ignoring
Here's the simplest-possible correct explanation...
2/3 of the time, your initial pick is wrong. The host will then show you the "goat" prize (the other being the good prize). Ergo, switching will get you the good prize 2/3 of the time. 1/3 of the time you'll lose the good prize. This is obviously better than sticking with the initial choice (which is right only 1/3 of the time).How much simpler does it need to be, to be comprehensible?
I get it now.
Some will never get it, no matter how it is explained.
d
There are two iron rules that dictate which door the host opens.
(1) There can't be a car behind it
(2) It can't be the door the contestant picked.
If the contestant picks a door with no car, then there is only one other door the host can choose. The host has no choice and can't affect the outcome.
If the contestant chooses a door with a car, then the host can choose either of two two doors that have no car, but which one he chooses doesn't seem to affect the outcome. His effect on the odds comes from the fact that he revealed one of the two doors with no car behind it.
How can the host manipulate the odds?
either
to
Correct. The host has no effect on the odds.
Part of the confusion occurs because people confuse permutations and combinations. In the situation where the contestant has chosen the good prize, the two bad prizes form a combination, not a permutation.
Wrong, on a purely statistical basis the first case is 50:50, BB, BG, GB, or GG. Two out of four meet the criteria. The second case is 50:50 Boy or Girl, One out of two meets the criteria.
Trevor.
The second case is: BB, BG, GB. The couple told you the GG case does not exist. Get it now? The goat problem has similar probability outcome changes.
David
No. You now know that the prize is not behind door $3, so your chance of winning in the, "second game" is 50-50. But you had to buy yourself this chance at the second game. You did this by switching doors.
I know it's unkind to tell people who agree with you that they're wrong, but... you're wrong. You really need to think this through carefully.
snipped-for-privacy@spam.com (Don Pearce) wrote in news: snipped-for-privacy@news.eternal-september.org:
Why? Because at the point of the final decision, that's the situation. How do the preceding 8 steps affect the final step?
Just as in flipping coins. Getting 5 heads in a row is 1/32. But getting the 5th head after already getting 4 is still 1/2.
That isn't the final situation. I will take this a step at a time.
There are three doors - one with a car, two with goats
I choose one. I have a 1 in 3 chance of being right
That means there is a 2 in 3 chance of the car being in the other two
I know for a fact that at least one of the other two is a goat.
That does not change the odds - it is still 2 in 3 that the car is in one of those
The host shows me one of the two - one he knows to contain a goat.
This is not new information, I knew there was a goat there, I still know there was a goat there.
The odds are still 2 in 3 that the car is in one of those two doors.
But now those 2 in 3 odds have been concentrated into the one remaining door of the two, which I will open because that is better than the 1 in 3 chance of it being my first choice.
d
eptember.org:
The big difference: In the Monty Hall problem there is only one "coin flip". Only one random choice is made -- the first choice of a door. In the coin flip situation, there are five coin flips, five random choices.
Now, in contrast, if the car and remaining goats were randomly shuffled after each goat door was revealed, then the situation would be different. But in the MHP problem the car does not move.
The really interesting thing is that, even if the car does not move, conditional probability theory says the odds have changed, and you should switch doors.
Right!
And understanding that games of pure chance have NO memory for any previous actions. It's just like the old question, what are the odds of tossing a coin 10 heads in a row? If you toss 9 heads in a row, what are the odds of tossing a 10th? (First you MUST assume the coin is untampered with, you cannot assume the same for a TV game show however!)
Trevor.
Or interpreting it from a view of TV game show reality rather than a purely statistical basis.
Trevor.
Nobody said it wasn't comprehensible. But it simply ignores the fact that game shows are NOT pure chance.
Trevor.
What has this to do with the question?
d
Oops you are quite correct, there are still 3 possibilities.
Trevor.
Nothing I guess for the specific case in question. As was pointed out there is only one possible set of events if a switch is offered for a one of 3 game of chance. I admit to confusing this with other TV games where the host can and does influence the outcome.
Trevor.
snipped-for-privacy@spam.com (Don Pearce) wrote in news:4dd0c3d7.381246241 @news.eternal-september.org:
Stop there. No, I didn't know there was a goat THERE. I knew there was a goat behind at least one of the door besides the one I chose, but I didn't know which one. Now a variable is removed from the equation.
Revealing a goat behind a door doesn't change the odds? Of course it does. Otherwise, revealing the car behind a door also wouldn't change the odds.
Once the host has revealed a goat, then there's an even chance that the car is behind one of the two remaining doors--and I have no information either way (unless you're counting the psychological factors) that the door I chose is or is not the correct one.
Not trying to be argumentative, but I still don't see the logic.
Sorry, but if you don't get it by now, you simply aren't going to. Give up and try something else.
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