ed84c Posted February 10, 2004 Share Posted February 10, 2004 This is concerning the maths problem where you are in a gameshow and three doors are available behind 2 there are goats but one is a car. You choose one but do not open it. The host opens one of the two that you did not choose and shows a goat. Should you open the door you chose or change your mind with the remaining one door? Well you should change door Please somebody explain the maths of this (No there is not a 1 in 2 chance on getting the car in the remaining 2 doors) Link to comment Share on other sites More sharing options...
YT2095 Posted February 10, 2004 Share Posted February 10, 2004 well it`s a one in 3 to start with. IF you are then shown the goat (and it wasn`t your choice) then it is a 1 in 2 maths alone dictates this, I assume you suspect trickery behind the doors, in which case it`s a non maths problem Link to comment Share on other sites More sharing options...
JaKiri Posted February 10, 2004 Share Posted February 10, 2004 You should change. The only definitive explanation is to look at the probability of each event. Event 1: You choose a goat. If you change, you get a car. Event 2: You choose the other goat. If you change, you get a car. Event 3: You choose the car. If you change, you get a goat. Therefore, there's a 2/3 chance of you getting the car if you change. Link to comment Share on other sites More sharing options...
Nevermore Posted February 20, 2005 Share Posted February 20, 2005 nevermind Link to comment Share on other sites More sharing options...
Kygron Posted February 20, 2005 Share Posted February 20, 2005 You should change. The only definitive explanation is to look at the probability of each event. Event 1: You choose a goat. If you change' date=' you get a car. Event 2: You choose the other goat. If you change, you get a car. Event 3: You choose the car. If you change, you get a goat. Therefore, there's a 2/3 chance of you getting the car if you change.[/quote'] Now assume that there's a second person in the audience (and a third but he doesn't count for this explaination) who has chosen the OTHER closed door before seeing the goat. Since he has chosen a door and one goat was shown to be in another door, we can then follow your steps to discover that he, too, was a much better chance of changing his mind in order to get the car. In this case there is a 2/3 + 2/3 = 133% chance that the car is behind one of the doors.... meaning that there's a 33% chance that there is a car behind BOTH doors. So I assure you game show players, the game IS fixed, and it's fixed IN YOUR FAVOR!!! ROFL Sorry, that was fun I believe you'll find that step 2 is impossible, since you've been shown the "other goat" and it's not one that you picked. This argument WOULD work, however, if the host randomly chose which door to open, giving you a 33% chance to see where the car is, wait, no then you'd just pick the car, nevermind Hm... you know... if the host randomly chose a door, and you randomly chose a door, and then the rules would say if you see the car but didn't pick it you lose, if you see it and did pick it you win, and if you don't see it you choose as originally stated.... then you'd only have a 4/9 chance (I think) of picking the car, and the game show would be alot more profitable while at the same time giving people the chance of seeing the car behind the door they picked without having to ask to open it... fun! lol sorry for rambling. Link to comment Share on other sites More sharing options...
Primarygun Posted February 20, 2005 Share Posted February 20, 2005 I've heard of this question just several months ago. http://www.ylmass.edu.hk/~mathsclub/ILC/Decision/index.html Link to comment Share on other sites More sharing options...
mcoy Posted February 20, 2005 Share Posted February 20, 2005 I've heard of this question just several months ago.http://www.ylmass.edu.hk/~mathsclub...sion/index.html well thanks a lot....maybe it will take me several more months to 'understand' and answer that..... anyway, for homer's sake my friends....(simp) the probability is struck down to 50/50 when the goat was shown. Its just a matter of hint from the host which one has the car. Link to comment Share on other sites More sharing options...
matt grime Posted February 20, 2005 Share Posted February 20, 2005 There are two possibilities here, that affect the choices. 1. If the host knew where the goats were, then the conditionals tell you to swap. 2. If the host picked a door at random, then don't change, or change, the odds are the same. Link to comment Share on other sites More sharing options...
mcoy Posted February 21, 2005 Share Posted February 21, 2005 the question is how do we know that the host picked the door in random.... well of course the host DO KNOW where it is, it happens most of the time, i mean, EVERYTIME. Link to comment Share on other sites More sharing options...
Primarygun Posted February 21, 2005 Share Posted February 21, 2005 well thanks a lot....maybe it will take me several more months to 'understand' and answer that..... You're welcome. Link to comment Share on other sites More sharing options...
matt grime Posted February 21, 2005 Share Posted February 21, 2005 the question is how do we know that the host picked the door in random....well of course the host DO KNOW where it is' date=' it happens most of the time, i mean, EVERYTIME.[/quote'] It should be declared in the problem, otherwise it is a malformed question. A lot of probability questions are like that. Here's another one like this: you are offered the chance to take one of two envelopes, one with 1 dollar bill in the other with a 100 dollar bill. After you take an envelope you can change you mind and swap, and keep on swapping as long as you desire until you decide upon one, at which point you open it. You can using posteriors in a seemingly correct way show that it the "expected gain" increases with out bound after every swap, which is obviously nonsense. Look at Devlin's Angle columns for the proper wording. Link to comment Share on other sites More sharing options...
boxhead Posted February 21, 2005 Share Posted February 21, 2005 This is concerning the maths problem where you are in a gameshow and three doors are available behind 2 there are goats but one is a car. You choose one but do not open it. The host opens one of the two that you did not choose and shows a goat. Should you open the door you chose or change your mind with the remaining one door? Well you should change door Please somebody explain the maths of this (No there is not a 1 in 2 chance on getting the car in the remaining 2 doors) do you think this is maths ? Link to comment Share on other sites More sharing options...
matt grime Posted February 21, 2005 Share Posted February 21, 2005 it is. it;s a well known case where knowing conditional probablity increases your chances of winning. Link to comment Share on other sites More sharing options...
Kygron Posted February 22, 2005 Share Posted February 22, 2005 1. If the host knew where the goats were' date=' then the conditionals tell you to swap. [/quote'] Would you clarify this a bit for us? Or at least tell me what was wrong with my explaination (the first part anyway, when I was being serious). I've assumed that once you see the goat, the choices become goat or car, and that's 50/50, right? Link to comment Share on other sites More sharing options...
matt grime Posted February 22, 2005 Share Posted February 22, 2005 Ok, the conditionals are easy if we assume the host knows where the goat is. There are 3 possibilities when you pick. Goat1, Goat2, Car. If you pick either goat1 or goat2 and after seeing the host open a door you swap you win. In the third option if you swap after being shown a goat you lose. So 2/3 of the time you win. The conditional is in the fact that of you pick goat1 he must show goat2, and vice versa. In the case where the host guesses a door (and we assume if he opens the car's door you can't swap), then if you pick goat1 he shows goat2 with prob. 1/2, so the probability of a win after swapping is 1/6, simililarly with picking goat1. in the third option you lose always with a swap. so swapping means you win 1/6+1/6=1/3 of the time. Link to comment Share on other sites More sharing options...
Kygron Posted February 23, 2005 Share Posted February 23, 2005 Now assume that there's a second person in the audience [b'](and a third but he doesn't count for this explaination)[/b] That's where I made my mistake, the third person DOES count. He would have been treated differently, thus his probability is undefined and allows the probabilities of the other two to add up to greater than 100%. thanks for clarifying! Link to comment Share on other sites More sharing options...
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