The movie 21 Please explain the math.

[pepsi]

[quote=vegasrenter]

I reserve the right to be wrong, given that some really smart guys see it the other way, but it seems pretty simple. There’s no difference between the remaining 2 closed doors.

[/quote]

It is not independent event, because the doors still closed are not random. One of them can not be opened for a reason.
In the 3 door example, the host has no choice when picking a door to show you the goat, when your first guess is wrong.
When your first guess is wrong (2/3 chance), the hose must opened (100% chance) a certain door that has goat behind it, then the remaining door must have the prize (100% chance).
So, 66% * 100% * 100% = 66%

[drunkle]

to put it another way, which would you rather have: 1 lottery ticket or the whole rest of the roll, complete with a well dressed and witty fellow to slavishly scratch all the tickets for you?

obviously, you’d take the roll because your chances are vastly improved.

[drunkle]

to put it another way, which would you rather have: 1 lottery ticket or the whole rest of the roll, complete with a well dressed and witty fellow to slavishly scratch all the tickets for you?

obviously, you’d take the roll because your chances are vastly improved.

[drunkle]

to put it another way, which would you rather have: 1 lottery ticket or the whole rest of the roll, complete with a well dressed and witty fellow to slavishly scratch all the tickets for you?

obviously, you’d take the roll because your chances are vastly improved.

[drunkle]

to put it another way, which would you rather have: 1 lottery ticket or the whole rest of the roll, complete with a well dressed and witty fellow to slavishly scratch all the tickets for you?

obviously, you’d take the roll because your chances are vastly improved.

[drunkle]

to put it another way, which would you rather have: 1 lottery ticket or the whole rest of the roll, complete with a well dressed and witty fellow to slavishly scratch all the tickets for you?

obviously, you’d take the roll because your chances are vastly improved.

[ucodegen]

Drunkle got one of the better explanations. It is easier to think of it with fewer objects than 1000 though. Simply the following.

We have goat1, goat2 and car. Goat1 and Goat2 are effectively equivalent.

Odds on first pick of picking either Goat1, Goat2 or Car are 33%.

1)If on first pick, you picked Goat1, the announcer will reveal Goat2 making the Car the other unselected pick. Odds of being in this state is 33% (see above). Switch has odds of 100% of success in this state (unless you need a goat)

2)If on first pick, you picked Goat2, the announcer will reveal Goat1 making the Car the other unselected pick. Odds of being in this state is 33% (see 2nd above). Switch has odds of 100% success in this state.

3)If on first pick, you picked Car, the announcer will reveal Goat1 or Goat2 randomly since they are both unselected. Odds of being in this state are 33% (see 3rd above). Switch has odds of 0% success in this state.

Therefore, there is a 66% chance of being in a state where switching choices will benefit and 33% chance where it won’t. It is important to note that the announcers actions are not random when in state 1 and 2.

It does seem counterintuitive. The key is that the announcer’s actions are not random in 66% of the cases.

[ucodegen]

Drunkle got one of the better explanations. It is easier to think of it with fewer objects than 1000 though. Simply the following.

We have goat1, goat2 and car. Goat1 and Goat2 are effectively equivalent.

Odds on first pick of picking either Goat1, Goat2 or Car are 33%.

1)If on first pick, you picked Goat1, the announcer will reveal Goat2 making the Car the other unselected pick. Odds of being in this state is 33% (see above). Switch has odds of 100% of success in this state (unless you need a goat)

2)If on first pick, you picked Goat2, the announcer will reveal Goat1 making the Car the other unselected pick. Odds of being in this state is 33% (see 2nd above). Switch has odds of 100% success in this state.

3)If on first pick, you picked Car, the announcer will reveal Goat1 or Goat2 randomly since they are both unselected. Odds of being in this state are 33% (see 3rd above). Switch has odds of 0% success in this state.

Therefore, there is a 66% chance of being in a state where switching choices will benefit and 33% chance where it won’t. It is important to note that the announcers actions are not random when in state 1 and 2.

It does seem counterintuitive. The key is that the announcer’s actions are not random in 66% of the cases.

[ucodegen]

Drunkle got one of the better explanations. It is easier to think of it with fewer objects than 1000 though. Simply the following.

We have goat1, goat2 and car. Goat1 and Goat2 are effectively equivalent.

Odds on first pick of picking either Goat1, Goat2 or Car are 33%.

