[Shoeless]

I came across this while watching the movie '21', and thought that it's extremely interesting. Here's a little introduction;

The Monty Hall problem is a probability puzzle loosely based on the American television game show 'Let's Make a Deal'. The name comes from the show's host, Monty Hall.

Suppose you're on a game show and you're given the choice of three doors. Behind one door is a car; behind the others, goats. The car and the goats were placed randomly behind the doors before the show. The rules of the game show are as follows: After you have chosen a door, the door remains closed for the time being. The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it. If both remaining doors have goats behind them, he chooses one randomly. After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door. Imagine that you chose Door 1 and the host opens Door 3, which has a goat. He then asks you "Do you want to switch to Door Number 2?" Is it to your advantage to change your choice?

Is there a higher probability of getting the car by switching doors, or is the probability of getting the car still the same at 50/50 even after the host has opened a door with a goat behind it?



(To give an idea of how tricky this problem is, vos Savant (1996:15) in her book 'The Power of Logical Thinking' quotes cognitive psychologist Massimo Piattelli-Palmarini as saying "... no other statistical puzzle comes so close to fooling all the people all the time" and "that even Nobel physicists systematically give the wrong answer, and that they insist on it, and they are ready to berate in print those who propose the right answer.")

[Krazy P]

The Monty Hall problem is a great example of the emerging science of behavioral economics. The default bias is huge (a current McDonald's ad demonstrates it quite well) and the immediacy bias (a preference for short term benefit over long term benefit).

All these new developments can be abused by companies (and routinely are) to take advantage of consumers who are just following human nature. Likewise they can be used by companies to enrich consumers. Very interesting stuff.

One way we are experimenting with default bias is to require automatic savings as a condition of borrowing for a car. Once the default bias is established, fewer than 20% will cancel their automatic savings (even though they have the ability to do so).

[TheLastMohican]

( Show Spoiler )

Your chances of getting the car are better if you switch doors.
Monty Hall already knew what was behind the doors, and there was no chance of him opening the door concealing the car. You had a one-in-three chance of picking the door concealing the car the first time around. Suppose that's what happened, and then Monty randomly chose one of the remaining doors, since they both concealed goats. Now if you switch doors, you lose the car. But two thirds of the time, you would have chosen a door concealing a goat, and Monty would have to choose the remaining door with a goat, leaving the car.
Therefore, there is a 2/3 chance that switching doors will win the car, and a 1/3 chance that switching doors will lose the car.

(Sorry for not expressing this mathematically...I imagined it rather than formulating a proper equation.)

[Max T]

Could TheLastMohican or someone else please tell me where I'm going wrong on this (I've read the spoiler)?

I'm thinking that the chance of picking the car has shifted from 1 in 3 before Monty Hall has opened a door to 1 in 2 afterwards (50/50), as you adjust probabilities based on new information (Monty's pick).


So isn't it 1 in 2 chance of winning the car, making the question of switching irrelevant?
Inverted, the question of which door you originally chose is irrelevant, because you're just waiting for Monty to eliminate one door (you don't know whether he chose randomly or with cunning).
You could play the game by asking "Monty- eliminate one of the doors"... and that gives you 1 in 2 chance.

Thanks for any help.

[Shoeless]

	Originally Posted by Max T To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	I'm thinking that the chance of picking the car has shifted from 1 in 3 before Monty Hall has opened a door to 1 in 2 afterwards (50/50), as you adjust probabilities based on new information (Monty's pick). So isn't it 1 in 2 chance of winning the car, making the question of switching irrelevant? Thanks for any help.

You are right with the first part, that you have 1/3 chance of choosing the door with the car. After Monty opens a door with a goat behind it, you are left with 2 doors; the one you chose and one that is unopened, one has a car behind it and the other has a goat. From this manner, it does seem like a 50/50 chance of getting the car, that switching your choice does not give you a higher probability of getting the car.

Instead of just explaining it all to you (since some others haven't seen this yet), try to imagine all the possibilities that you can encounter.

[Luthor Rex]

Thought it better I make this a spoiler.

( Show Spoiler )

I found this easier when I didn't think about it in numbers.

There are three doors with two goats and one car.

Round 1: You get to choose the first door, but since there are more goats then cars you almost certainly picked a goat.

Round 2: Monty gets the second choice and since he know's what's what he will ceertainly pick a goat.

Round 3: As it stands you almost-certainly picked a goat, and Monty certainly picked a goat. Almost-certain + certain = get the hell out of Dodge and pick the other door.

[LionsPride]

Okay, I follow now. It makes more sense, to me at least, if we make the odds worse in the first round for visualization sake.

If there were 100 doors, 99 goats and 1 car. You'd have a 1 in 100 shot at the car. The chances you picked wrong are most likely. If Monty then had to open 98 doors to leave you with a choice of two doors, one with the goat and one with the car, then there is a more likely chance that there is a car behind the door you didn't choose. By not switching, you are sticking with the idea that your 1 in 100 guess was right. It's possible, that you were right the first time, but it's far more likely the car is behind the other door.

