Monty Hall Problem
I’m writing this to clarify my thoughts on the topic. The problem, while seemingly simple, is often source of confusion, amusement, and occasionally even arguments. I have recently encountered fresh debate on the topic on Hacker News that started quite innocently (as conversations there often do).
For a comprehensive (though lengthy) introduction to the problem and its various forms see Monty Hall problem on Wikipedia.
Here is re-creation of a scribble I made while thinking about the problem.
Each layer represents what we know about possible states of the universe as we progress through the problem.
Obviously at the beginning (layer 0) we know nothing (with probability 1).
Layer 1 emerges as we are presented with tree doors, car behind one of them. Each case has probability of 1/3.
Layer 2 represents our choice of a door. As we don’t know anything, we choose randomly. Now our multiverse has 9 possible states (still uniform probability, now 1/9). If we ask what is the probability of us having chosen the correct door, we just sum the probabilities of all matching states (red border) and arrive at probability 1/3 (3 * 1/9). (As an exercise try thinking what would have happened if we chose non-uniform strategy here.)
Alternative thinking about layer two is, that by choice we have collapsed the multiverse (grouped by the collapse).
Layer 3 is when host opens one of remaining doors that has goat (there might be one or two such doors, if there are two host chooses randomly… As an exercise think about what would it mean if host did not choose uniformly). Red states are the ones where we win by keeping our door, green ones where we win by switching. Interestingly both global (whole layer) and all collapsed multiverses imply winning probabilities for staying and switching 1/3 and 2/3 respectively!
Last layer is an attempt at demonstration for one of cases where confusion is coming from. The question whether the car is behind left or right door is different from whether it is the door we choose or the other one.
If you look at uncollapsed multiverse and ask about probabilities for left versus right, it is uniform (1/2 for each of remaining door). But once you have chosen a door suddenly asymmetry emerges. Part of our multiverse have collapsed. If our door is the left one of remaining ones, then it has only 1/3 probability of hiding a car.
The question is not whether it is behind left or right door (in the uncollapsed multiverse), but whether it is behind left or right door given our choice of door.
Perhaps confusion here is similar to Gambler’s_fallacy where people assume that after head flipping another head is less probable not realising that by first flip part of multiverse available to them has collapsed and now it is 1/2 again.
Could it be that the way the question is phrased is the issue? An elegant and concise answer is only possible due to the presence of symmetries in the problem. If those were broken the answer might not be as simple as “always switch for 2/3 chance to win the car” but more like (if you chose door 1 and host showed you door 2 then switch for chance P, but if 3 then stay for chance Q, if you choose door 2…).
Interesting insights from the thread:
- By choosing to switch you are actually choosing two doors instead of one.
- Try thinking how would things work with 100 doors.
Ideas:
- Interactive/scriptable version where it would be possible to play with different strategies of both host and player and see how final probabilities change.