1)If on first pick, you picked Goat1, the announcer will reveal Goat2 making the Car the other unselected pick. Odds of being in this state is 33% (see above). Switch has odds of 100% of success in this state (unless you need a goat)

2)If on first pick, you picked Goat2, the announcer will reveal Goat1 making the Car the other unselected pick. Odds of being in this state is 33% (see 2nd above). Switch has odds of 100% success in this state.

3)If on first pick, you picked Car, the announcer will reveal Goat1 or Goat2 randomly since they are both unselected. Odds of being in this state are 33% (see 3rd above). Switch has odds of 0% success in this state.

Therefore, there is a 66% chance of being in a state where switching choices will benefit and 33% chance where it won’t. It is important to note that the announcers actions are not random when in state 1 and 2.

It does seem counterintuitive. The key is that the announcer’s actions are not random in 66% of the cases.

[ucodegen]

Drunkle got one of the better explanations. It is easier to think of it with fewer objects than 1000 though. Simply the following.

We have goat1, goat2 and car. Goat1 and Goat2 are effectively equivalent.

Odds on first pick of picking either Goat1, Goat2 or Car are 33%.

1)If on first pick, you picked Goat1, the announcer will reveal Goat2 making the Car the other unselected pick. Odds of being in this state is 33% (see above). Switch has odds of 100% of success in this state (unless you need a goat)

2)If on first pick, you picked Goat2, the announcer will reveal Goat1 making the Car the other unselected pick. Odds of being in this state is 33% (see 2nd above). Switch has odds of 100% success in this state.

3)If on first pick, you picked Car, the announcer will reveal Goat1 or Goat2 randomly since they are both unselected. Odds of being in this state are 33% (see 3rd above). Switch has odds of 0% success in this state.

Therefore, there is a 66% chance of being in a state where switching choices will benefit and 33% chance where it won’t. It is important to note that the announcers actions are not random when in state 1 and 2.

It does seem counterintuitive. The key is that the announcer’s actions are not random in 66% of the cases.

[ucodegen]

Drunkle got one of the better explanations. It is easier to think of it with fewer objects than 1000 though. Simply the following.

We have goat1, goat2 and car. Goat1 and Goat2 are effectively equivalent.

Odds on first pick of picking either Goat1, Goat2 or Car are 33%.

1)If on first pick, you picked Goat1, the announcer will reveal Goat2 making the Car the other unselected pick. Odds of being in this state is 33% (see above). Switch has odds of 100% of success in this state (unless you need a goat)

2)If on first pick, you picked Goat2, the announcer will reveal Goat1 making the Car the other unselected pick. Odds of being in this state is 33% (see 2nd above). Switch has odds of 100% success in this state.

3)If on first pick, you picked Car, the announcer will reveal Goat1 or Goat2 randomly since they are both unselected. Odds of being in this state are 33% (see 3rd above). Switch has odds of 0% success in this state.

Therefore, there is a 66% chance of being in a state where switching choices will benefit and 33% chance where it won’t. It is important to note that the announcers actions are not random when in state 1 and 2.

It does seem counterintuitive. The key is that the announcer’s actions are not random in 66% of the cases.

[citydweller]

Here’s an easy way to prove that that the odds of winning the car are higher if you switch doors after the reveal.

Assuming that you will always switch doors after the reveal, then the ultimate outcome is determined as soon as you make your first choice.

In other words, if your first choice was a goat, then you are going to end up with the car. The chance of picking a goat on your first choice is 66%.

[citydweller]

Here’s an easy way to prove that that the odds of winning the car are higher if you switch doors after the reveal.

Assuming that you will always switch doors after the reveal, then the ultimate outcome is determined as soon as you make your first choice.

In other words, if your first choice was a goat, then you are going to end up with the car. The chance of picking a goat on your first choice is 66%.

[citydweller]

Here’s an easy way to prove that that the odds of winning the car are higher if you switch doors after the reveal.

Assuming that you will always switch doors after the reveal, then the ultimate outcome is determined as soon as you make your first choice.

In other words, if your first choice was a goat, then you are going to end up with the car. The chance of picking a goat on your first choice is 66%.

[citydweller]

Here’s an easy way to prove that that the odds of winning the car are higher if you switch doors after the reveal.

Assuming that you will always switch doors after the reveal, then the ultimate outcome is determined as soon as you make your first choice.

In other words, if your first choice was a goat, then you are going to end up with the car. The chance of picking a goat on your first choice is 66%.

[Author]

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