If you do not know if the host left the last cup by knowledge or luck, it is still in your best interst to switch because if the feat was accomplished by knowledge, then you are 99 times better off by switching and if it was by luck, you are no worse off as the chance that the coin is under either cup is 50/50. However, if I knew for a fact that it was by luck, even though the true odds are 50/50, I would be inclined to keep my original choice because I would feel that the host was able to accomplish this feat because the coin was under my cup. Joe.
Correct. This is a rough explanation how you can calculate your advantage in case of uncertainty about Hostâs knowledge or luck. Supose you have the 3-doors case. Your prior belief is that you are (for example) 80% certain the Host opens the doors by luck. Take the natural logarithm of the two competitive hypothesis p: Host opens doors by luck (1-p) : Host opens doors by knowledge N(0) = Ln(p/(1-p)) = ln((0.8/0.2) = 1.39 Now supose you chose door nbr 1 and the host opens door nbr 2 and this door is empty Update your beliefs about hostâs knowledge or luck Delta = LN(P(door is empty conditional Host opens doors by luck)/ P(door is empty conditional Host opens doors by knowledge) ) Delta = LN( (2/3) / 1 ) = -0.41 N(1) = N(0) + Delta = 0.98 Posterior p = 1 / ( 1 + exp (N(1)) = 0.73 So now you are only 73% certain that the Host opens doors by luck. Probability the car is behind door 1 = 0.73 * (1/2) + 0.27 * (1/3) = 0.455 Probability the car is behind door 3 = 0.73 * (1/2) + 0.27 * (2/3) = 0.545 In the case of the 100 doors game after the Host opens 98 empty doors it is almost certain that he did it by knowledge and the advantage of switching is huge.
What if you are sure he did it by luck. But he in fact picked 98 cups in a row without a coin. Do you switch in this particular instance?
I ran some computer simulations of this problem when the host knows and when the host is guessing (where it culls out the result if he shows the coin, because it is assumed that he showed a cup with no coin underneath). The results are that if the host knows, switching gives you a 2/3 chance of picking the coin. If the host is guessing, switching gives you a 1/2 chance of picking the coin, just like numerous posters have stated. Apologies to everyone who corrected me. I understand intuitively the original problem, now I need to understand intuitively what is going on with this knowing/guessing wrinkle. If I were a contestant I would always switch anyway, even if I suspected the host was just guessing. Because the host might not be, as gbos stated.
The cases where the host randomly selects a cup and the coin is underneath are cases where if the host had known, he would have picked the other cup, and the player would have won by switching. These cases are culled out of the possible outcomes if you know that the host is guessing.
I think the confusion about this problem is that some people look at the probability at the moment after the host has opened a door with a goat. At that particular time, the chance of picking the door with the car is 1/2, and swithing or not doesn't improve player's chances. But, if you analyze the entire experiment, starting with the initial door picking and the host opening one door, then swithing makes sense, because the host - if and only if he knows where the car is - will always open a door with a goat, thus eliminating some of the possible runs of the experiment. It's all on wikipedia.
incredible is that the posters were given the extreme example of 99 additional doors and are still in a state of denial that you should switch. not surprising since when the original monte hall problem was stated years ago mathematicians who were embarrassed for giving the wrong answer were in the same state of denial. there is a lesson in here.