Coming off of my last Switcheroo puzzle, here's another similar one in which you have to figure out the maximum number of options for the riddle.

The Monty Hall problem goes as follows: you're on a game show, and you can pick one of three doors. One has a car, and two have goats. There's an equal probability of any prize being behind any door. After selecting one door, the host reveals a goat behind one of the other doors, and offers you the option to switch. Should you? The answer, again in spoilers for those who don't know:

is to switch. Most people assume there's a 50-50 chance that your door has the car, and thus you shouldn't switch. They don't realize that by switching, you increase your odds to 2/3: since there are three doors, two of which contain goats, there's a 66% chance that you did not select the car.

My question is as follows: What if, instead of there being three doors, there were more? There's still only one car, and the rest of the doors contain goats. Each time, the host reveals a door with a goat, and you have the option to switch. The process repeats itself until you're left with one door. How many doors can there be for the logic above,

that you should switch every time,

to still apply? To prevent the answer from being infinity, say that once you switch from a door, you're not allowed to pick it ever again, unless you've already picked all of the remaining doors.

  • $\begingroup$ One thing that most people assume about the problem, is the the host is actively trying to make sure the contestant doesn't get the car. What if the host doesn't know? $\endgroup$
    – user16469
    Jun 8, 2016 at 21:05
  • $\begingroup$ @Zymus "the host reveals a goat" - the host's only actions are explicitly specified in the puzzle, his motivation doesn't matter. $\endgroup$
    – Blorgbeard
    Jun 8, 2016 at 21:53
  • 1
    $\begingroup$ This variation was considered recently at Mathematics: math.stackexchange.com/questions/1782952/… $\endgroup$ Jun 8, 2016 at 22:17

2 Answers 2


The optimal strategy is to

stay until the last switch, then switch at the last possible moment. This way you only lose if you picked the winning door initially, giving you a $\frac{N-1}{N}$ chance of victory.

Some intuition to the why this is the best you can do:

Every time Monty opens a door, you can say the probability of the door he opened is distributed to the remaining doors, and the probability of each of those doors of being the winning door increases (or stays the same), being that the probability for doors you're not on increase more.

At the end, you want the largest amount of probability to be concentrated in a single door, which happens if the remaining door has the least amount of probability. As all probabilities only increase, the least you can get for a single door is the initial probability of $\frac{1}{N}$, which is achieved by sticking to one door for as long as you can.

  • 3
    $\begingroup$ I think there's an important distinction. Switching always gives you a better chance that the car is under the new door you select, so looking only at one round at a time, Marius seems to be correct (e.g. suppose the host could stop the game at any time). However your strategy does seem to optimize for choosing the car on the final move. $\endgroup$
    – p.s.w.g
    Jun 8, 2016 at 20:05
  • 3
    $\begingroup$ Indeed, this goes to show that locally optimal choices are not always globally optimal. We are concerned with the final choice for this specific puzzle of course, as "The process reveals itself until you're left with one door" $\endgroup$
    – ffao
    Jun 8, 2016 at 20:54
  • 1
    $\begingroup$ I've always thought the Monty Hall problem was one of the best probability puzzles because it leads to so many interesting and often counter-intuitive results, yet it's concepts are simple enough that non-mathematicians (like myself) are able to grasp them. $\endgroup$
    – p.s.w.g
    Jun 8, 2016 at 21:21
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    $\begingroup$ I figured some people might not be convinced, so I wrote this jsfiddle to simulate and evaluate various strategies. $\endgroup$
    – p.s.w.g
    Jun 9, 2016 at 6:48

You should



There is a 67% chance of a goat and 33% chance of a car. You're likely to get a goat. So, when Monty opens the other goat door, you still probably have a goat, because you picked the door before the other goat door opened. Then, since you probably have a goat, the other door is probably the car. You should switch every time. By the way, I will always stick because I think goats are cute.


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