You need to read *and* comprehend it. The key bit is that the host
*always* opens a losing door, if they don't and randomly open one of the doors you havn't picked you are back to 50:50 (I think...).
When you make your first choice you have 1/3 chance of being correct, the other two have 2/3 chance of being correct.
Now the host eliminates one of the doors and you still have 1/3 chance of yours being correct and the host still has 2/3 chance of his being correct, you haven't done anything to change that.
So you swap to the 2/3 chance.
I can't agree with that! There are *two* doors he didn't open - yours and one other. Why should the *other* one have a higher chance than yours of being the winner?
Which is why appeals to common sense generally get people into trouble.
The same birthday statistical fallacy works incredibly well once there are more than 23 people in the room it is a suckers bet.
So far so good. But you have to ennumerate the cases independently
1/3 : Win 2/3 : LoseThe host then removes one losing play and so there is a degeneracy factor on the remaining one that he didn't take away. You don't know for sure that it is the right answer but it is twice as likely to be so.
It would make the example much more obvious if it was one win out of 10 and you choose one and then the host takes away 8 non winning ones.
You know from the structure of the game that you can't do any worse by swapping and in fact you can actually do better. After the host has taken away one losing door you know that the one he has left has a 2/3 chance of winning against the 1/3 for the one you are holding.
You have gained additional key information from the actions of the host. He cannot choose the winning door so in the 2/3 of initial cases where you have lost he leaves you with a certain win if you swap.
Originally it was mainly about land ownership and weighs and measures.
Yes. Mathematical proofs are probably some of the most enduring immutable constructs of humanity. Pythagoras's theorem still works today as do all of Euclid's laws of geometry on the plane.
Mathematics is precise, unambiguous and deduces logically what it can from the adopted starting axioms. It is possible to choose other starting axioms and get new results which might or might not relate to reality. Geometry in non-Euclidean spaces like a sphere where the angles in a triangle no longer add up to 180 degrees for instance.
He's simply not taking everything into account, much like Prof Meadows didn't with the cot death stuff.
See here for more info on that:
But he deliberately opens a loosing door.
Chance doesn't enter into it at that stage, He is picking with knowledge, rather than luck.
I suspect that the point is that the door you chose originally *affects* the door the host then chooses. Thus the variables are not independent and the simple calculation is wrong.
In a slightly different way it's similar to the old joke that, because the chance of a bomb on the plane is one in a million, then if I take one on myself then the chance of two is one in a million million so I'm safer.
Ah, that's the problem, the door was loose and so opened all by itself.
yes - if you have initially picked a non winning door the host is telling you what the other non winning door is, they are changing the information available to you.
So far so good and enumerating them (your initial choice at LHS)
You Pool Prob T F F 1/3 F T F 1/3 F F T 1/3
But you are not starting from scratch. The fact that you have chosen one door when the host make hismove prevents the host from choosing it and so he must take away a losing door from the other two.
After the host's move the game is changed - reduced to
T F 1/3 F T 1/3 F T 1/3
Which simplifies to
T F 1/3 F T 2/3
I am amazed that Erdos didn't accept it until he'd seen a simulation!
But they are *NOT* equally likely! The structure of the problem means that swapping doubles your chances of winning the prize.
You have forgotten that if the host removes one losing door in the 2/3 of cases where you have lost then he gifts you a certain win.
So your final odds with this additional piece of information are
1/3 to win by holding onto what you have or 2/3 by swapping.
No, the fact that the coin was heads this time tells you *nothing* about what it will do next time.
They are not abstract arguments, they are precise arguments that deal with reality.
There is no "current view" of mathematical truth. As has already been stated, I can't sit down and today "prove" that Pythagoras was wrong or that the angles in a triangle on a flat surface don't, in fact, add up to 180deg.
Yes you do.
You took a 1 in 3 bet that you would select a door which would get you to the next round. You didn't bet it had the car behing it. You be that the host wouldn't open the door.
So you won the first round, and it doesn't matter what is behind the door because you are still in the game.
Now you are in another game - there are two doors and one has a car behind it.
I note that: " Paul Erd?s, one of the most prolific mathematicians in history, remained unconvinced until he was shown a computer simulation confirming the predicted result "
Also that it only applies under the condition that the host knows the door you picked and never opens it.
"The argument depends on the assumptions, made explicit in more extended problem descriptions given by Selvin (1975a) and by vos Savant (1991a), that the host always opens a different door to the door chosen by the player and always reveals a goat by this action"
O.K. - so you haven't won the first round because the host will never open your door.
