The movie 21 Please explain the math.

[ucodegen]

Oy vey..

It is assumed that the contestant is making a correct statement by saying “The contestant does decide to change from door number 1 to door #2. He states by changing from original chosen door #1 to door #2 his odds of winning the car increased to 66.7%.” He is not. He is showing a complete lack of understanding in probability. The gameshows do this to up the ‘excitement’ because otherwise the game is played out. It allows them to clock another 5+ minutes on the same stupid choice with everyone else yelling and screaming. What it does tell you is that the host knows where the car is and a good poker player may be able to bluff it out of the game show host by watching the hosts responses to potential choices.

Basically, on the initial selection: 1 correct out of 3, making the odds 33% that any particular choice is right.

After the choice, one of the potentials is removed by uncovering the goat. This makes 1 out of 2 possible or 50% any particular choice is right.

Whether the choice is;
1) He has already chosen one and gets to modify his choice
2) There is a new choice between two brand new doors…

It is the same thing. The underlying thing is that revealing what is under one of the doors does not alter what is behind any of the other doors. Also remember that the sum of probabilities on any single choice have to sum to 1 and you do not get to carry over probabilities from a previous run on to a new choice, in particular when a new piece of information is added (revealing of the goat).

I dare the doubters of this to write a program that allows you to pick one of three.. reveals one that it knows is not it and asks you to choose to keep your choice or switch. Track the frequency that the original choice is right versus switching was the right choice. The revealed choice in the first selection is not counted because it was not made by the person doing the selection.

And for a final nail:
Why would a gameshow give you a choice that could suddenly improve your odds dramatically and so easily, particularly since they have to foot the bill?

[ucodegen]

The whole problem is that you added info to the system which altered the game. It also causes the probability calcs to be completely reshuffled. You can’t carry over probability calcs when info gets added to the system. Example:

I have 10 doors. The prize is behind one: The odds of winning are 10%

I then open the door with the prize and then close it.
Are the odds of winning 10% per door, or 100% on the door I opened and then closed with 0% everywhere else. The information resets the calcs.

[ucodegen]

The whole problem is that you added info to the system which altered the game. It also causes the probability calcs to be completely reshuffled. You can’t carry over probability calcs when info gets added to the system. Example:

I have 10 doors. The prize is behind one: The odds of winning are 10%

I then open the door with the prize and then close it.
Are the odds of winning 10% per door, or 100% on the door I opened and then closed with 0% everywhere else. The information resets the calcs.

[ucodegen]

The whole problem is that you added info to the system which altered the game. It also causes the probability calcs to be completely reshuffled. You can’t carry over probability calcs when info gets added to the system. Example:

I have 10 doors. The prize is behind one: The odds of winning are 10%

I then open the door with the prize and then close it.
Are the odds of winning 10% per door, or 100% on the door I opened and then closed with 0% everywhere else. The information resets the calcs.

[ucodegen]

The whole problem is that you added info to the system which altered the game. It also causes the probability calcs to be completely reshuffled. You can’t carry over probability calcs when info gets added to the system. Example:

I have 10 doors. The prize is behind one: The odds of winning are 10%

I then open the door with the prize and then close it.
Are the odds of winning 10% per door, or 100% on the door I opened and then closed with 0% everywhere else. The information resets the calcs.

[ucodegen]

The whole problem is that you added info to the system which altered the game. It also causes the probability calcs to be completely reshuffled. You can’t carry over probability calcs when info gets added to the system. Example:

I have 10 doors. The prize is behind one: The odds of winning are 10%

I then open the door with the prize and then close it.
Are the odds of winning 10% per door, or 100% on the door I opened and then closed with 0% everywhere else. The information resets the calcs.

[Eugene]

omg…

on the initial selection: 1 correct out of 3, making the odds 33% that any particular choice is right.

Exactly. So why do you think that the chance to have a car behind that door suddenly goes up from 33% to 50% when game show host reveals the goat?

Let’s call the initial choice door #1.

If you do this game 100 times, in 33 cases the car will be behind door #1, in 67 cases it will be behind door #2 or #3.

