[RESPONS] math problem
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20 Nov 2007, 17:39
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Journals
Earlier today someone wrote a journal about those 3 doors and a car, and claimed that you got 2/3 chance to get the car if you change your pick after the first empty door is chosen.
BUT, if 2 people choose one door each, and you open a 3rd door, which is empty. Who got most chance to get the car? 2/3 for one of em? I guess 50/50 or?!
Discuss
BUT, if 2 people choose one door each, and you open a 3rd door, which is empty. Who got most chance to get the car? 2/3 for one of em? I guess 50/50 or?!
Discuss
Edit: the solution is that you have a bigger chance if you change, but u ment smth else i think?
explain more about your "cars" and 'picks'
Tbh this should be pisseasy.
1/2 @ 2nd choice
It's kinda stupid isnt it oO ?
Where's the problem in this brainteaser?
€: Ah k i got the problem :D
in the journal earlier today he saw one door and he was aloud to choose another therefore the probability is 2/3.
Are we talking about "you"'s chances or about the chances of the two others.
Because he wrote: 2/3 for one of em?
maar je hebt volkomen gelijk ofc, het zou doesn't work moeten zijn
what do u mean now?
Are there 3 people ?
If one door is eliminated from the beginning which u know is empty [eliminited by yourself "and you open a 3rd door, which is empty.] then there are only 2 doors left which gives those 2 a 50/50 possibility...
And where the fuck is the problem now ???
Dude get brain plz.
when 1 empty one is out your chance to get it is 1/2
So you chose 1 door and the moderator has to close down 998 doors. But he is not allowed to close down the one with the car behind it so no matter behind which of those 999 doors the car would be it would still stay there.
Now it is pretty obvious that though the doors get opened it is very unlikely that the car will be behind the door you have chosen in the beginning.. 0,1% to be precise.
Not much will change if you cut down the amount of doors to 3 =)
You can do an easy test at home:
-take a deck of cards, select 1 card which will be the 'keycard'
-shuffle the cards and take one at random
-let someone else check all the other cards and sort everything out but 1 card, but he is not allowed to take out the 'keycard'
Now take a guess which of the cards with be the 'keycards' :)
/edit this is more an explenation to the first journal, this one is pretty pointless ;)
The most commonly voiced objection to the solution is that the past can be ignored when assessing the probability, as I said, there is a diffrence between math theory and reality theory.
The point is that he is not allowed to close down the door with the car behind it.
So basicly the moderator gets 999 doors behind 1 of whom is a car with a possibility of 999/1000. We agree until here I guess.
After that if behind 1 of these 999 doors is a car (which is 99,9%) he will open all the other doors
SO that means if door 2,3,4,5,6...1000 will have a car behind it at first you would have to chose the other door in the end.
ONLY if the car in fact is behind the door you named in the beginning it is good for you to stay with it.
I can't explain it any simplier, sorry :)
Those other 998 doors is closed, and therefore doesn't matter for the one choosing.
How hard can it be to understand that he actually only have 2 doors to choose from, one that have a 100% chance of having a car and the other a 0% chance. The past doesn't realy matter when it comes to reality theory.
one of them has a higher probability to be a winning door than the second. which would you choose?
It's 50/50 in this case as there is only two options with equal numbers to be the right one, 1 win 1 lose, there can be no /3.
I really don't get what you're trying to say. If you got a choice on whether to pet a domestic cat or to pet a lion in his cage at the zoo, would you choose the lion? I bet most people would choose the cat! But wait! It's a choice between two things, so it must be 50-50. But is it really 50-50? Judging after the number of people that pet the lions at zoos, I'd guess the odds for the lion are slightly less than 50%.
It's the same with the doors. You get to make a choice. Keep the door, or change it. If you keep it, you got 33% chances to win. If you change it, you got 66% chances to win.
- randomly chooses where the car is
- randomly makes a first pick
- eliminates one door as after the rules
After that it:
1) keeps it's pick
2) changes it's pick
3) randomly chooses between the two "new" choices
The results are as follows:
1) Got the car in 33% of the cases.
2) Got the car in 66% of the cases.
3) Got the car in 50% of the cases.
Why does it transfer to one door? It doesn't transfer anywhere! It stays in the GROUP represented by the two doors you have not picked initially. Why would it transfer out of that group?
It don't work that way, I tried with 3 pieces of lego with my friend where two of them was red and one blue, around 54% of the times I would get it if i stayed.
I have run it over 10000 iterations, and that gave out the percentages i've stated earlier.
If you NEVER change your choice, what use does it have to elliminate one of the remaining two? That means you can skip this ellimination. Thus, you choose one out of three. That gives for a 33% probability.
Since you are sure either your first pick or the other door has the car, the probabilities must add up to 100%. That makes for the probability for switching doors (100-33=67, let's say 66 for simplicity)
Stop arguing about this... You already start looking like having no understanding at all about statistics and prbability theory.
Have you read the wikipedia page?
It's about the probability that the CAR is behind the initial choice, or behind the alternative choice.
The 'player' selects a door.
The host checks all the other doors and
-in case behind of one of them is a car he opens the rest of the doors
-in case there is no car behind those doors he opens all doors but 1 at random
SIDE NOTE:
Here's a psychology hint: if you actually get the chance to look at Monty opening one of his two doors, you can expect Monty to pause to think when he's in front of two bad doors. Had only one door been bad, he'd open it automatically.
and that is very unlikely to have relevance in reality.
NOT
And you apear not to have understood the article you posted.