Another math thread: The Monty Hall problem

[Sayvara]

EDIT again: While this problem is related to the one posted in this trhead [http://www.escapistmagazine.com/forums/read/18.73797], they are not the same problem.


Since math seems to be the latest fad here at OT, let me bring you some fun math history. Curiously enough, many respectable mathematicians fell for this problem and gave the wrong answer. Will you be smarter? Can you solve...

The Monty Hall Problem

The name comes from the host Monty Hall, who hosted the game show "Let's Make A Deal", where this problem first made an appearance.

You are a contestant in a game show. The game show host presents you with three doors. Behind one of the doors is a grand prize, say a nice new car. Behind two of the other doors are a goat, signifying a loss.

You are asked to pick one door, which you do. You are not shown what is behind it.

The game show host will then open one of the other two doors and there will be a goat behind that particular door.

Now you are given a choice: you may stick with your original door, or you may switch to the other unopened one.

Assuming you really want to win the great prize, what should you do now?


1) It doesn't matter if you switch or not, both give equal chance of winning.
2) You should stick with your original choice because that gives the biggest chance of winning.
3) You should switch to the other unopened door because that gives the biggest chance of winning.

Here is the solution:

Spoiler

You benefit the most from switching to the other door. There is a 2/3 chance that the prize is behind the other door, and not a 50% chance. Why? Becuase the game show host did not pick a door to open on random. He will always show you a door with a goat behind it. That means that in two out of three cases, the prize will be behind the other unopened door than the one you picked.

http://en.wikipedia.org/wiki/Monty_Hall_problem

If you have already heard of this problem, please do not spoil it for those that want to jog their brains a bit.

/S

 

[Beowulf DW]

Incidentally, we already went over this problem Finite Math just last week.

Don't worry, I won't spoil it, even though the temptation is extreme.

 

[Dr Spaceman]

Ah, the Monty Hall problem. It took about an hour for a professor of mine to explain the answer to a class I took in college. It's kinda complicated, but jeez, not that complicated!

 

[tiredinnuendo]

In strictly math terms, there are reasons why you should switch.

Though with such a small choice group it hardly matters.

- J

 

[Lukeje]

In strictly math terms, there are reasons why you should switch.

Though with such a small choice group it hardly matters.

- J

I'd say that with such a small choice group it matters all the more?

Spoiler

You DOUBLE your probability of winning by switching

 

[tiredinnuendo]

Spoiler

Only on paper. In reality, this kind of thing would be obvious if we were dealing with a million doors and he opened 999,998 of them for you, then you'd know that switching was the superior choice.

Logically, you'd pick the right one on the first try 1/3 of the time, and while those odds aren't great, they're not all that terrible either. I'd go with my gut over the math with such a small focus group. I had to do this same test in highschool with the shell game and there was only a slight difference in success for those that switched. Less than 5%.

- J

 

[Sayvara]

Spoiler

Logically, you'd pick the right one on the first try 1/3 of the time, and while those odds aren't great, they're not all that terrible either. I'd go with my gut over the math with such a small focus group. I had to do this same test in highschool with the shell game and there was only a slight difference in success for those that switched. Less than 5%.

Was the shell that was shown to you picked on random?

I did a computer simulation of this... I got the answer I stated above.

/S

 

[WhitemageofDOOM]

Spoiler

Only on paper. In reality, this kind of thing would be obvious if we were dealing with a million doors and he opened 999,998 of them for you, then you'd know that switching was the superior choice.

Logically, you'd pick the right one on the first try 1/3 of the time, and while those odds aren't great, they're not all that terrible either. I'd go with my gut over the math with such a small focus group. I had to do this same test in highschool with the shell game and there was only a slight difference in success for those that switched. Less than 5%.

- J

Actually, logically the right choice is to switch. "Going with your gut" is irrational.
50% vs. 33% chance is not a small change. your looking at about a 17% higher chance out of a hundred, or comparatively a 50% higher chance.

 

[tiredinnuendo]

Spoiler

Logically, you'd pick the right one on the first try 1/3 of the time, and while those odds aren't great, they're not all that terrible either. I'd go with my gut over the math with such a small focus group. I had to do this same test in highschool with the shell game and there was only a slight difference in success for those that switched. Less than 5%.

Was the shell that was shown to you picked on random?

I did a computer simulation of this... I got the answer I stated above.

/S

This is exactly my point. Did you notice the part where I said "In reality"? We're talking about a game show with a real life host. When we did the shell test, we had a real life person holding the shells. As others point out "going with your gut" means following the instinctive information that you're picking up from the other party. Your ability to read people can skew the odds in your favor.

On paper or in a computer, yes, switch. In real life, it doesn't matter so much with this small sample pool.

- J

 

[goodman528]

I haven't read your solution yet, but I think it's probably wrong...

