## Help Me Understand What I’m Doing Wrong (The Monty Hall Problem)

April 6th, 2009First, if you don’t know the Monty Hall problem already, go ahead and skip this problem. It’s really a question aimed at people who know the “correct” answer, which is nonintuitive.

Ok, so I heard about the Monty Hall problem a long time ago. At the time, I made the same, presumably wrong, guess that most people make. Then I read the logic behind the correct answer, and it took a while, but eventually made sense.

Much later, I was thinking about the Monty Hall problem, and couldn’t remember the logic, so to refresh my memory, I drew out a chart of the permutations of the problem. Here it is:

So here’s the problem: I didn’t make this to disprove the conventional, non-intuitive answer. In fact, I did it expecting it to jog my memory about WHY the non-intuitive answer is true. But, from what I can tell, if you look at every equally likely possibility of the Monty Hall problem, of which there are 12 possible, keeping the same answer will get you the right door 6 times, and switching the answer will get you the right door 6 times.

So, what did I do wrong? I’m not asking for an explanation about WHY the non-intuitive answer is true. The net has tons of those. But for them to be correct, the permutation chart has to come up with more hits for “switch” than “keep the same”, and it doesn’t, so unless EVERYONE is wrong (which I doubt), the problem must be in my permutation chart. So what did I mess up on?

April 7th, 2009 at 11:42 pm

You start with 1/3 chance of correctly guessing the first time, then he opens a (wrong) door. The problem starts at that point. Your chart does not match the problem. The first two rows should actually be combine to a single row. Same for 6/7 and 11/12. The other (wrong) door is already open. Basically, starting with the wrong door 2/3 of the time means you are switching to the remaining (right) door. Starting with right door 1/3 of the time means you are switching to a remaining wrong door. Think about it.

April 8th, 2009 at 6:19 am

Ah, ok, I got it. The chart is correct, but it assigns each row an equal likelihood of happening, which is untrue. Instead, there is a 33% likelihood of me picking any door, so if there are multiple possibilities which result from picking a door, I need to subdivide my 33%. In other words, IF it’s behind Door 1, I have a 33% chance of guessing Door 1, 33% of guessing door 2, and 33% chance of guessing Door 3. If, as the chart shows, guessing Door 1 results in two possible outcomes, while guessing Doors 2 and 3 result in one possible outcome each, I have to divide the two possible outcomes from my 33% chance guess of Door 1 into 16.5% and 16.5%, respectively.

Cool, thanks.

April 8th, 2009 at 11:23 pm

Hmmm… The aren’t two outcomes from guess door 1. You pick the door; and then whether or not to switch. The choice of which door gets opened is not part of your decision making process, and it doesn’t change the outcome. You either picked a wrong door (66.6%) or a right door (33.3%) originally. After the host opens a door, the choice is whether to stay or not with your original choice.

Maybe it would help to change the problem a little. Let’s say there are 500 doors to choose from, and only one of the doors has “the prize”. You pick one door of the 500, and you have 1/500 chance of being right first time. Then the host opens 498 “wrong” doors, then offers you the choice to keep your original choice or to switch to the remaining door. It should be obvious that its better to switch. The host just changes the problem from 1/500 chance to 50% chance problem. The 3-door problem is the same, with less doors.

April 9th, 2009 at 7:11 pm

Ok, “outcome” is the wrong word, perhaps. There are two “possibilities”. The chart shows all the possible situations; it’s not predicated on what you can or can’t choose. I agree, it doesn’t change the outcome, but the whole point of the chart is to show all the possibilities, including the parts out of your control and knowledge, and including stuff that in the end doesn’t chage the outcome. But your first answer answered what was wrong with the chart, and it makes sense now, so I’m good.

April 15th, 2009 at 2:36 am

Trent beat me to it. I was going to say you’ve got all the permutations but then you’ve forgot the probabilities. The initial probability of 6 permutations is only 1/3.

Choose Monty Stay Switch

1/3- choose car- Monty reveals goat a – win car- win goat b

Monty reveals goat b – win car- win goat a

1/3- choose goat a-Monty reveals goat b – win goat a- win car

1/3- choose goat b-Monty reveals goat a- win goat b- win car