196 replies to this topic

[Shanks]

why is there so much talk of probability??!! Its making my head spin. Anyway, here is my reasoning and am not using probability here!

[Door 1] has goat
[Door 2] Picked by jennifer
[Door 3]

Jennifer should definitely switch since if the car is behind 2, she will get a dinner. But if the car is behind 3, she gets car wax. Now why would someone give you a car wax unless and until you have a car! the car is in 3.

[xamdam]

Spoiler for I got

If Jennifer follows the strategy of always switching, she will win the Jag about 2/3 of the time.

Here is what will happen if she sticks with the "always switch" strategy:

About 1/3 of the time, unfortunately, Jennifer's first door will be the Jag door. She will lose, because
when Monty points out one of the goats, she will switch and select the door with the other goat.

About 2/3 of the time, fortunately, Jennifer's first door will be one of the goat doors. She will win, because
when Monty points out the other goat, she will switch and select the door with the Jag.

[Carpe Mofo]

It's 50/50. The problem in your logic is, after he shows a goat, your still figuring in the door that we now know has goat as a possibility, she can no longer choose that door which brings the possible selection down to two doors and since she can only choose one, that means she is choosing one door out of two possible doors, meaning a one in two chance... It can ONLY be a two out of three chance if all three doors are unknown and she can choose two of them. I'm not trying to be a troll here, but the level of stupidity required for adults not to understand 7th grade math is astounding.

[caprirs302]

It's 50/50. The problem in your logic is, after he shows a goat, your still figuring in the door that we now know has goat as a possibility, she can no longer choose that door which brings the possible selection down to two doors and since she can only choose one, that means she is choosing one door out of two possible doors, meaning a one in two chance... It can ONLY be a two out of three chance if all three doors are unknown and she can choose two of them. I'm not trying to be a troll here, but the level of stupidity required for adults not to understand 7th grade math is astounding.

You have just proven that it is you that doesn't understand 7th grade math, and have thus been labeled a "troll."

To sum it up for you, you originally have a 1/3 chance of picking a winner, which means you have a two thirds chance of picking a loser. Odds are you picked a loser. If the host shows you the OTHER loser, then that means that it is a good bet that the third door is actually the winner, and you should switch. You would be a fool, or a troll, if you did not.

[JoeSKull]

I have spent a lot of time thing about this problem and I have read and fully understood the answer from many sources.

So let me put it to you like this, unless you are the smartest mathematician to ever ponder this question then there is no way that you are correct in saying that it is 50/50. Since mathematicians have actually assessed this problem and concluded that the answer is that you should switch and win 2/3 times. If you do not believe me then look it up or look at one of the many links already provided. Or read and understand what it is that the rest of these people are actually saying since you are obviously not.

Good day sir.

[spindlethin]

I've been struggling with this question forever, but I've finally (and grudgingly) come to realize that the probability really does change if you switch your choice of doors. The number of doors doesn't matter, but the more there are, the more evident the logic is. So, based on a model of 100 doors:

1. You choose a door. you have a 1% chance of choosing the correct door.

2. All doors except the one you chose and one other are revealed. There are now two possibilities:
A - You picked the correct door the first time, at a 1% probability
B - You didn't pick the correct door the first time, at a 99% probability.


The only way that the other door can be the wrong door is if you picked the correct door the first time. The trick to this puzzle is that the options of the second choice are based on the decisions of the first choice - they are not independent events.

[tonym144]

you know...you can sit here and write out all the math you want, but in the end if there's 3 doors and you already know 1 to have the goat, then one door has a goat and 1 door has a car. That's a 50/50 chance. If there were 20 doors and I took away 18, you'd still be left with 2 and 1 would be the winning door. It's always going to be 50/50. It's pure logic. Mathematics may not agree with logic, but logic doesn't care. Logic is as logic does.

[k-man]

you know...you can sit here and write out all the math you want, but in the end if there's 3 doors and you already know 1 to have the goat, then one door has a goat and 1 door has a car. That's a 50/50 chance. If there were 20 doors and I took away 18, you'd still be left with 2 and 1 would be the winning door. It's always going to be 50/50. It's pure logic. Mathematics may not agree with logic, but logic doesn't care. Logic is as logic does.

tonym144, you might want to read other posts in this thread and some more information on the web. There are links in this thread to some other resources related to this topic if you didn't find arguments in this thread convincing. Or you can just google "monty hall problem".
Your mistake is that you are thinking that if there are only 2 possible outcomes then they must have equal probability. That's not logical at all. If you buy a lottery ticket there are 2 outcomes - you win or you don't win. The probability is obviously not 50/50.

[bonanova]

you know...you can sit here and write out all the math you want, but in the end if there's 3 doors and you already know 1 to have the goat, then one door has a goat and 1 door has a car. That's a 50/50 chance. If there were 20 doors and I took away 18, you'd still be left with 2 and 1 would be the winning door. It's always going to be 50/50. It's pure logic. Mathematics may not agree with logic, but logic doesn't care. Logic is as logic does.

tonym144, it's a fun puzzle and one that's debated a lot because we like to follow our intuition instead of doing all that messy math.

Spoiler for Here's an intuitive path to the other answer.

When you pick a door at the outset, that door gives you a 1/3 chance of winning.
Ask yourself how that probability could change by being shown a goat.
Remember Monty knows what's behind the other doors - he's not guessing about what he shows you.
And if your door has a 1/3 chance, the other 2/3 chance, instead of being split between the other two doors, now has to totally be with the third door, doesn't it?

[aaronbcj]

I will still stick to #2.
Earlier i had a blind guess of winning a car prob:1/3
Now i have prob:1/2

My hunch tell me however:: Had i selected a door with goat on first choice, the host would have happily opened it to let all know, the game would end there. They were always more interested in saving a jaguar for a goat.