Spoiler

"For your first decision the chance of picking the car is 1/3, for you second decision if you switch then your chance of picking the car becames 1/2." That's what I've heard as the solution when I was shown this problem before.

However, I disagree with this line of thinking, because probabilities always have to add up to 1. So, if your chance of winning the car if you switched was 1/2 now, that means your chance of winning the car if you don't switch must also be 1/2 (there are no other choices). The way I see it the 4 possible situations are as follows [probability of winning car in brackets]:

Picked Car, Switched [1/3 x 1/2]
Picked Goat, Switched [2/3 x 1/2]
Picked Car, Didn't Switch [1/3 x 1/2]
Picked Goat, Didn't Switch [2/3 x 1/2]

The 1/2 doesn't imply anything about your decision making preferrences, it just means there are 2 doors to choose from, you have to pick only one of them. Notice how the probability of all 4 possible situations adds up to 1.

 

[Sayvara]

I haven't read your solution yet, but I think it's probably wrong...

Whose solution? Yours or mine?

Spoiler

"For your first decision the chance of picking the car is 1/3, for you second decision if you switch then your chance of picking the car becames 1/2." That's what I've heard as the solution when I was shown this problem before.

However, I disagree with this line of thinking, because probabilities always have to add up to 1. So, if your chance of winning the car if you switched was 1/2 now, that means your chance of winning the car if you don't switch must also be 1/2 (there are no other choices). The way I see it the 4 possible situations are as follows [probability of winning car in brackets]:

Picked Car, Switched [1/3 x 1/2]
Picked Goat, Switched [2/3 x 1/2]
Picked Car, Didn't Switch [1/3 x 1/2]
Picked Goat, Didn't Switch [2/3 x 1/2]

The 1/2 doesn't imply anything about your decision making preferrences, it just means there are 2 doors to choose from, you have to pick only one of them. Notice how the probability of all 4 possible situations adds up to 1.

Very nice argument... the only thing I'm missing here is a conclusion. Is switching beneficial or not?

/S

 

[Sayvara]

On paper or in a computer, yes, switch. In real life, it doesn't matter so much with this small sample pool.

Heh, in real life, with a shell-game, you'll lose 100%. It's a fraud.

/S

 

[The_root_of_all_evil]

According to Goodman... it makes it switching equal to non-switching.

The problem comes from Monty deliberately changing the odds at the last choice.

If it's strictly Car/Goat, then it's 50%; but as he's altering the odds at that point by eliminating a 'wrong' choice; then it's actually better to switch.

Similar to the heads/tails arguments; Coming up heads is always 49.9999% (Minute chance of sides); but if you're taking into account the number already thrown as the chance, that would change the probability.

So, 3 heads in a row is 12.5% but throwing a third head after two heads is still 50%.

You can see this paradox in action by considering the Prisoner's Dilemma [http://en.wikipedia.org/wiki/Prisoners_dilemma] where the probability is based on your perception of the probability, rather than the actual one.

You can see the answer easily by having 2 cars and 2 goats, with Monty eliminating one car after you pick. Obviously it's safer to keep at this point; despite it being a fair choice to begin with.

 

[Lord Krunk]

According to Goodman... it makes it switching equal to non-switching.

That's what I was thinking. Unless the things behind the doors move, then it's still an equal chance, right?

 

[The_root_of_all_evil]

According to Goodman... it makes it switching equal to non-switching.

That's what I was thinking. Unless the things behind the doors move, then it's still an equal chance, right?

It would be if it was only the Car/Goat from the start. However, the switch chance is lowered by Monty's foul choice.
If he could remove any of the goats, including the one you may have chosen: Then it would be fair.

 

[Logan Westbrook]

What if you wanted the goat?

 

[The_root_of_all_evil]

What if you wanted the goat?

It's true...he does look a little butch...

*ahem*

Interestingly enough, using that idea; you're actually better keeping because you have 2/3 chance of getting it right first time and then 50% the second time; so best to go with your first bet.

 

[asiepshtain]

It's amazing how complicated this little problem is, and how even when you know the solution it still looks wrong.

In-fact, once, I constructed a computer program that ran a simulation of the situation, picking whether to switch or not on random and counting the results. Even at a low number of 1000 runs the ratio was exactly as predicted by the solution ( the one the OP hints at ).
So anyone who thinks the OP and the solution are wrong, trust him (and me), it's been tested.

 

[Sayvara]

What if you wanted the goat?

I covered that already.

Behind one of the doors is a grand prize, say a nice new car. Behind two of the other doors are a goat, signifying a loss.

...

Assuming you really want to win the great prize, what should you do now?

/S

 

[Sir Broccoli]

I once explained this to a friend of mine, and he didn't believe me (even though he's really good at maths) Anyway, we bet over it and we both made a computer simulation (a lot of people seem to do that )
I won