
Originally Posted by slim
Nothing influences your chances of having picked a winning door. What does influence your decision to switch or not at the end is that the host knows where the car is located and will never open that door.
So when you start you have a 1 in 1000 chance of picking the correct door. Therefore there is a 999 in 1000 (99.9%) chance that the car is behind one of the closed doors. Now the host will open 998 doors that all have goats behind them. Do you think it is 50/50 whether the car is behind your door or the unopened door? I hope you don't! The key is that the host knows where the car is and will only open goat doors.
If the host had no idea where the car was and randomly opened doors, it would be a different story.
Right, I see your point, however I'm not sure the model can be scaled down to work with only 3 doors. The fact is that the host will always show exactly 1 goat and leave 1 door (beside your chosen) unopened leaving you with a 50/50 chance of getting it right. Now, if it was 4 or more doors I believe your model is applicable as you'll get 2 goats against 1 unopen door leaving you with a 66% chance if you decide to switch. Similarly, you'll have 75% chance if switching on a quiz with a total of 5 doors, 80% with 6 etc.

Premium Member
Originally Posted by dulence
Right, I see your point, however I'm not sure the model can be scaled down to work with only 3 doors. The fact is that the host will always show exactly 1 goat and leave 1 door (beside your chosen) unopened leaving you with a 50/50 chance of getting it right. Now, if it was 4 or more doors I believe your model is applicable as you'll get 2 goats against 1 unopen door leaving you with a 66% chance if you decide to switch. Similarly, you'll have 75% chance if switching on a quiz with a total of 5 doors, 80% with 6 etc.
It's not 50/50 because the three possible outcomes are:
You Have ______________________________ Remaining Box Has
You have the goat______________________ The other box has car
You have the goat______________________The other box has car
You have car__________________________The other box has a goat
By switching you increase your chance of having a car from 33% to 66%.

Originally Posted by s0095063
It's not 50/50 because the three possible outcomes are:
You Have ______________________________ Remaining Box Has
You have the goat______________________ The other box has car
You have the goat______________________The other box has car
You have car__________________________The other box has a goat
By switching you increase your chance of having a car from 33% to 66%.
That nailed it, thanks.

No, it still works with three doors.
I'm not sure how you are calculating the 66% in a 4 door game, 75% in a 5 door game, etc... can you explain that in detail and I will try to show you where you are going wrong?
Try looking at it this way:
3 doors labelled A, B, C  one has a car, 2 have goats. Let's put the car behind door A.
Here are the possible scenarios:
You choose Door A  you win if you don't switch.
You choose Door B  you win only if you switch.
You choose Door C  you win only if you switch.
Once the host has shown you that Door B (or door C) has a goat, you will win 66% of the time by switching.
Another way to look at it is like this:
3 doors A, B, C and only the host knows where the car is located. You pick door A.
Right now there is a 33% chance that the car is behind your door and a 66% chance that the car is behind either door B or C. Treat the phrase 'either door B or C' as a single entity, a single door if you will.
There are 2 things you know about this entity: there is a 66% chance that it holds the car and 1/2 of the entity is empty.
The host shows you that one half is empty... SO!? You already knew that! You have no new information about that single entity. It still has a 66% chance of being where the car is located. Because whether the goat is behind C or behind B, the host will show you the one with the goat.

Originally Posted by AmeliesDad
3) It is known that in the general population a certain drug is taken my 0.5% of people. A company devises a test that is 99% sensitive and 99% specific (only 1% of positive tests are false positives and the same is true of negative results). A staff member in your office is tested positive but protests his innocence, what is the probability that he is a user?
Probably wrong but my logic...
two possible cases:
takes drug and positive: 0.005*0.99 = 0.00495
doesn't take drug and negative: 0.995*.01 = 0.00995
So probability that he is a drug taker is 0.00495/(0.00995+0.00495) ~ 1/3
Originally Posted by AmeliesDad
4) If you continually flip a coin until the sequence HTT arises and then start over you can average your results to find the average number of flips it takes to appear. Then you do the same for the sequence HTH. Which of theses statements is true.
a) The average will be the same
b) The average number for HTT will be lower
c) The average number for HTH will be lower
Not very statistical but i asssume HTT will be lower.
Assume you have got the first 2 correct (HT). In both cases you need just one more right to win. But in the first case if you lose (get a H) you will have started another possible winning streak and only need 2 more to win. In the second case if you lose (get a T) you need 3 more right to win.

2 immaculate answers from Granthrax. You been studying stats or something?

Originally Posted by AmeliesDad
2 immaculate answers from Granthrax. You been studying stats or something?
Yey . I did a bit of stats at Alevel but I'm pretty interested in logic. I suppose these sorts of questions are pretty logical.
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