Monty Hall problem

[Torc]

The solution to this is actually simple, there is a goat behind all three doors.

They just switch the goat behind the unchosen door with the car.

Thus the probability for the mug, sorry contestant, is always 0/3

[homer13j]

What? No year's supply of Turtle Wax?

[FunkyDexter]

I didn't bother following your algorithm because anything that proves 1/3 + 1/3 = 1 is clearly wrong.

[Maven]

I didn't bother following your algorithm because anything that proves 1/3 + 1/3 = 1 is clearly wrong.

too bad nobody ever says.... hey maven... how about not doing that xor and then it'll display crap correctly (IE: car = 1). and then I go... wow doh on my part... I should stop doing this crap so late.. and they go... np...

[FunkyDexter]

lol - you shouldn't need us to point your mistakes, you evil genius, you.

I once spent an evening arguing with my nephew (a horribly good coder BTW) who thought he'd come up with a fool proof system for playing roulette. He went away the next day and him and his boss spent the whole day programming a simulation (not much else to do I guess) which didn't give them the results they predicted but did show them making a profit. Only after another evenings argument did he realise that reseeding the rnd function every time you call it in VB6 tends to skew your results.

[wy125]

I didn't look too much at your code but I did follow what you were saying. You correctly noted that the host intentionally chooses to open a door where it is known that that car is not located. This makes it favorable to switch doors when the host offers. The fact that the probabilities are not equal in this scenario is what makes this problem so difficult. Most people see the probability as equal. If the host had simply chosen a door at random to open (ignoring the location of the car), revealing a goat, then the probability that the car is behind the door you chose is 1/2 and switching would do you no good (if this were really happening the host would reveal the car 1/3 of the time and the game would be over right there). This is the scenario most people imagine and why most people get it wrong and that's precisely why, as you say, 'the host is ignored as a variable' -- it makes the problem less obvious, which is the point of telling these problems in the first place.

[wy125]

I didn't look too much at your code but I did follow what you were saying. You correctly noted that the host intentionally chooses to open a door where it is known that that car is not located. This makes it favorable to switch doors when the host offers. The fact that the probabilities are not equal in this scenario is what makes this problem so difficult. Most people see the probability as equal. If the host had simply chosen a door at random to open (ignoring the location of the car), revealing a goat, then the probability that the car is behind the door you chose is 1/2 and switching would do you no good (if this were really happening the host would reveal the car 1/3 of the time and the game would be over right there). This is the scenario most people imagine and why most people get it wrong and that's precisely why, as you say, 'the host is ignored as a variable' -- it makes the problem less obvious, which is the point of telling these problems in the first place.

BTW, your prob. of 1/3 and 1/3 isn't so crazy. You probably counted all trials, even those that resulted in the host revealing the car. You should get 1/3 win by keeping same door, 1/3 win by switching, and 1/3 host selects car and game over. But if you limit the trials to those where the host doesn't reveal the car, then it's 50/50 keeping the door or switching.