21 The Movie - Math Question (1 Viewer)

[Timmypg]

I'm pretty good at math (I have a degree in accounting) but for the life of me I can't understand this scene: http://www.youtube.com/watch?v=cXqDIFUB7YU

To me by opening door 3 and revealing a goat the odds on the car went up for both doors to 50-50. Please explain.

 

[Mr. Pickles]

:facepalm:

 

[TheIronSheik]

Can't watch the video, but I believe you are referring to Bayes' Theorem.

 

[fatguyinalittlecoat]

http://forums.footballguys.com/forum/index.php?showtopic=452514&hl=+monty%20+hall

One of several threads where this was discussed. i think there are some really long ones that got purged.

 

[Short Corner]

I'm pretty good at math

 

[Chaka]

Because, math!

 

[TheIronSheik]

Because, math!

#### math.

 

[Officer Pete Malloy]

Nerds

 

[sn0mm1s]

Not sure if trolling but basically do the same thing but with many more doors - it is easier to see how this works.

Suppose you have 100 doors, 99 with goats and 1 with a car.

You choose the first door - your chance of winning the car is 1/100

Now, before the first door is opened, the game show host opens all the other doors (revealing 98 goats) but 1.

You now have the choice of switching doors. Your chance of winning isn't 50/50 if you switch - it is 99/100. The only way you lose was if you chose the correct door initially.

 

[Officer Pete Malloy]

Not sure if trolling but basically do the same thing but with many more doors - it is easier to see how this works.

Suppose you have 100 doors, 99 with goats and 1 with a car.

You choose the first door - your chance of winning the car is 1/100

Now, before the first door is opened, the game show host opens all the other doors (revealing 98 goats) but 1.

You now have the choice of switching doors. Your chance of winning isn't 50/50 if you switch - it is 99/100. The only way you lose was if you chose the correct door initially.

Listen to this guy...he went to BHS.

 

[Timmypg]

Not sure if trolling but basically do the same thing but with many more doors - it is easier to see how this works.

Suppose you have 100 doors, 99 with goats and 1 with a car.

You choose the first door - your chance of winning the car is 1/100

Now, before the first door is opened, the game show host opens all the other doors (revealing 98 goats) but 1.

You now have the choice of switching doors. Your chance of winning isn't 50/50 if you switch - it is 99/100. The only way you lose was if you chose the correct door initially.

Ok I gotcha now, thanks!

 

[belljr]

It's always 50/50 either you win or you dont

 

[Fennis]

now explain The Jokers Wild to me.

 

[xulf]

VARIABLE CHANGE.

 

[bostonfred]

Science #####es

 

[Dinsy Ejotuz]

It's always 50/50 either you win or you dont

It would be if the person eliminating the other 98 doors was doing so randomly. But since they know which door the prize is behind they aren't eliminating randomly -- they're eliminating doors that don't have the prize (unless you already picked it).

So the odds that the two doors have the prize are 1/100 (the one you picked) and 99/100 (the one they didn't eliminate). Which totals a 100% chance the prize is behind one of the two doors -- as it should.

 

[belljr]

It's always 50/50 either you win or you dont

It would be if the person eliminating the other 98 doors was doing so randomly. But since they know which door the prize is behind they aren't eliminating randomly -- they're eliminating doors that don't have the prize (unless you already picked it).

So the odds that the two doors have the prize are 1/100 (the one you picked) and 99/100 (the one they didn't eliminate). Which totals a 100% chance the prize is behind one of the two doors -- as it should.

so 50 50

 

[shadyridr]

-5^2

 

[Sarnoff]

I'm not sure about that scene in the movie "21". Did they include it thinking that everyone by now knows the Monty Hall problem, so him being one of the people that "gets it" shows that he's smart? Or did they think that it's something very few people understand and that by having him basically recite a paragraph from the wikipedia page would make him seem smart?

In either case, in the world of the film, I find it hard to believe that an MIT professor would think that his classroom full of MIT mathletes wouldn't have heard it before.

 

[Dexter]

It's always 50/50 either you win or you dont

It would be if the person eliminating the other 98 doors was doing so randomly. But since they know which door the prize is behind they aren't eliminating randomly -- they're eliminating doors that don't have the prize (unless you already picked it).

So the odds that the two doors have the prize are 1/100 (the one you picked) and 99/100 (the one they didn't eliminate). Which totals a 100% chance the prize is behind one of the two doors -- as it should.

This is a good example.

 

[rodg12]

Not this again!