Of course, when you are working with the 1 in 3 odds, enough people would have guessed right the first time that the statistical benefit of switching doors wouldn't become well known to the public.

[TheLastMohican]

	Originally Posted by Max T To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	I'm thinking that the chance of picking the car has shifted from 1 in 3 before Monty Hall has opened a door to 1 in 2 afterwards (50/50), as you adjust probabilities based on new information (Monty's pick).

There are two possible sequences of events to consider:
1) You choose the door concealing the car. Monty then randomly picks one of the other doors, since both conceal goats.
2) You choose a door concealing a goat. Monty then picks the remaining door concealing a goat, and has no other choice.

Sequence 1 will happen one third of the time, and sequence 2 will happen two thirds of the time. In sequence 2, you will have left the door you picked (with the goat) and the door Monty left (with the car). Therefore switching doors will win the car two thirds of the time.





TheLastMohican added to this post, 16 minutes and 37 seconds later...

I googled this and found a quite extensive Wiki page with detailed solutions.

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[Max T]

Cheers for the explanations (even better than wiki's).
Hell, with my thinking I would not have even reached that round to win a car.


To reciprocate (and show I'm not completely dumb! ('though the following is simpler than Monty Hall's)), recently I was reading about the coin toss paradox (heads and tails).

The Gamblers fallacy is in thinking that, after 6 heads in a row, the next toss is more likely to be a tail. But it's still 50/50 chance- the coin has no memory of its last toss.
Judging probability based on propensity applies here: the event's likelihood is based on the event's system (50/50). Las Vegas makes a living off people making this mistake on pure luck games (e.g. roulette).

The Ludic fallacy is in thinking that, after 99 heads in a row, the next toss is still 50/50 heads or tails (i.e. misapplying the above's explanation). It's now more likely to be tails, because with numerous coin tosses, the more likely you are to see it land evenly on either heads or tails (the law of large numbers).
Judging probabability based on frequency now applies here: the event's likelihood is based on the population at large- you toss a coin 100 times and it'll roughly be 50/50 (+/- say 15) heads and tails. Toss it a million times and it virtually will be 50/50 for a symmetrical coin.

[TheLastMohican]

	Originally Posted by Max T To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	The Ludic fallacy is in thinking that, after 99 heads in a row, the next toss is still 50/50 heads or tails (i.e. misapplying the above's explanation). It's now more likely to be tails, because with numerous coin tosses, the more likely you are to see it land evenly on either heads or tails (the law of large numbers).

This doesn't seem quite right to me. As you said before, the coin has no memory of its previous toss(es), so it still has at least a 50% chance of coming up heads again. I say "at least" because after 99 heads in a row I would be fairly certain it is a two-headed coin.
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TheLastMohican added to this post, 13 minutes and 34 seconds later...

Googled this too:

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It's actually more likely to land heads up, according to the Ludic Fallacy. He compares to betting on black for a roulette wheel that has landed on red 99 times in a row. It isn't representative of true mathematical theory; it's about real-life cases including rigged probabilities.

[Max T]

Originally Posted by TheLastMohican To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.

This doesn't seem quite right to me. As you said before, the coin has no memory of its previous toss(es), so it still has at least a 50% chance of coming up heads again. I say "at least" because after 99 heads in a row I would be fairly certain it is a two-headed coin. To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts. TheLastMohican added to this post, 13 minutes and 34 seconds later... Googled this too: To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts. It's actually more likely to land heads up, according to the Ludic Fallacy. He compares to betting on black for a roulette wheel that has landed on red 99 times in a row. It isn't representative of true mathematical theory; it's about real-life cases including rigged probabilities.

The wiki entry for Ludic fallacy is confusing as it a. includes probability that coins/ roulette wheels are fixed and b. leaves a question unanswered.
Ignore probability of a fixed coin/ roulette wheel.

We know that all the coins ever tossed will be evenly distributed 50% heads and 50% tails- that's the population.
Within this population, a sample of 50 coin tosses might throw 10 tails and 40 heads.
But as the sample increases in size- 100... 1,000... 10,000, the ratio of heads to tails will converge to 50:50 as it more closely mimics the population. That's the law of large numbers.


1000 tosses all heads is like an ultra-rare 6 sigma event so the probability of the next toss being tails is heightened- not due to any memory because there isn't any, but due to the probability of 1001 heads being even rarer still.

Ludic fallacy is misapplying mathematical models to real life. Yes the first few tosses will go either way, but toss enough coins and they will become virtually perfectly evenly distributed 50:50.

[TheLastMohican]

	Originally Posted by Max T To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	1000 tosses all heads is like an ultra-rare 6 sigma event so the probability of the next toss being tails is heightened- not due to any memory because there isn't any, but due to the probability of 1001 heads being even rarer still.

