2 remaining options at the beginning.

I cannot resist sharing my sister-in-law's story. Her college boyfriend played the lottery all the time. When she (an operations research major) tried to explain the odds, he stubbornly continued, reasoning: "It's a 50/50 chance. You either win or you lose."

[ergative]

This is usually framed in a sneaky way, which is what causes a lot of confusion. Monte Hall does not open a door at random. If he did, the odds would be 50:50.
Instead, he opens one of the two doors with the booby prize. That non-random act adds information.

The other way to think of this is the "Deal or no Deal" version. If you get to the end, 23 of 25 cases have been opened. Howie then offers you a choice: stick with your initial selection, or swap with the remaining one.

This truly is a 50/50 decision. There's no information that you can have about which is higher, and nobody has had a thumb on the scale anywhere during the elimination rounds. (Well, I assume that the game isn't rigged.)

Note that the last cases in Deal or no Deal don't necessarily contain the million dollars -- that's almost never the case (so to speak). So, there was a 1/25 for each of the 2 remaining options at the beginning.

I cannot resist sharing my sister-in-law's story. Her college boyfriend played the lottery all the time. When she (an operations research major) tried to explain the odds, he stubbornly continued, reasoning: "It's a 50/50 chance. You either win or you lose."

When I was 15 I thought I would be all cute and transgressive by insisting on that kind of reasoning. The reason I thought it was cute to talk that way was because I knew it was wrong. And I was 15.

[jimbogumbo]

This is indeed known as Monty's Dilemma, and there was a very readable explanation as to why 50% is not correct given that Monty reveals a door without the prize in the The Mathematics Teacher. There is something called the Gambler's Fallacy which is linked here: https://en.wikipedia.org/wiki/Gambler's_fallacy

[saffie]

There's a walkthrough of the Monty Hall problem on the board at about the 27:45 point in the video here:

https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-042j-mathematics-for-computer-science-fall-2010/video-lectures/lecture-18-probability-introduction/

[the_geneticist]

The two complications are: Montey KNOWS what is behind each door and will ALWAYS reveal a goat.  He's not going to open a door with a car behind it.  So, it's not that the math is complicated so much as there are conditional outcomes.  Plus, there are some psychological factors are play.  Most folks like to stick with their first choice because they chose it.  Switching makes you look uncertain/wrong, even though in this situation it's the most likely path to a reward (assuming that you really do want a car more than a goat!).  A clever player would see it as a guaranteed prize no matter what - I mean some folks really like goats & you don't have to pay as much in capital gains taxes on them compared to a car.

[apl68]

I mean some folks really like goats & you don't have to pay as much in capital gains taxes on them compared to a car.

A good dairy goat is well worth having.

[clean]

The classic gambler's fallacy, which author Grimes also elucidates as being the case where you flip 20 coins in a row, all coming up heads, and then the fallacious gambler assumes that the next flip is more likely to be tails, because, presumably, of the law of averages, seems different from the Monte Hall problem above.   

Flipping coins is an independent event.  Switching doors is not independent.

In the olden days (before Vegas got burned and figured it out), you could gain an advantage by watching the blackjack game and adjusting your bet when the deck turned to your favor.  What happens Before effects what happens after.

I would argue that the additional information of showing the door is similar to knowing that most of the 10 value cards are gone from the deck, but that there are plenty of low value cards remaining (which means the dealer is less likely to bust, so lower your bets).

[kaysixteen]

Vegas has more or less eliminated professional card counters, yes.

I get the argument about the 2/3 chance, when stated for *before* any doors have been selected and opened.   There could have been three doors at the start, or 3000, but when only two remain, and the contestant knows that one has a car, and the other a goat, and he is given a chance to (essentially) rechoose, either by keeping his original choice or switching, this particular choice situation is a new situation, and not related to the prior one.   In this case, it is obviously true that the contestant has a 50/50 chance of being right.   I am just simply not convinced that any other answer is legitimate here, again, for this particular *new* situation.

When Grimes was elucidating the problem,  giving the same 2/3 answer that apparently vos Savant had given, he noted that numerous PhDs, including various STEM ones, had reamed out vos Savant for her answer.   He specifically noted one guy, whose name escapes me but who apparently is some sort of award-winning science heavyweight, who initially agreed with me but eventually, after several weeks, came around to her view.   I am not going to claim anything resembling that sort of expertise, obviously, or even competence, as of course math is my weak subject, and as a classicist I have never had to use significant math, but it does seem to me that we are dealing with two distinct scenarios here-- the first one, before any doors have been opened, seems clearly to be the 2/3 one, but when the new scenario develops, well it is a 50/50 choice then, irrespective of the previous scenario, and this new choice situation cannot be anything other than a 50/50 choice.

