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In high school our math teacher introduced a problem that could not be solved, which involved a TV-show game where the contender would choose one of 3 doors, an

by z3t4 7y ago

In high school our math teacher introduced a problem that could not be solved, which involved a TV-show game where the contender would choose one of 3 doors, and if there was something behind it, the contender would win it. But the show host would remove one of the doors without something behind it, and ask the contender if he/she would like to switch door. I simply simulated the game on a computer one million times which revealed that changing the door after the host removed a empty one had a 66% probability of winning. (rather then 50%) but the teacher wouldn't believe me...

drewmol 7y ago
It's known at the Montey Hall Problem[0] or another variation is The Three Prisoners Problem[1] (IMO [1] is an easier way to conceptualize the probability of different outcomes). It can be understood that the contestant choosing to switch doors has a probability of winning equivalent to that them being allowed to choose two of the three doors in the first place. By not switching they are choosing one door (1/3) but by switching (since one of the empty doors is removed) they are effectively choosing both the other two doors (2/3). [0]https://en.m.wikipedia.org/wiki/Monty_Hall_problem https://en.m.wikipedia.org/wiki/Monty_Hall_problem [1]https://en.m.wikipedia.org/wiki/Three_Prisoners_problem https://en.m.wikipedia.org/wiki/Three_Prisoners_problem
fastball 7y ago
You used Monte Carlo method to solve the Monty Hall problem.
kortilla 7y ago
The Monty hall problem (what you’ve described) doesn’t require simulations. What are the odds you guessed it right on your first pick? (1/3) That’s the only time switching answers will not reveal the prize. The rest of the time there is a losing door and a winning door and the host is forced to reveal the losing door so switching wins 2/3 of the time. It’s that simple.
sunstone 7y ago
This problem is tricky with three doors but if you imagine playing the same game but with 100 doors instead of three, then the result is very clear.
shepherdjerred 7y ago
https://en.wikipedia.org/wiki/Monty_Hall_problem?wprov=sfti1 https://en.wikipedia.org/wiki/Monty_Hall_problem?wprov=sfti1
etrk 7y ago
> "a problem that could not be solved" The solution is counter-intuitive, but the problem can be solved with some elementary probability theory.
verytrivial 7y ago
Yeah, the trick with this one is the game show host knows where the prize is. In the case where you picked correctly first (1:3) the switching to the other door will be a loss. But in the case where you picked wrong (2:3) the host will eliminate the non-prize door, so a win is guaranteed if you switch. So, 1/3 * 0 + 2/3 * 1 = 66% of winning for the always switch strategy. Shame on your teacher!
nl 7y ago
It's very unclear from your description, but I suspect "could not be solved" means "could not be proved", in which case of course a simulation is not a proof.
- fjfaase 7y ago
  This and similar math puzzles are trivial to prove compared to the math problems like the one discussed in the article. It is only that they are counter intuitive that makes it hard for people to believe them. If you change the three door to hundred door problem and have the host open 98 doors, it is rather obvious that you should switch to the one door he has not opened besides the door that you did choose at the start.
  - nl 7y ago
    I have no dispute with that. I'm saying that since the OP appears to believe that a simulation proves it (which of course it doesn't), there's a decent chance other parts of the story also are incorrect. Because yes, this puzzle is quite well known for confounding intuition. Presumably a math teacher knows that when they introduced the question.
computerfriend 7y ago
This is a famous problem and the solution is not contested.
- brazzy 7y ago
  I'd rather say that the solution is constantly contested by people who can't shake off their intuition.
  - retsibsi 7y ago
    To be fair, the problem is often badly posed. The host always reveals a goat, and this is crucial information, but people often fail to make it clear. If the host chose randomly and just happened to reveal a goat, the solution would be the 'intuitive' one.
    - aardshark 7y ago
      No, I don't think that's accurate. In the scenario you pose, sticking and switching have an equal reward, so both are valid solutions.
      - retsibsi 7y ago
        > sticking and switching have an equal reward Right, and this is the intuitive solution. Nobody argues that switching reduces your chance of success; the split is over whether it increases it to 2/3 or leaves it at 1/3. (The problem statement ends with 'Is it to your advantage to switch your choice?' The right answer under the intended assumption about the host is 'yes', but the intuitive answer is 'no, it makes no difference'.)
        retsibsi 7y ago
        Sorry, I shouldn't have said 'increases it to 2/3 or leaves it at 1/3'. If he picks randomly and reveals a goat, the probability that your original choice of door is a winner is now 1/2, as is the probability that the remaining door is a winner.
qubex 7y ago
This is the Monty Hall Problem, and though counter-intuitive, is well understood: https://en.wikipedia.org/wiki/Monty_Hall_problem https://en.wikipedia.org/wiki/Monty_Hall_problem

