| Yet more of the door problem |
| The assumption that there is an equal chance that the car is behind each door is essential to all the conclusions I have come to so far (see page 1 and page 2 of this problem). If we abandon this assumption your choices will obviously depend upon the probabilites for each door. If you are unaware of these then there is very little you can do to improve your chances of winning. If you are aware of the bias we can remodel the situation. Let's state the problem again since it has been so long since we originally saw it. |
| You have won a gameshow and are now on the last round to choose your prize. There are three doors behind two doors is a dustbin and behind the third door is a car. The host tells you that the probability the car is behind the first door is X that the probability it is behind the second door is Y and behind the third door is Z. You choose door 3. The host opens a door and reveals a dustbin. He then gives you the opportunity to switch doors. Should you? |
| For now we will assume that the host knows which door the car is behind and upon opening the first door makes sure he doesn't reveal the car. The headache that will be caused by the problem if the host doesn't know which door the car is behind will be dealt with later. |
| Redrawing the original probability table |
| Door number: 1 2 3 Arrangment 1: C B B Arrangment 2: B C B Arrangment 3: B B C |
| Remember C represents car and B represents bin. |
| We are once again assuming that the host knows which door the car is behind so if the car is behind door 1 or 2 the host opens the other door. The three possible arrangments are 1 with probability X, 2 with probability Y and 3 with probability Z. |
| We can see that, if you switch doors you will win in possiblities 1 and 2, which have a combinded probability of X+Y. If you don't switch doors you will win in possibility 3, which has a probability of Z. |
| It is easy to conclude from this that you should switch doors if X+Y > Z (X+Y is bigger than Z) |
| Given the host knows what door the car is behind, which door should I choose to maximize my chances of winning? |
| You should switch doors if X+Y > Z. We know that since the car has to be behind one door X+Y+Z = 1. To increase our chances of winning we therefore need the difference between X+Y and Z to be as much as possible. Therefore I believe (and you may question me on this by e-mailing me) that the best strategy to make a win most likely (provided the host knows what door the car is behind and delebritly doesn't open that door first) is to select the door with the least probability of having the car behind it, then after the first door has been opened swop. This will make X+Y (the probability that you will find the car when you swap) as large as possible. I will give a numerical example below. |
| Door number: 1 2 3 Arrangment 1: C B B Arrangment 2: B C B Arrangment 3: B B C Prob. car: 3/8 2/4 1/8 |
| If I select least likely door/ I select door 3. Probability car behind door I selected = 1/8. Probability I will find car if I switch doors after first door has been opened = 3/8 + 2/4 (if you don't understand why I'm allowed to add see page 1) = 3/8 + 4/8 = 7/8 (see adding fractions below if you don't understand this) |
| If I select most likely door/ I select door 2. Probability car behind door 2 = 2/4 = 4/8. (values put over 8 to make easily comparable) Probability I will find car if I switch doors after first door has been opened = 3/8 + 1/8 = 4/8 |
| If I select middle probability door/ I select door 1. Probability car behind door 1 = 3/8. Probability I will find car if I switch doors after first door has been opened = 1/8 + 2/4 = 1/8 + 4/8 = 5/8 |
| Taking the greatest probability for each selection; selecting the least likely door provides the largest probability, 7/8, while most likely door provides least likely probability, 4/8. To have maximum chances of winning in this situation you should select the least likely door then switch. |