[smallcleanrat]

In the "new scenario" you are carrying over your choice from the first scenario and given the option to stick with it or not.

Based on the way the game is played, when the choices are narrowed down to two doors (and one must hide a goat and the other a car), what is behind the other door does depend on your initial choice.

If your first guess was wrong, you should switch. If your first guess was right, you should stick with your first choice. With 3000 doors, your first guess is much more likely to be wrong
than right. So it is much more likely that, after the doors are eliminated, the other door hides the prize.

Why do you keep saying that the 50-50 odds are 'obvious' or 'self-evident?' If there is this much disagreement, how obvious could it be? It's certainly not obvious to me.

If your original choice is probably wrong, than your choice at the end should be to drop it and choose the other (which must be right if your original choice is wrong).

[arcturus]

If you have no additional information - as is the case with Deal or No Deal - when narrowed down to two choices it is a 50/50 chance as to which case has the higher dollar amount. However, in the Monty Hall case, you now have a *prior*, which is that the door that was openned was known to not have the car. Thus, the odds are no longer a strict 50/50 chance between the two doors. See the detailed examples written out above. It is more favorable to change your choice to the other door.

[clean]

It is an easy experiment.

Get some cards.  Use 1 face card and 2 number cards.  Deal them into 3 spots (A, B and C).  Your pick a spot.  Turn over card one card. (IF that card is the face card, then that one does not count.)  Now do 100 trials and NOT change your answer.  Then do 100 trials where you do.

IF you have a youngin in the house, play with them.

Keep track of the results and see if it is roughly 50/50 or not.

[Puget]

It is an easy experiment.

Get some cards.  Use 1 face card and 2 number cards.  Deal them into 3 spots (A, B and C).  Your pick a spot.  Turn over card one card. (IF that card is the face card, then that one does not count.)  Now do 100 trials and NOT change your answer.  Then do 100 trials where you do.

IF you have a youngin in the house, play with them.

Keep track of the results and see if it is roughly 50/50 or not.

Nope, this is the wrong experiment. The right experiment is for someone else who knows what the cards are to turn over a non-face card before you decide to stay or switch. That's the thing people keep trying to explain here and other people keep not getting-- Monty knows where the goats and car are, and always reveals one of the goats-- it is NOT a random draw between the two remaining doors, that's why it provides additional information. Really, it isn't complicated as soon as you understand that.

[onthefringe]

I feel like one of the issues with this problem is the disconnect between how the math/stats problem is presented and how the real game worked.

If the "host" KNOWS what doors have which prizes, and the host ALWAYS opens a door after the contestant picks, and he NEVER open the door the contestant did, and the door the host opens ALWAYS has a goat, then yes, the contestant should always switch, and their probability of winning if they switch jumps from 1/3 to 2/3.

Kay — see if this helps (it's just a more explicit version of ergative's explanation above). 3 doors (1, 2, and 3) with 2 goats and one car give you three possible scenarios — call them A, B, and C, shown in the grid below (ie scenario A has goats behind doors 1 and 2 and a car behind door 3).


   A  B  C
1  G  G  C
2  G  C  G
3  C  G  G

If the contestant chooses door 1 they have a 1/3 chance of getting a car (they only get it if the REAL scenario is "C").Then Monty opens a door. He KNOWS whether the setup is A, B, or C.

If the REAL scenario is A, Monty HAS to open door 2 in order to reveal a goat. If the real scenario is B, Monty HAS to open door 3. If the real scenario is C, Monty can open either door. After he opens a door, there are two scenarios where switching would get the contestant a car, and only one where switch gets the contestant a goat. Switching gives the contestant a 2/3 chance of getting the car.

That's not actually how Let's make a Deal worked though, so real contestants were operating under different constraints, making the stats problem one of those "assume a spherical cow on a frictionless surface" problems.

[mamselle]

We could ask this guy...

   https://www.looper.com/418312/what-it-actually-means-to-win-a-car-on-the-price-is-right/

Just appeared in my newsfeed.....

M.

[kiana]

Interestingly, I just saw a variation posted in a math help group I'm in.

There are four doors, one of which contains a prize. You choose door A. The host opens one of the other four doors that is incorrect. If you switch to one of the two closed doors, what is the probability of success?