Monty Hall Logic
Date: 03/09/2001 at 10:35:24 From: George Dinwiddie Subject: The Monty Hall problem The Dr. Math FAQ (see The Monty Hall Problem, http://mathforum.org/dr.math/faq/faq.monty.hall.html ), states that "We will assume that there is a winning door and that the two remaining doors, A and B, both have goats behind them. There are three options: The contestant first chooses the door with the car behind it. She is then shown either door A or door B, which reveals a goat. If she changes her choice of doors, she loses. If she stays with her original choice, she wins. The contestant first chooses door A. She is then shown door B, which has a goat behind it. If she switches to the remaining door, she wins the car. Otherwise, she loses. The contestant first chooses door B. She is then is shown door A, which has a goat behind it. If she switches to the remaining door, she wins the car. Otherwise, she loses. " There are, in fact, four options. The first one described is actually two separate options: 1A. the contestant chooses the car, is shown a goat behind A or 1B. the contestant chooses the car, is shown a goat behind B There are three choice points, not just two. There is a logical fallacy in suggesting there is any connection between the two contestant choice points. There is a mathematical fallacy in here also. At the first choice, I have a 1/3 chance of choosing the right door. Let's presume I choose door A. If the prize is behind door C, Monty has a 100% chance of opening door B. If the prize is door B, he has a 100% chance of opening door C. If the prize is in door A, he has a 50% chance of opening B and a 50% chance of opening C. He cannot open either the door I've chosen or the door with the prize. This is left out of the published "analysis." So, we are left with these possibilities: A B C 1 prize open empty 2 prize empty open 3 empty prize open 4 empty open prize Hmmm... I now have a 50% probability of having the right door. The mathematical fallacy is in considering the first two possibilities as a single possibility when they are, in fact, distinct. There are three choice points, not just two. The host has a choice also, but that choice is sometimes 100% (if I've chosen an empty door) and sometimes 50% (if I've chosen the correct door).
Date: 03/09/2001 at 11:12:53 From: Doctor Anthony Subject: Re: The Monty Hall problem Your analysis is interesting but perhaps you would comment on the following. There is 2/3 probability that you initially choose a wrong door; if you always change, you thereafter must get the correct door. Whereas there is 1/3 probability that you initially choose the correct door, and if you always stick to your first choice your probability of success remains at 1/3. In the original setting you have initially 1/3 probability of having selected the right door and 2/3 probability of having selected the wrong door. The various possibilities are as follows: 1st choice correct - then changed Prob of success = 1/3 x 0 = 0 1st choice correct - did not change " = 1/3 x 1 = 1/3 1st choice incorrect - then changed Prob of success = 2/3 x 1 = 2/3 1st choice incorrect - did not change " = 2/3 x 0 = 0 Total probability of success if you change = 0 + 2/3 = 2/3 Total probability of success if do not change = 1/3 + 0 = 1/3 So there is a better probability of success if you always change. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/
Date: 03/09/2001 at 11:25:56 From: Doctor Rick Subject: Re: The Monty Hall problem Hi, George. I saw Dr. Anthony's response; perhaps I can add to it. You say yourself that I have a 1/3 probability of choosing the correct door first. But then you argue that because Monty can make two choices if I have chosen the correct door, and only one choice if I haven't, therefore there is a 1/2 probability that I have chosen the correct door. You are therefore saying that the probability that I have chosen the correct door changes depending on what Monty is going to do after I have made the choice! No, there is a 1/3 probability that I have chosen the correct door, regardless of what Monty does next. Suppose Monty flips a coin to decide which door to open next if he has two choices. The possibilities are then (if the correct door is C): 1. I choose door C, Monty opens door A (probability 1/3 * 1/2 = 1/6) 2. I choose door C, Monty opens door B (probability 1/3 * 1/2 = 1/6) 3. I choose door B, Monty opens door A (probability 1/3) 4. I choose door A, Monty opens door B (probability 1/3) Analyzing the game in terms of four choices doesn't change the remainder of the argument, because the probabilities are the same as in the published solution. Your fallacy is in assuming that, if you can enumerate N possibilities, the probability of each must be equal. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/
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