I don't understand how this can be the case. Before you started tossing the coin, you could say that there was an extremely small chance of getting 1000 heads in a row, but once it has actually happened, that chance is irrelevant. Once the coin has landed heads up 1000 times, you still have a 50% chance of getting heads one more time, regardless of how many times the coin has already been tossed. It's just very unlikely that you will ever get there in the first place.

[Max T]

Originally Posted by TheLastMohican To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.

I don't understand how this can be the case. Before you started tossing the coin, you could say that there was an extremely small chance of getting 1000 heads in a row, but once it has actually happened, that chance is irrelevant. Once the coin has landed heads up 1000 times, you still have a 50% chance of getting heads one more time, regardless of how many times the coin has already been tossed. It's just very unlikely that you will ever get there in the first place.

Well, the probability that I could be wrong has increased since my first post (what do they say about dumb people keeping quiet to not confirm their stupidity)!

But to see my perspective, completely ignore the issue of coins having no memory and its corrolary (sp?) that after the 1000th head, the chance of 1000th heads is irrelevant.
That thinking leads to the pure math theory of always 50/50.

I'm not being flippant (flip- toss- geddit?) but you could literally get 10 coins, toss them all 10 times to get 100 head/tail results... and as you toss more and more and count 'em, you'll get closer to 50/ 50 split (from an initial 10 coin toss of say 1 head and 9 tails).

If you agree with that, then you'll see that 1000 tosses with say 212 tails and 788 heads will marginally more likely produce a tail next toss than a head. Marginally only, but balance tips in favour of tails next- say 52/ 48.

(This is right, isn't it?).

[TheLastMohican]

	Originally Posted by Max T To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	I'm not being flippant (flip- toss- geddit?) but you could literally get 10 coins, toss them all 10 times to get 100 head/tail results... and as you toss more and more and count 'em, you'll get closer to 50/ 50 split (from an initial 10 coin toss of say 1 head and 9 tails).

It's true that the higher the number of tosses, the greater the probability of an approximately 50/50 split...in terms of percentages of tosses. If you toss ten times you are more likely to have a perfect 50/50 split than if you toss 10,000 times. On the other hand, you also much more likely to have 40% or 60% of heads or tails in the former case, because it just takes one coin either way to make the difference.

However, the probability of the final total changes with each toss, while the probability of each toss remains 50/50. If you toss the first four of ten coins and all of them come up heads, then you have a greater probability of a final total of seven heads than of making up the difference with five out of the next six tosses coming up tails. But if you go on to get tails for the next four tosses, you go back to an even 5 heads-5 tails split as the most likely result. The probability that you had at the beginning does not stay constant as the situation changes.

BTW, don't call yourself dumb. I think we all have an intuitive idea of the gambler's fallacy that's hard to shake. I too have an instinct that when I have some bad luck, the bright side is that it won't happen to me again any time soon.
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[Moriarty]

I'd really like to see a simple computer program that could simulate this scenario. Something with:

The computer could play the role of Monty, i.e., know which door contains the prize, etc..
An adjustable number of "doors" to choose from to help visualize the problem over a varying base..
a simple clickable or menu based interface to choose the doors

I'd probably spent a few hours playing with it and compiling the results.
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[TheLastMohican]

	Originally Posted by Moriarty To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	I'd really like to see a simple computer program that could simulate this scenario.

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Have fun.

[Moriarty]

You rock!

[TheLastMohican]

	Originally Posted by Moriarty To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	You rock!

Thanks. So does Google.

[Max T]

	Originally Posted by Moriarty To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	I'd probably spent a few hours playing with it and compiling the results. To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.

I spent a few minutes playing, to see how many tries it would take to match the theoretical:
- 1 in 3 (33%) chance of winning if you stay with your selected door (7 correct out of 21 tries) and
- 2 in 3 (66%) chance of winning if switching to the last remaining door (32 correct out of 48 tries).

[Kisai]

Originally Posted by LionsPride To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.

Okay, I follow now. It makes more sense, to me at least, if we make the odds worse in the first round for visualization sake. If there were 100 doors, 99 goats and 1 car. You'd have a 1 in 100 shot at the car. The chances you picked wrong are most likely. If Monty then had to open 98 doors to leave you with a choice of two doors, one with the goat and one with the car, then there is a more likely chance that there is a car behind the door you didn't choose. By not switching, you are sticking with the idea that your 1 in 100 guess was right. It's possible, that you were right the first time, but it's far more likely the car is behind the other door. Of course, when you are working with the 1 in 3 odds, enough people would have guessed right the first time that the statistical benefit of switching doors wouldn't become well known to the public.

You know, I've been confounded by trying to understand the Monty Hall problem many times. This explanation is the one that makes me understand. Thank you very much.