The movie 21 Please explain the math.

[drunkle]

it’s much easier to think about with higher numbers. ie., 1000 doors; your odds of picking the winner is 1:1000. the odds of the car being behind any one of the remaining doors is 999:1000. eliminating all of the other incorrect doors means that the chance of the car being in the door you didn’t pick is 999:1000.

[vegasrenter]

“it’s much easier to think about with higher numbers. ie., 1000 doors; your odds of picking the winner is 1:1000. the odds of the car being behind any one of the remaining doors is 999:1000. eliminating all of the other incorrect doors means that the chance of the car being in the door you didn’t pick is 999:1000.”

In this case, your odds BEFORE the game starts is pretty good; 999/1000. BUT once the 998 doors are open, the odds of the first choice door or the switch choice door holding the car are the same: 1/1000 of the original number, or 1/2 of the remaining unknown doors. Switching makes no difference. Knowledge of what was behind the 998 open doors does not create any knowledge of what is behind the remaining 2 closed doors.

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.

[vegasrenter]

“it’s much easier to think about with higher numbers. ie., 1000 doors; your odds of picking the winner is 1:1000. the odds of the car being behind any one of the remaining doors is 999:1000. eliminating all of the other incorrect doors means that the chance of the car being in the door you didn’t pick is 999:1000.”

In this case, your odds BEFORE the game starts is pretty good; 999/1000. BUT once the 998 doors are open, the odds of the first choice door or the switch choice door holding the car are the same: 1/1000 of the original number, or 1/2 of the remaining unknown doors. Switching makes no difference. Knowledge of what was behind the 998 open doors does not create any knowledge of what is behind the remaining 2 closed doors.

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.

[vegasrenter]

“it’s much easier to think about with higher numbers. ie., 1000 doors; your odds of picking the winner is 1:1000. the odds of the car being behind any one of the remaining doors is 999:1000. eliminating all of the other incorrect doors means that the chance of the car being in the door you didn’t pick is 999:1000.”

In this case, your odds BEFORE the game starts is pretty good; 999/1000. BUT once the 998 doors are open, the odds of the first choice door or the switch choice door holding the car are the same: 1/1000 of the original number, or 1/2 of the remaining unknown doors. Switching makes no difference. Knowledge of what was behind the 998 open doors does not create any knowledge of what is behind the remaining 2 closed doors.

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.

[vegasrenter]

“it’s much easier to think about with higher numbers. ie., 1000 doors; your odds of picking the winner is 1:1000. the odds of the car being behind any one of the remaining doors is 999:1000. eliminating all of the other incorrect doors means that the chance of the car being in the door you didn’t pick is 999:1000.”

In this case, your odds BEFORE the game starts is pretty good; 999/1000. BUT once the 998 doors are open, the odds of the first choice door or the switch choice door holding the car are the same: 1/1000 of the original number, or 1/2 of the remaining unknown doors. Switching makes no difference. Knowledge of what was behind the 998 open doors does not create any knowledge of what is behind the remaining 2 closed doors.

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.

[vegasrenter]

“it’s much easier to think about with higher numbers. ie., 1000 doors; your odds of picking the winner is 1:1000. the odds of the car being behind any one of the remaining doors is 999:1000. eliminating all of the other incorrect doors means that the chance of the car being in the door you didn’t pick is 999:1000.”

In this case, your odds BEFORE the game starts is pretty good; 999/1000. BUT once the 998 doors are open, the odds of the first choice door or the switch choice door holding the car are the same: 1/1000 of the original number, or 1/2 of the remaining unknown doors. Switching makes no difference. Knowledge of what was behind the 998 open doors does not create any knowledge of what is behind the remaining 2 closed doors.

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.

[drunkle]

no, the original 1/1000 odds that your first pick was correct (or conversely, 999/1000 that you picked incorrectly) remains the same even after elimination simply because there were that many choices at the start. that is, your 1/1000 chance does not improve to 1/2 with the elimination of incorrect choices.

if you make a choice *after* elimination, then it’s 50 50 because the door you may have chosen prior to elimination may be eliminated. but since you chose prior, it’s 1/1000 that you pick the right one and 999/1000 that you didn’t. 999/1000 odds that you chose incorrectly, given the removal of 998 doors, it’s almost a certainty that the remaining door that you didn’t chose is the winnar.

[drunkle]

no, the original 1/1000 odds that your first pick was correct (or conversely, 999/1000 that you picked incorrectly) remains the same even after elimination simply because there were that many choices at the start. that is, your 1/1000 chance does not improve to 1/2 with the elimination of incorrect choices.

if you make a choice *after* elimination, then it’s 50 50 because the door you may have chosen prior to elimination may be eliminated. but since you chose prior, it’s 1/1000 that you pick the right one and 999/1000 that you didn’t. 999/1000 odds that you chose incorrectly, given the removal of 998 doors, it’s almost a certainty that the remaining door that you didn’t chose is the winnar.

[drunkle]

no, the original 1/1000 odds that your first pick was correct (or conversely, 999/1000 that you picked incorrectly) remains the same even after elimination simply because there were that many choices at the start. that is, your 1/1000 chance does not improve to 1/2 with the elimination of incorrect choices.

if you make a choice *after* elimination, then it’s 50 50 because the door you may have chosen prior to elimination may be eliminated. but since you chose prior, it’s 1/1000 that you pick the right one and 999/1000 that you didn’t. 999/1000 odds that you chose incorrectly, given the removal of 998 doors, it’s almost a certainty that the remaining door that you didn’t chose is the winnar.

[drunkle]

no, the original 1/1000 odds that your first pick was correct (or conversely, 999/1000 that you picked incorrectly) remains the same even after elimination simply because there were that many choices at the start. that is, your 1/1000 chance does not improve to 1/2 with the elimination of incorrect choices.

if you make a choice *after* elimination, then it’s 50 50 because the door you may have chosen prior to elimination may be eliminated. but since you chose prior, it’s 1/1000 that you pick the right one and 999/1000 that you didn’t. 999/1000 odds that you chose incorrectly, given the removal of 998 doors, it’s almost a certainty that the remaining door that you didn’t chose is the winnar.

[drunkle]

no, the original 1/1000 odds that your first pick was correct (or conversely, 999/1000 that you picked incorrectly) remains the same even after elimination simply because there were that many choices at the start. that is, your 1/1000 chance does not improve to 1/2 with the elimination of incorrect choices.

if you make a choice *after* elimination, then it’s 50 50 because the door you may have chosen prior to elimination may be eliminated. but since you chose prior, it’s 1/1000 that you pick the right one and 999/1000 that you didn’t. 999/1000 odds that you chose incorrectly, given the removal of 998 doors, it’s almost a certainty that the remaining door that you didn’t chose is the winnar.

[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%

[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%

[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%

[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%

[Author]

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