﻿ John Petro | Monte Hall Problem revisited yet again

# Monte Hall Problem revisited yet again

Here is the Monte Hall problem. There once was a game show called "Let's Make a Deal", the famous emcee was a gentleman named Monte Hall. One of the games a contestant, we'll call her Alice, could play was the following. Alice was shown three doors on the stage, behind one was a grand prize, say a Porsche automobile. Behind the other two were goats. Alice was allowed to pick a door and whatever was behind it, that was her prize to keep.  Monte knew the correct door. After Alice made her selection but before opening the door and revealing Alice's 'prize' to her Monte would open one of the other doors, one that had a goat (recall he knows the winning door, and regardless of what door Alice chose, one of the two remaining doors would hold a goat, so he could always do this). At this point, before revealing the contents of Alice's original selection and to add an air of suspense, Monte would give Alice a chance to change her selection. The question we address here is, should she do accept his offer to switch? To many people it seems that it is "six of one, half dozen of the other", that changing does not change her odds of winning. To the surprise of these people it does matter, and she should always accept the offer and change her selection to the remaining unopened door. This will double her probability of getting the valued prize from the original 1/3 to now 2/3.

If you are one who does not believe this, then how about the following variation. Suppose instead of three there are 10 doors, behind one of which is the Porsche and the other nine hide goats. Alice picks a door. Her probability of being correct (choosing the door with the Porsche) is 1 out of 10. The probability that behind her door is a goat and the Porsche is behind one of the non-picked doors is 9 out of 10. Monte knows the correct door and it is always the case that at least eight out of the nine unpicked doors hide goats. He opens eight doors revealing goats. Now there are two doors remaining that are unopened, the one Alice chose originally, and the one remaining unpicked and unopened door. In the 9 out 10 chance that Alice's original choice held a goat, that 9th and final goat is behind the door Alice has selected. Monte has done Alice the favor of eliminating all the bad remaining doors from the nine unselected and left only the door with the Porsche unopened. A switch by Alice in these 9 of 10 cases would be from goat to Porsche. Only in the 1 out of 10 chance that Alice was right to begin with would switching be a mistake. Thus, her probability of getting the Porsche goes from 1 out 10 to 9 out of 10 if she blindly switches. A similar argument applies in the original three door case as well.

Or here's another way to look at it. Suppose (in the original 3 door case) that after you choose your door but before opening it to show you what you've won, Monte gives you the option to swap your chosen door for BOTH of the unchosen doors. In other words, you get what's behind both of the doors you did not originally select. Isn't it a no-brainer now that if your chance of picking the door with the car was originally 1/3 (which, unless you have clairvoyance or some other inside information is what your odds are), that with probability 2/3 the car is behind one of the two remaining unchosen doors and Monte is offering you that along with a goat (and you can donate the goat to a farm or children's petting zoo). This is the same as the original game, except that Monte is eliminating from the two unchosen doors the one with the goat. By choosing to switch you are effectively taking both of the unchosen doors. If this is still not clear, imagine the scenario with 10 doors as described above and after you've picked one door out of 10 Monte says you can trade that choice for everything behind the remaining 9 doors.