| When the host is unaware of which door the car is behind in a varying probability situation. |
| Ok, I admit it, I lied before. I implied that this section would be more interesting (complicated) than the last when it fact is will be a lot simpler. |
| Firstly let us review what we learned on page 2. We discovered that if the host was unaware of which door the car was behind when he opened the first door, it would make no differences to your chances of winning if you switched doors or not. Logically this should be a similar situation. |
| Once the first door has been opened if the car hasn't been revealed both probabilities (of finding the car) for the alternative two doors increase. These will increase in proportion to their original probabilites of the car being found behind them. What on earth does that mean? If we look back at page 2 we see thar the two non chosen doors both had an original probability of the car being found behind them as 1/3. When the first door is opened this increases to 1/2. In both probabilites are the same. The probability that the car is behind door one is X, the probability that it is behind door two is Y and the probability that it is behind door three Z. If we eliminate one of the doors say one at random then we can say that the total probability that the car is behind door two after the opening is Y(original probability) + Y/(Z+Y) x X and the probability that it is behind door 3 is Z + Z/(Z+Y) x X. Let us add these two together. |
| Y+XY/(Z+Y) + Z + ZX(Z+Y) as we are adding fractions we need a common denominator =(YZ + YY)/(Z+Y) + YX/(Z+Y) + (ZZ + ZY)/(Z+Y) + ZX(Z+Y) = (YZ + YY + YX + ZZ + ZY +ZX) / (Z+Y) = (Y(Z+Y+X) + Z(Z+Y+X) ) / (Z+Y) = ((Y+Z) (Z+Y+X)) / ((Y+Z) = Z+Y+X The probability (of finding the car) is the same as all the original probabilities added together or 1. |
| So provided the car isn't found in the first opening (when you would have no option to change doors anyway). The two probabilities of finding the car that you were left with both increase by amount proportional to their original proability. This means that the door which was most likely before will be most likely again. |
| I will run calculations for an example of equal probabilities. The 1/3 problem on page two. |
| The first door has been opened at random and the door hasn't been found. The probability of this event is 2/3. From now on this will be igoned. |
| The equation I gave earlier is New probability car behind door two = Y+XY/(Z+Y) New probability car behind door three = Z + ZX(Z+Y) Remember X, Y and Z were all originally 1/3 New probability car behind door two = 1/3 + (1/9) / (2/3) = 1/3 + 3/18 = 6/18 + 3/18 = 9/18 = 1/2 New probability car behind door three = 1/3 + (1/9)/(2/3) = 1/3 + 3/18 = 6/18 + 3/18 + 9/18 = 1/2 Which agrees with what I had on the previous page. |
| To make accuate calculations we would have to consider which of the two doors the host would choose to open if he was aware of the probability of finding the car behind each one. His decision may not be random as he tries to change the participants chance of winning. However to find the best option (which is what this problem is about) all this additional information is unessential. The door that was most likely to contain the car before the door was opened will still be the door most likely to contain the car when the door has been opened (provided the car hasn't be revealed). Therfore I belive it is fair to conclude and I welcome e-mails if you disagree with me, that when the host is unaware, which door the car is behind it is best to go first for the most likely door (to minimize chances of car being revelaed when first door is openend) then stick to this choice. |