And
""Switching" only fails to give the car when the player had initially picked the door hiding the car, which only happens one third of the time."
This is the bit that is counter intuitive.
Whilst the 'proof' seems to make logical sense, something somewhere in the back of my mind is still yelling "No!".
You pick a door. You know that whatever you pick, a goat will be revealed and discarded, and the real choice is in the second round where it seems to be a straight
50:50.I guess the computer simulation cannot lie but it is a weird proposition.
Urgh!
You make a choice at 1 in 3 odds of finding the car. So perhaps the intermediate reveal makes no difference to your initial choice, still 1 in 3 if you stick.
I wonder if you tossed a coin at the second stage instead of automatically switching what that would do to your chances.
Counter intuitive to suggest it would make them worse but it could well do so.
Anyway, I quite like goats :-)
Cheers
Dave R
Duh!
Penny finally dropped.
"Simply put, if the contestant picks a goat (to which two of the three doors lead) the contestant will win the car by switching as the other goat can no longer be picked, while if the contestant picks the car (to which one door leads) the contestant will not win the car by switching. So, if contestants switch, they will win the car if they originally picked a goat and they will not win if they originally picked the car. As they have a 2 in 3 chance of originally picking a goat, they have a 2 in 3 chance of winning by switching."
In fact the aim of the contest is to pick a goat in the first round, because you know if you do that and switch you have two chances out of three of winning.
Now off to do some real work.
Cheers
Dave R
Please tell me where I'm going wrong in the following:
I make my choice out of three doors, odds 1/3.
The host removes a losing door, leaving two doors, one a winner, the other a loser.
I exit the room and am replaced by another contestant who is asked to choose one of the remaining two doors. The odds on each door would then appear to be 1/2 or 50:50.
If the new contestant has an evens chance of picking the correct door why, if I walk back in, do I not have the same odds as him?
Are you telling me that a mathematical proof that has been accepted has never subsequently been proved to be wrong?
Ever?
Not even once??
Probably drifting into the difference between a theorem and a proof, though :-)
Cheers
Dave R
I diasagreed. Then I wrote a Monte-Carlo simulation, and I no longer disagree!
I pick a door at random, and it has a one-in-three chance of being a winning door.
The host opens a door, and shows no car.
At this point I have three options:
(1) Stay with my choice. It still has a 1:3 chance of winning.
(2) Pick again at random. I am now picking from 2 doors, one of which has a car, so I have a 1:2 chance of winning. That's improved my odds.
(3) Switch. My original choice has a 1:3 chance of winning. The two doors left must between them have a winner. Therefore the other door has a 2:3 chance.
I became convinced of this when I worked out that, given two remaining doors, the easiest way to score a switch was to say
if (!doors[choice]) ++switchscore;
Or in English, if the door I had previously chosen is wrong then I win.
Andy
How do you know? if you know you've won you would refuse to swap, if you know you've lost you would always swap, but you have no possible way of knowing which.
The game never ends after you pick your door, it only ends after you've accepted or refused opportunity to swap, they aren't rounds as such.
The host never opens your door, only one of his that is a loser (he may have one loser so no choice, or two losers and an irrelevant choice).
No, it is the same game, you are trying to reinvent the rules to make life seem easier, rather than apply the rules as they are stated.
Certainly it's a tricky one, I doubt many understand when they first encounter it, I didn't, it wouldn't be interesting if everyone did.
Bing, you've got it!
[other points left uncommented on]
He, he. Glad I'm not the only one who has trouble with lose and loose. Just remember you don't want a Moose on the loose. B-)
When you first make a play, the odds are against you 3 to 1. Statistically, over many plays the odds are you will pick the wrong door.
When the presenter opens a second door to reveal an empty box/room the odds are still 3 to 1 and as you are most likely to have chosen the wrong door, you swap.
polygonum wrote: [snip]
Similarly, I used to set the cut off point for tests used to diagnose T. gondii infections. By determining IgM responses (recent infection) it is possible to tell if a pregnant woman has an active infection. However the tests aren't perfect and theres not a clear separation between positive and negative responses.
The consequence of a positive test is a recommendation for an abortion so the outcome of the test is critical.
The solution is to use different cut offs depending upon the prevalence of T. gondii infection in the country where the test is done. There's a low incidence if the parasite in the UK do the cut off is biased towards false negatives. The probability is that the person is going to be negative anyway because of the low infection rate. In holland and France the majority of the population is infected so tests are biased towards false positives because the probability is that the person will be infected.
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