The movie 21 Please explain the math.

[Coronita]

[quote]I don’t know why some of you people are arguing against the 2/3 answer. It IS the correct answer, whether you understand the logic or not.

People who study probabilities for a living have said this is the answer. All you amateur naysayers should just capitulate and move on…:-)
[/quote]

LOL…I chuckle a bit here, because I had the same feeling.But, thankfully, a lot of you aren’t responsible for engineering things we depend on. Just kidding.

Anyway, don’t mean anything by it. Take a probability class, and then take a stochastic processes class. Much of this is required especially for information theory and communication systems. I guess it’s also applicable to finance, but I don’t know much about that subject.

[equalizer]

Here’s a very simple explanation. You pick door 1 and since there is a 1/3 chance of picking car, then there must be 2/3 chance that car is behind door 2 or 3. When Monty shows you door 2, he will never show you the car. Thus, he just spilled the beans, that is the odds of the car behind door 2 or 3 is still 2/3, but since he showed it is not door 2, the odds that it is just behine door #3 is 2/3.

Here’s the definitive explanation and simulation from nerds at UCSD math dept. Monty knows where the car is and that is why the odds from switching is 2/3. (He never open up the door with the car)

Simulation:
http://math.ucsd.edu/~crypto/Monty/monty.html
explanation:
http://math.ucsd.edu/~crypto/Monty/montybg.html

[equalizer]

Here’s a very simple explanation. You pick door 1 and since there is a 1/3 chance of picking car, then there must be 2/3 chance that car is behind door 2 or 3. When Monty shows you door 2, he will never show you the car. Thus, he just spilled the beans, that is the odds of the car behind door 2 or 3 is still 2/3, but since he showed it is not door 2, the odds that it is just behine door #3 is 2/3.

Here’s the definitive explanation and simulation from nerds at UCSD math dept. Monty knows where the car is and that is why the odds from switching is 2/3. (He never open up the door with the car)

Simulation:
http://math.ucsd.edu/~crypto/Monty/monty.html
explanation:
http://math.ucsd.edu/~crypto/Monty/montybg.html

[equalizer]

Here’s a very simple explanation. You pick door 1 and since there is a 1/3 chance of picking car, then there must be 2/3 chance that car is behind door 2 or 3. When Monty shows you door 2, he will never show you the car. Thus, he just spilled the beans, that is the odds of the car behind door 2 or 3 is still 2/3, but since he showed it is not door 2, the odds that it is just behine door #3 is 2/3.

Here’s the definitive explanation and simulation from nerds at UCSD math dept. Monty knows where the car is and that is why the odds from switching is 2/3. (He never open up the door with the car)

Simulation:
http://math.ucsd.edu/~crypto/Monty/monty.html
explanation:
http://math.ucsd.edu/~crypto/Monty/montybg.html

[equalizer]

Here’s a very simple explanation. You pick door 1 and since there is a 1/3 chance of picking car, then there must be 2/3 chance that car is behind door 2 or 3. When Monty shows you door 2, he will never show you the car. Thus, he just spilled the beans, that is the odds of the car behind door 2 or 3 is still 2/3, but since he showed it is not door 2, the odds that it is just behine door #3 is 2/3.

Here’s the definitive explanation and simulation from nerds at UCSD math dept. Monty knows where the car is and that is why the odds from switching is 2/3. (He never open up the door with the car)

Simulation:
http://math.ucsd.edu/~crypto/Monty/monty.html
explanation:
http://math.ucsd.edu/~crypto/Monty/montybg.html

[equalizer]

Here’s a very simple explanation. You pick door 1 and since there is a 1/3 chance of picking car, then there must be 2/3 chance that car is behind door 2 or 3. When Monty shows you door 2, he will never show you the car. Thus, he just spilled the beans, that is the odds of the car behind door 2 or 3 is still 2/3, but since he showed it is not door 2, the odds that it is just behine door #3 is 2/3.

Here’s the definitive explanation and simulation from nerds at UCSD math dept. Monty knows where the car is and that is why the odds from switching is 2/3. (He never open up the door with the car)

Simulation:
http://math.ucsd.edu/~crypto/Monty/monty.html
explanation:
http://math.ucsd.edu/~crypto/Monty/montybg.html

[sdduuuude]

To those who don’t realize that switching is the right decision, it is extremely difficult to convince them.

The only way is to simulate it, really. Try the game about 1000 times. Switch half the time and don’t switch the other half. You’ll see you win more often if you switch.

Another way to look at it …

Start with 100 doors. Choose 1 door.
Now – I am going to take away 98 of the 99 remaining doors, with the promise that none of the 98 I eliminate contain the prize.

If you play this game a few times, you’ll soon realize that in 1 out of 100 cases, you will chose the right one in the first place, which means that if you switch you will be switching to the prize in 99 out of 100 cases.

I understand people who think it is 50/50 but they are wrong.