 

[Hooper31]

Nobody like math. Its pretty much an axiom.

 

[Anarchy99]

I'm not the brightest bulb in the pack when it comes to math. But for the problem in the movie, don't the odds go up because Kevin Spacey PICKS which door to open, ie, it is not a random outcome? If the player picks Door #1, that leaves Door #2 and Door #3. Spacey can't pick to open the door with the new car in it and reveal the car. Do the odds change at all if the doors are opened randomly vs. being opened by knowing what is behind each door?

 

[belljr]

I'm not the brightest bulb in the pack when it comes to math. But for the problem in the movie, don't the odds go up because Kevin Spacey PICKS which door to open, ie, it is not a random outcome? If the player picks Door #1, that leaves Door #2 and Door #3. Spacey can't pick to open the door with the new car in it and reveal the car. Do the odds change at all if the doors are opened randomly vs. being opened by knowing what is behind each door?

yes they are now 50 50

 

[Short Corner]

I'm not the brightest bulb in the pack when it comes to math. But for the problem in the movie, don't the odds go up because Kevin Spacey PICKS which door to open, ie, it is not a random outcome? If the player picks Door #1, that leaves Door #2 and Door #3. Spacey can't pick to open the door with the new car in it and reveal the car. Do the odds change at all if the doors are opened randomly vs. being opened by knowing what is behind each door?

yes they are now 50 50

or 0-0

 

[Doubtingthomas]

I'm not sure about that scene in the movie "21". Did they include it thinking that everyone by now knows the Monty Hall problem, so him being one of the people that "gets it" shows that he's smart? Or did they think that it's something very few people understand and that by having him basically recite a paragraph from the wikipedia page would make him seem smart?

In either case, in the world of the film, I find it hard to believe that an MIT professor would think that his classroom full of MIT mathletes wouldn't have heard it before.

it is an easily understood problem explained simply so us non mensa can relate.

 

[jon_mx]

Does this problem still work if you have one large and two medium pizzas?

 

[JaxBill]

I heard that was half the movie that 42 was.

 

[moleculo]

here's the assumption that trips me up: the game-show host decides to open a door to reveal a goat. Does he always do that, or does he have a choice not to?

If you had selected a goat with your initial choice, why would the game show host not just reveal that door?

If you know the game show host HAS to open a door to reveal a goat after your choice, of course you should switch.

 

[jon_mx]

Not sure if trolling but basically do the same thing but with many more doors - it is easier to see how this works.

Suppose you have 100 doors, 99 with goats and 1 with a car.

You choose the first door - your chance of winning the car is 1/100

Now, before the first door is opened, the game show host opens all the other doors (revealing 98 goats) but 1.

You now have the choice of switching doors. Your chance of winning isn't 50/50 if you switch - it is 99/100. The only way you lose was if you chose the correct door initially.

It is a good method to exaggerate a problem to illustrate the point. When dealing with only three, it does not seem as intuitive as it does here.

 

[joffer]

Does this problem still work if you have one large and two medium pizzas?

Depends where you define the center of your coordinate system.

 

[jon_mx]

here's the assumption that trips me up: the game-show host decides to open a door to reveal a goat. Does he always do that, or does he have a choice not to?

If you had selected a goat with your initial choice, why would the game show host not just reveal that door?

If you know the game show host HAS to open a door to reveal a goat after your choice, of course you should switch.

He either reveals the goat or the medium prize, he never reveals the grand prize. Part of the reason is sometimes the contestant will choose the door with the goat. The host does not reveal the door you pick.

 

[jon_mx]

Does this problem still work if you have one large and two medium pizzas?

Depends where you define the center of your coordinate system.

It is usually a wise to pick the center of the pizza which is orbiting the other two.

 

[spOOfy]

bell doing some nice fishing in here

 

[Fennis]

I'm not sure about that scene in the movie "21". Did they include it thinking that everyone by now knows the Monty Hall problem, so him being one of the people that "gets it" shows that he's smart? Or did they think that it's something very few people understand and that by having him basically recite a paragraph from the wikipedia page would make him seem smart?

In either case, in the world of the film, I find it hard to believe that an MIT professor would think that his classroom full of MIT mathletes wouldn't have heard it before.

You know its a tough class because of all the charts and formulas covering the chalkboards.

The kids must be lined up to take his class, "...and then in the fifth class he holds up an ink picture, polls the class, and whoever explains why some of the class sees an old lady and the rest see a young woman gets an A."