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General Chat Thread, Monty Hall Problem (mathematics) in General; There's definitely no 50:50 odds - because whilst you are down to one car, one goat, the chance of the ...
  1. #16


    tom_newton's Avatar
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    There's definitely no 50:50 odds - because whilst you are down to one car, one goat, the chance of the door you are switching to being a goat is reduced, because in two out of 3 situations (ie those wher eyou haven't picked the car in the 1st place - doesn't matter that you don't know) Monty (with knowledge) has explicitly avoided that door because its a car.

    So, two thirds of the time, you will pick Monty's avoided car. One third of the time, you will already have the car, and you'll pick monty's discarded random goat.

    I think this problem is skewed by the fact that the first 1-in-3 random choice, people believe they might have the car, and switching is intuitively disadvantageous.

  2. #17
    Jake's Avatar
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    So in effect, its only a 2/3 IF the host knows where the car is. If the host doesnt know where the car is, then its 50/50 as per the wikipedia table

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    Quote Originally Posted by tom_newton View Post
    Having said that, you might want a goat. You cant milk a car.
    with the rate milk is going up the goat will probably have better resale value

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    Quote Originally Posted by SeanVin View Post
    Here’s another maths problem that is counter intuitive. Suppose you drive one lap of a 1 mile race track at an average speed 60mph. How fast would you have to drive on your second lap to complete both laps at an average speed of 120 mph?
    180?

    And, in the original problem, the host does know where the car is - he will always show one of the remaining goats. I first came across this in "The Curious Incident of the Dog int he Nighttime", quite a nice novel.


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    The best way to put it (in my opinion) is, pretend that there are 100 doors. You choose one, then monty hall gets rid of all the other doors except the one you chose and one other, one of which has to be the door with the car behind it. You can see there that it would be in your best interest to swap your choice because the door you chose has much less chance of being the correct door then the one that monty hall leaves you with.

    That is effectively what happens, even though there are only three doors

  6. #21
    SeanVin's Avatar
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    That is a very good expination Bat looks like I was completely wrong. Having had a little think it would appear you are actually betting on what Monty is left with after your first choice, there is a 2/3 chance of him being left with a goat and a car in which case he must leave the car.

    Andy the answer to the race track problem is not 180mph!

  7. #22

    SYSMAN_MK's Avatar
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    Quote Originally Posted by SeanVin View Post
    Here’s another maths problem that is counter intuitive. Suppose you drive one lap of a 1 mile race track at an average speed 60mph. How fast would you have to drive on your second lap to complete both laps at an average speed of 120 mph?
    Trick question my friend. Can't be done. It's all down to time, and you will have none left after the first lap.

  8. #23
    somabc's Avatar
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    Quote Originally Posted by SYSMAN_MK View Post
    Trick question my friend. Can't be done. It's all down to time, and you will have none left after the first lap.
    edit: I was wrong it's not 240mph

    LaTeX Code: \\frac {\\bar v}{2L} = \\frac {2 v_1}{2L} = t_1
    Last edited by somabc; 16th July 2008 at 03:36 PM.

  9. #24

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    Ahh but Average Speed = Distance / Time. And your calculations don't include the time.

  10. #25
    SeanVin's Avatar
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    Quote Originally Posted by SYSMAN_MK View Post
    Trick question my friend. Can't be done. It's all down to time, and you will have none left after the first lap.
    Your absolutely right, it would take you a minute to do two laps at an average speed of 120Mph but it already took you a minute to do the first lap. Most people come up with 180mph reasoning that 180+60 is 240 and dividing it by the two laps giving 120mph. What they have done is taken the average of two averages.

    In my RAF days we used to produce monthly statistics for the number of engines damaged by foreign objects (FOD, a significant problem for jet engines). At the end of the year the management produced their annual FOD report with an exceptionally low number of incidences per 1000/flying hrs. It looked all wrong to us, we had calculated a much higher figure. It took us a while to work out what was going on, but as you can probably guess, they had averaged out our monthly figures that were themselves averages. We pointed this out and gave them the correct data. We were of course totally ignored, it wouldn’t have reflected well on them if they had published the higher figure.

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