[sdduuuude]

To those who don’t realize that switching is the right decision, it is extremely difficult to convince them.

The only way is to simulate it, really. Try the game about 1000 times. Switch half the time and don’t switch the other half. You’ll see you win more often if you switch.

Another way to look at it …

Start with 100 doors. Choose 1 door.
Now – I am going to take away 98 of the 99 remaining doors, with the promise that none of the 98 I eliminate contain the prize.

If you play this game a few times, you’ll soon realize that in 1 out of 100 cases, you will chose the right one in the first place, which means that if you switch you will be switching to the prize in 99 out of 100 cases.

I understand people who think it is 50/50 but they are wrong.

[sdduuuude]

To those who don’t realize that switching is the right decision, it is extremely difficult to convince them.

The only way is to simulate it, really. Try the game about 1000 times. Switch half the time and don’t switch the other half. You’ll see you win more often if you switch.

Another way to look at it …

Start with 100 doors. Choose 1 door.
Now – I am going to take away 98 of the 99 remaining doors, with the promise that none of the 98 I eliminate contain the prize.

If you play this game a few times, you’ll soon realize that in 1 out of 100 cases, you will chose the right one in the first place, which means that if you switch you will be switching to the prize in 99 out of 100 cases.

I understand people who think it is 50/50 but they are wrong.

[sdduuuude]

To those who don’t realize that switching is the right decision, it is extremely difficult to convince them.

The only way is to simulate it, really. Try the game about 1000 times. Switch half the time and don’t switch the other half. You’ll see you win more often if you switch.

Another way to look at it …

Start with 100 doors. Choose 1 door.
Now – I am going to take away 98 of the 99 remaining doors, with the promise that none of the 98 I eliminate contain the prize.

If you play this game a few times, you’ll soon realize that in 1 out of 100 cases, you will chose the right one in the first place, which means that if you switch you will be switching to the prize in 99 out of 100 cases.

I understand people who think it is 50/50 but they are wrong.

[sdduuuude]

To those who don’t realize that switching is the right decision, it is extremely difficult to convince them.

The only way is to simulate it, really. Try the game about 1000 times. Switch half the time and don’t switch the other half. You’ll see you win more often if you switch.

Another way to look at it …

Start with 100 doors. Choose 1 door.
Now – I am going to take away 98 of the 99 remaining doors, with the promise that none of the 98 I eliminate contain the prize.

If you play this game a few times, you’ll soon realize that in 1 out of 100 cases, you will chose the right one in the first place, which means that if you switch you will be switching to the prize in 99 out of 100 cases.

I understand people who think it is 50/50 but they are wrong.

[sdduuuude]

By the way, this logic provides some insight into poker as well.

If you are playing 10-handed no-limit hold ’em, for example, and one guy bets, then 8 guys fold pre-flop, and you have a mid-quality hand, you should fold. Even though it comes down to you against him, the fact that he is likely the best of 9 hands makes it very unlikely you have a better starting hand.

However, if there are only 3 players, one guy bets and the other guy folds, you should probably stay in as your maginal hand has a much better chance against the best of 2 random starting hands as opposed to the best of 9 random starting hands.

The act of folding in this case is like the host taking away the door with the goat.

[sdduuuude]

By the way, this logic provides some insight into poker as well.

If you are playing 10-handed no-limit hold ’em, for example, and one guy bets, then 8 guys fold pre-flop, and you have a mid-quality hand, you should fold. Even though it comes down to you against him, the fact that he is likely the best of 9 hands makes it very unlikely you have a better starting hand.

However, if there are only 3 players, one guy bets and the other guy folds, you should probably stay in as your maginal hand has a much better chance against the best of 2 random starting hands as opposed to the best of 9 random starting hands.

The act of folding in this case is like the host taking away the door with the goat.

[sdduuuude]

By the way, this logic provides some insight into poker as well.

If you are playing 10-handed no-limit hold ’em, for example, and one guy bets, then 8 guys fold pre-flop, and you have a mid-quality hand, you should fold. Even though it comes down to you against him, the fact that he is likely the best of 9 hands makes it very unlikely you have a better starting hand.

However, if there are only 3 players, one guy bets and the other guy folds, you should probably stay in as your maginal hand has a much better chance against the best of 2 random starting hands as opposed to the best of 9 random starting hands.

The act of folding in this case is like the host taking away the door with the goat.

[sdduuuude]

By the way, this logic provides some insight into poker as well.

If you are playing 10-handed no-limit hold ’em, for example, and one guy bets, then 8 guys fold pre-flop, and you have a mid-quality hand, you should fold. Even though it comes down to you against him, the fact that he is likely the best of 9 hands makes it very unlikely you have a better starting hand.

However, if there are only 3 players, one guy bets and the other guy folds, you should probably stay in as your maginal hand has a much better chance against the best of 2 random starting hands as opposed to the best of 9 random starting hands.

The act of folding in this case is like the host taking away the door with the goat.

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