If game show host always shows you the goat and never misses, he knows which door has the goat behind it. You will have 33 cases where the car is behind #1 and 67 cases when the car is behind the only other door that hasn’t been opened yet. Chance to win when switching: 67/100.

If game show host opens the door at random: in 33 cases he will reveal the car and all will be over; in 33 cases the car will be behind #1; and in 33 cases it will be behind the other unopened door.

[Eugene]

omg…

on the initial selection: 1 correct out of 3, making the odds 33% that any particular choice is right.

Exactly. So why do you think that the chance to have a car behind that door suddenly goes up from 33% to 50% when game show host reveals the goat?

Let’s call the initial choice door #1.

If you do this game 100 times, in 33 cases the car will be behind door #1, in 67 cases it will be behind door #2 or #3.

If game show host always shows you the goat and never misses, he knows which door has the goat behind it. You will have 33 cases where the car is behind #1 and 67 cases when the car is behind the only other door that hasn’t been opened yet. Chance to win when switching: 67/100.

If game show host opens the door at random: in 33 cases he will reveal the car and all will be over; in 33 cases the car will be behind #1; and in 33 cases it will be behind the other unopened door.

[Eugene]

omg…

on the initial selection: 1 correct out of 3, making the odds 33% that any particular choice is right.

Exactly. So why do you think that the chance to have a car behind that door suddenly goes up from 33% to 50% when game show host reveals the goat?

Let’s call the initial choice door #1.

If you do this game 100 times, in 33 cases the car will be behind door #1, in 67 cases it will be behind door #2 or #3.

If game show host always shows you the goat and never misses, he knows which door has the goat behind it. You will have 33 cases where the car is behind #1 and 67 cases when the car is behind the only other door that hasn’t been opened yet. Chance to win when switching: 67/100.

If game show host opens the door at random: in 33 cases he will reveal the car and all will be over; in 33 cases the car will be behind #1; and in 33 cases it will be behind the other unopened door.

[Eugene]

omg…

on the initial selection: 1 correct out of 3, making the odds 33% that any particular choice is right.

Exactly. So why do you think that the chance to have a car behind that door suddenly goes up from 33% to 50% when game show host reveals the goat?

Let’s call the initial choice door #1.

If you do this game 100 times, in 33 cases the car will be behind door #1, in 67 cases it will be behind door #2 or #3.

If game show host always shows you the goat and never misses, he knows which door has the goat behind it. You will have 33 cases where the car is behind #1 and 67 cases when the car is behind the only other door that hasn’t been opened yet. Chance to win when switching: 67/100.

If game show host opens the door at random: in 33 cases he will reveal the car and all will be over; in 33 cases the car will be behind #1; and in 33 cases it will be behind the other unopened door.

[Eugene]

omg…

on the initial selection: 1 correct out of 3, making the odds 33% that any particular choice is right.

Exactly. So why do you think that the chance to have a car behind that door suddenly goes up from 33% to 50% when game show host reveals the goat?

Let’s call the initial choice door #1.

If you do this game 100 times, in 33 cases the car will be behind door #1, in 67 cases it will be behind door #2 or #3.

If game show host always shows you the goat and never misses, he knows which door has the goat behind it. You will have 33 cases where the car is behind #1 and 67 cases when the car is behind the only other door that hasn’t been opened yet. Chance to win when switching: 67/100.

If game show host opens the door at random: in 33 cases he will reveal the car and all will be over; in 33 cases the car will be behind #1; and in 33 cases it will be behind the other unopened door.

[jficquette]

Ucodegen,

I agree with you however the games shows want people to win so that people will watch. The merchandise is donated by the manufacturers.

John

[jficquette]

Ucodegen,

I agree with you however the games shows want people to win so that people will watch. The merchandise is donated by the manufacturers.

John

[jficquette]

Ucodegen,

I agree with you however the games shows want people to win so that people will watch. The merchandise is donated by the manufacturers.

John

[jficquette]

Ucodegen,

I agree with you however the games shows want people to win so that people will watch. The merchandise is donated by the manufacturers.

John

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