 This topic has 220 replies, 18 voices, and was last updated 15 years, 10 months ago by sdduuuude.

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July 27, 2008 at 6:05 PM #248134July 27, 2008 at 8:29 PM #247972equalizerParticipant
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.htmlJuly 27, 2008 at 8:29 PM #248126equalizerParticipantHere’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.htmlJuly 27, 2008 at 8:29 PM #248132equalizerParticipantHere’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.htmlJuly 27, 2008 at 8:29 PM #248193equalizerParticipantHere’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.htmlJuly 27, 2008 at 8:29 PM #248199equalizerParticipantHere’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.htmlJuly 27, 2008 at 8:57 PM #247985sdduuuudeParticipantTo 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.
July 27, 2008 at 8:57 PM #248142sdduuuudeParticipantTo 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.
July 27, 2008 at 8:57 PM #248146sdduuuudeParticipantTo 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.
July 27, 2008 at 8:57 PM #248208sdduuuudeParticipantTo 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.
July 27, 2008 at 8:57 PM #248214sdduuuudeParticipantTo 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.
July 27, 2008 at 9:04 PM #248000sdduuuudeParticipantBy the way, this logic provides some insight into poker as well.
If you are playing 10handed nolimit hold ’em, for example, and one guy bets, then 8 guys fold preflop, and you have a midquality 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.
July 27, 2008 at 9:04 PM #248156sdduuuudeParticipantBy the way, this logic provides some insight into poker as well.
If you are playing 10handed nolimit hold ’em, for example, and one guy bets, then 8 guys fold preflop, and you have a midquality 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.
July 27, 2008 at 9:04 PM #248160sdduuuudeParticipantBy the way, this logic provides some insight into poker as well.
If you are playing 10handed nolimit hold ’em, for example, and one guy bets, then 8 guys fold preflop, and you have a midquality 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.
July 27, 2008 at 9:04 PM #248223sdduuuudeParticipantBy the way, this logic provides some insight into poker as well.
If you are playing 10handed nolimit hold ’em, for example, and one guy bets, then 8 guys fold preflop, and you have a midquality 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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