by z3t4 7y ago

drewmol 7y ago

It's known at the Montey Hall Problem[0] or another variation is The Three Prisoners Problem[1] (IMO [1] is an easier way to conceptualize the probability of different outcomes). It can be understood that the contestant choosing to switch doors has a probability of winning equivalent to that them being allowed to choose two of the three doors in the first place. By not switching they are choosing one door (1/3) but by switching (since one of the empty doors is removed) they are effectively choosing both the other two doors (2/3). [0]https://en.m.wikipedia.org/wiki/Monty_Hall_problem https://en.m.wikipedia.org/wiki/Monty_Hall_problem [1]https://en.m.wikipedia.org/wiki/Three_Prisoners_problem https://en.m.wikipedia.org/wiki/Three_Prisoners_problem

fastball 7y ago

You used Monte Carlo method to solve the Monty Hall problem.

kortilla 7y ago

The Monty hall problem (what you’ve described) doesn’t require simulations. What are the odds you guessed it right on your first pick? (1/3) That’s the only time switching answers will not reveal the prize. The rest of the time there is a losing door and a winning door and the host is forced to reveal the losing door so switching wins 2/3 of the time. It’s that simple.

sunstone 7y ago

This problem is tricky with three doors but if you imagine playing the same game but with 100 doors instead of three, then the result is very clear.

shepherdjerred 7y ago

https://en.wikipedia.org/wiki/Monty_Hall_problem?wprov=sfti1 https://en.wikipedia.org/wiki/Monty_Hall_problem?wprov=sfti1

etrk 7y ago

> "a problem that could not be solved" The solution is counter-intuitive, but the problem can be solved with some elementary probability theory.

verytrivial 7y ago

Yeah, the trick with this one is the game show host knows where the prize is. In the case where you picked correctly first (1:3) the switching to the other door will be a loss. But in the case where you picked wrong (2:3) the host will eliminate the non-prize door, so a win is guaranteed if you switch. So, 1/3 * 0 + 2/3 * 1 = 66% of winning for the always switch strategy. Shame on your teacher!

nl 7y ago

It's very unclear from your description, but I suspect "could not be solved" means "could not be proved", in which case of course a simulation is not a proof.

fjfaase 7y ago
This and similar math puzzles are trivial to prove compared to the math problems like the one discussed in the article. It is only that they are counter intuitive that makes it hard for people to believe them. If you change the three door to hundred door problem and have the host open 98 doors, it is rather obvious that you should switch to the one door he has not opened besides the door that you did choose at the start.
- nl 7y ago
  I have no dispute with that. I'm saying that since the OP appears to believe that a simulation proves it (which of course it doesn't), there's a decent chance other parts of the story also are incorrect. Because yes, this puzzle is quite well known for confounding intuition. Presumably a math teacher knows that when they introduced the question.

computerfriend 7y ago

This is a famous problem and the solution is not contested.

brazzy 7y ago
I'd rather say that the solution is constantly contested by people who can't shake off their intuition.
- retsibsi 7y ago
  To be fair, the problem is often badly posed. The host always reveals a goat, and this is crucial information, but people often fail to make it clear. If the host chose randomly and just happened to reveal a goat, the solution would be the 'intuitive' one.
  - aardshark 7y ago
    No, I don't think that's accurate. In the scenario you pose, sticking and switching have an equal reward, so both are valid solutions.
    - retsibsi 7y ago
      > sticking and switching have an equal reward Right, and this is the intuitive solution. Nobody argues that switching reduces your chance of success; the split is over whether it increases it to 2/3 or leaves it at 1/3. (The problem statement ends with 'Is it to your advantage to switch your choice?' The right answer under the intended assumption about the host is 'yes', but the intuitive answer is 'no, it makes no difference'.)
      - retsibsi 7y ago
        Sorry, I shouldn't have said 'increases it to 2/3 or leaves it at 1/3'. If he picks randomly and reveals a goat, the probability that your original choice of door is a winner is now 1/2, as is the probability that the remaining door is a winner.

qubex 7y ago

This is the Monty Hall Problem, and though counter-intuitive, is well understood: https://en.wikipedia.org/wiki/Monty_Hall_problem https://en.wikipedia.org/wiki/Monty_Hall_problem