???
No it isn't. Think, originally when you chose the door, your chance of getting it correct was 1/100. But because the host already eliminated 98 doors for you, your chance of getting it right if you switch, jumps to 99/100... but if you stayed, your chances are still 1/100. Is this really that hard to conceptualize?
what the f*ck is wrong with you people, have you never taken any sort of probability course in your lifetime? this is not flawed logic, you have flawed thinking. do a f*cking test if you still think you're right.
ugh.
ugh.
kranky
Elite Member
- Oct 9, 1999
- 21,019
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Let's stay with the 100-door example. You say there's a 50/50 chance either way. Would you then agree to take the "never switch" strategy? There is only one prize. You pick one of 100 doors. We both know there are 99 empty doors, so you won't mind if I show you 98 of them. That changes nothing, right?
Now there are two doors left out of the 100. The one you picked, and the one I didn't show you yet. You took the "never switch" strategy (because you said there would be a 50/50 chance either way, and therefore it doesn't matter if you switch or not). Can you see how your odds didn't magically grow to 50/50 just because I showed you 98 empty doors? You would be saying that you had a 50/50 chance of being right from the start, and we know that's wrong. You had only a 1 in 100 chance.
Why wouldn't the original door picked also have a 2/3 chance of being right since 1/3 of the doors has now been proven to be the wrong one?
You have two doors that each have 2/3 chance of being right?
You have two doors that each have 2/3 chance of being right?
Originally posted by: LeRocks
hey! that's what i said!
just agreeing
edit need to stop fat-fingering.
Thinking back over all of my statistics classes, I don't remember any point where previous random "drawings" affected a current "drawing".
When you get down to 2 doors, there are only 2 options: pick door a, or pick door b. You have a choice between one or the other, not between one out of 100. All that matters is the probability at that point, not the probability of you getting to that point.
When you get down to 2 doors, there are only 2 options: pick door a, or pick door b. You have a choice between one or the other, not between one out of 100. All that matters is the probability at that point, not the probability of you getting to that point.
Seriously its obvious for the 100 example. It would only be 50/50 if you actually. Changed where the prize was.
If after removing 98 doors they changed the location of the prize randomly placing it between the door remaining doors
then its 50/50. Otherwise your first choice is still 1/100. You may only have two choices. But if you stick with the
original door you are saying that out of 100 doors you guessed it to be correct. Where as the other door is basically
saying its not this door, so 99/100.
Another way to think of it. If you don't change doors lits like removing all 99 doors at once. There is no way you can say
thats 50/50 chance. All choosing the other doors says is that you think its not this door.
If you still don't believe it, right a program, script, function that does this probability and run it for this example or 1000 doors
Run it out for 1 million times. Have it stick with the same door and change. Find out how many times it gets it right and gets it
wrong. You'll see.
If after removing 98 doors they changed the location of the prize randomly placing it between the door remaining doors
then its 50/50. Otherwise your first choice is still 1/100. You may only have two choices. But if you stick with the
original door you are saying that out of 100 doors you guessed it to be correct. Where as the other door is basically
saying its not this door, so 99/100.
Another way to think of it. If you don't change doors lits like removing all 99 doors at once. There is no way you can say
thats 50/50 chance. All choosing the other doors says is that you think its not this door.
If you still don't believe it, right a program, script, function that does this probability and run it for this example or 1000 doors
Run it out for 1 million times. Have it stick with the same door and change. Find out how many times it gets it right and gets it
wrong. You'll see.
sure, but the weight assigned to the door you didn't pick is 2/3, and the weight assigned to the door you did pick is 1/3.
Why can't the door picked also have 2/3 chance since 1/3 of the original choices are now gone? That seems a bit flawed to assign 2/3 chance to an unpicked door and not to the picked door.
correct - you had a 1-in-100 chance. now, you have a 1 in two chance. you can't pick the doors that have already been opened, can you? prior selections don't affect the current selection, once you make it there.
you had a 2/3 chance you were wrong. this does not change simply because the number of ways you could have been wrong is reduced. it simply increases the probability that any particular remaining way of being wrong was the case.
also, if both doors had 2/3 chance, you would have 4/3 total probability, which is impossible.
Ok...but the now known door removes 1/3 from the equation so there's a 1/3 chance you're wrong and a 1/3 chance you're right. You can't throw the other 1/3 back into the equation. It would be more accurate to say your remaining chances are 1/2 but you still have the original 1/3 probability.
that makes no sense to me. every time you have the option of choosing, it is a new case. new weighting must be applied based on the probability at that point, disregarding any previous probabilities. if there are two doors available, and you have to choose one (the situation you would be in at that point), there is a 50-50 chance you will pick the correct door. it doesn't matter how many doors have been opened prior.
no you must throw the 1/3 back, the total probability must sum to 1.
look at it this way... the set of doors you did not choose has 2/3 probability. by removing a door from that set, you don't remove any probability, you just remove the number of elements in that set, meaning that when it comes time to distribute the probability of that set amongst it's elements, each remaining element will get more
sure you can throw it back in - you have a 1/3 + (1/2)*(1/3) chance you are right, and a 1/3 + (1/2)*(1/3) chance you are wrong. 50% either way.
no it is not a new case, because you have existing information on the doors. if you were to mix up the two doors such that you didn't know which one was which, then yes, you would have 1/2 chance each. but you know which door is which.
It does still equal 1...it's just that 1/3 of that one is something you would not pick...it's now a known loser door. If you toss that one door out, you're left with two and your chances are now 1/2. You can't have 2/3 with only two possible choices.
but those elements don't matter any more. that's equivalent to saying that you have a lower probability of rolling a six just because you just rolled a six. each case is independent.
What about cheapbidder's scenario?
3 doors
2 people
You pick door #1
Other person picks #2
Host shows door #3 is empty
If you both switch doors, you'd both have 2/3 chance .. making it 4/3 total
the only existing information you have is this:
there are two doors. one has a prize. one does not. they have equal opportunity to have the prize - it's not like one door has been opened and you have been shown what's behind it.
But when you rolled a dice, each event is not dependent on the past... but in this instance, IT IS dependent on what the host does. Because he eliminates a door for you.
Lets take a very extreme example... there are 1 million doors.
You pick one, your chances are 1/1000000.
The host eliminates 999998 other doors until 2 doors are left... one you choose, and another closed one.
You're telling me the one you chose has the same odds as the other one?
kranky
Elite Member
- Oct 9, 1999
- 21,019
- 156
- 106
How about this? There are 100 doors, and you pick one. Let me have the other 99. The host shows that 98 of my doors are empty (what do I care? I knew 98 of them would be empty anyway), and now it's down to your door and the one I have left. Do we have an equal chance of being right?
You are being fooled by the fact that some of the doors are shown to you before you get to switch. That changes nothing about your original chances of being right.
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