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Discussion in 'Jokes and Humour' at netrider.net.au started by nearlyempty, Apr 29, 2005.

  1. I was shown this puzzle recently & I found it very interesting.

    You're on a game show & the game show host shows you three doors. Behind two of the doors are a goat, but the third conceals a car. If you pick the door with the car, you win the car, but if you pick a door with a goat, you lose.

    You pick a door & the game show host then opens one of the two remaining doors, revealing a goat.

    He then gives you one final option, you can stick with the door you have already chosen, or you can swap to the last remaining door.

    What should you do?
  2. stick to the one you've chosen cos its obviously the one with the car. takes a much bigger door to get a car in than a goat (unless its some kind of radioactive supergoat) and you'd have to be pretty dumb to pick one of the small doors :LOL:
  3. There's always one, & it's always got a lovely bunch.
  4. so whats the real answer then???
  5. Man, I'd be happy with the goat...
  6. Yep. Instinct tells you that it doesn't make any difference, so you may as well stay put, but you'd be wrong.

    For those that don't want to follow the link, you should always switch doors because the chances of you picking a door with a goat first time around are higher than if you had picked the door with the car.

    Swapping doors when asked, switches the odds back in your favour.

    Well, I thought it was interesting anyway, but then I also like cheese & HP sandwiches.
  7. whoah :shock:

    i'll stick with my theory thanks...
  8. Unless you consider the second step an entirely new circumstance (as the previous one has been resolved) which makes it 50:50 so staying put doesnt really change your odds at all.

    Personally, I agree with the bit about HP and cheese sandwiches :D
  9. Swapping doors does not change the odds in your favour.
    Originally you had a 33% chance of winning the car (1 in 3). One favourable outcome from three options.
    If one of the doors is opened and revealed a goat the odds have changed to a 50% chance (1 in 2) of winning the car. One favourable outcome from two options.
    Changing the doors now (after one door was opened) does not change the odds, you still have a 50% chance, that is one favourable outcome from two options.
    You might have picked the car at the beginning before the door was opened.
    The changing of doors is known as an exclusive event, it has no bearing on the odds of the outcome, only the ultimate outcome in that you might change to the door that has the car behind it.
    A further example of this is, if I throw a dice and get a six, what's the probabilty of throwing a six the next time? It's still one in six as throwing a six the second time is not dependent upon what I threw the on the first occasion. However, if the question is rephrased as what's the probabilty of throwing two sixes in a row compared with throwing a single six then the odds change from one in six to one in thirty-six.
  10. Nodz, you sound like an engineer... from one eng to another... well said.
  11. Man, Americans really do think weird!
    Following the procedure as described in the link, there is an unstated assumption that the host is trying to thwart you (wrong!), but then again (if the procedure is always the same) he then acts to improve your odds by eliminating a losing option!
    The answer is: your first choice is always completely irrelevant, and your second choice is always 50:50.
    What paradox?
  12. No, no, no, you've all got it wrong. Eat the goats, sell the car and go for a ride on your damn motorbike!!!

  13. The trouble is, you're treating the second choice in separation, instead of considering the problem as a whole.
    Given two doors, yes you have a 50-50 chance of picking the correct one. But you are not choosing between two doors, you're choosing between three.
    Your odds of choosing the correct door first time around is 33%, which means that you have a 66% of getting a wrong door. When one of the wrong doors is then eliminated, there is a greater probability that the remaining door conceals the car.

    Oops! edited. Just read your dice example - It's the same thing.
  14. Like nodz said, by the time you get to the last two doors, there is a fifty fifty chance.

    Listen carefully and follow your nose, this might be the only way to somehow sway those odds.
  15. Having done Statistics at University I am of the opinion that statistics is the biggest load of bullsh#t of an excuse for a mathematic field in existance. It's just someone's lame attempt at trying to justify crap that no-one cares about and give someone a job
  16. Yeah but...
    The second operation renders the first one irrelevant. The host always removes a losing option (at least according to the author). If this is the case you will always be left with a simple choice of 2.
  17. The thing is, the question isn't "Behind which of the two doors is the car", it is "should you swap or stick".

    The host doesn't remove the losing door until you have made your first choice.
  18. Assuming host knows where car is located.

    The actual puzzle states that you choose a door, a door is then opened by the host, you can then either open the door you chose or you can swap.

    I was trying to make the point that once the door is opened by the host, you either open the door you chose first or swap doors but this does not change the odds, you still have a one in two chance.
  19. But... The host knows.

    One time in three (If you chose the car first up) he will reveal one of the goats. Two times in three (when you have chosen a goat) he will reveal the only remaining goat. S two times in three, swapping will get you the car.