Everybody knows the famous Monty Hall problem; way too much ink has been spilled over it already. Let's take it as a given and consider the following variant of the problem that I thought up this morning.

Suppose Monty has three apples. Two of them have worms in them, and one doesn't. (For the purposes of this problem, let's assume that finding a worm in your apple is an undesirable outcome). He gives three "contestants" one apple each, then he picks one that he knows has a worm in his apple and instructs him to bite into it. The poor contestant does so, finds (half of) a worm in it, and runs off-stage in disgust.

Now consider the situations of the two remaining contestants. Each one has a classical Monty Hall problem facing him. From player A's perspective, one "door" has been "opened" and revealed to have a "goat"; using the same logic as before, he should choose to switch apples with player B.

The paradox is that player B can use the same logic to conclude that he should switch apples with player A. Therefore, each of the two remaining contestants agree that they should switch apples, and they'll both be better off! Of course, this can't be the case. Exactly one of them gets a worm no matter what.

Where is the flaw in the logic? Where does the analogy between this variant of the problem and the classical version break down?

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Adrian Petrescu
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    The players are not playing the Monty Hall problem: they did not each choose an apple at random from three possibilities, they were *given* an apple. Even if you assume each player "chose" an apple, the first player had a 1-in-3 chance of getting a worm-free apple and is indeed playing a classic MH, but the second player isn't: he does not have three free choices with two goats and one car; he has two choices, which are contingent on player's 1 choice. The third player is even worse off, as he has no choice at all. So at least one of A and B is not in the situation of the MH player. – Arturo Magidin Sep 17 '11 at 21:01
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    Gotcha; such subtle distinctions between random choice and randomly assigning a mapping of players to apples is pretty counter-intuitive to me, but I get it. Thanks, Arturo. (Your comment strikes me as a valid answer, you could post it as such). Cheers! – Adrian Petrescu Sep 17 '11 at 21:08
  • So is tzs's, for that matter. – Arturo Magidin Sep 17 '11 at 21:12

6 Answers6


I completely agree with Henning Makholm: the important difference between this problem and the classic Monty Hall problem is not whether the apples are chosen by the players or assigned to them — in fact, that makes absolutely no difference, since they have no information to base any meaningful choice on at the point where the apples are given to them.

Rather, the key difference is that, in the classic Monty Hall problem, the player knows that Monty will never open the door they choose. Similarly, if one of the players in this problem knew that they wouldn't be asked to bite into the first apple, they'd be better off switching apples with the other remaining player. But of course, if the apples are assigned randomly, it's impossible for more than one of the players to (correctly) possess such knowledge: if two of the players knew they'd never be chosen to go first, and the third one got the wormless apple, Monty would have no way to pick a player with a wormy apple to go first.

Anyway, you don't really have to believe my reasoning above; as with the classic Monty Hall problem, we can simply enumerate the possible outcomes. Of course, I'm making here a few assumptions which weren't quite explicitly stated by the OP, but which seem like reasonable interpretations of the problem statement and match the classic Monty Hall problem:

  • Each of the players is equally likely to get the wormless apple.
  • Monty will always choose a player with a wormy apple to go first.
  • Of the two players with wormy apples, both are equally likely to be chosen to go first.
  • All the players know all of the above things in advance.

Given these assumptions, there are six possible situations the might occur, with equal probability, at the point where the two remaining players are asked to switch:

  1. $A$ has the wormless apple, $B$ went first $\to$ $A$ and $C$ remain.
  2. $A$ has the wormless apple, $C$ went first $\to$ $A$ and $B$ remain.
  3. $B$ has the wormless apple, $C$ went first $\to$ $B$ and $A$ remain.
  4. $B$ has the wormless apple, $A$ went first $\to$ $B$ and $C$ remain.
  5. $C$ has the wormless apple, $A$ went first $\to$ $C$ and $B$ remain.
  6. $C$ has the wormless apple, $B$ went first $\to$ $C$ and $A$ remain.

From the list above, you can easily count that, for each player, there are four scenarios where they remain, and in two of those they have the wormless apple. So it makes no difference whether they switch or not.

But what if one player, say $A$, knew that they'd never be chosen to go first? Then, if $B$ or $C$ got the wormless apple, Monty would have to choose the other one of them to go first. Thus, scenarios 4 and 5 above become impossible, while 3 and 6 become twice as likely. Thus, if, say, $A$ and $B$ remain, they know that they have to be in scenarios 2 or 3 — and of those, the one in which $B$ has the wormless apple (3) is now twice as likely as the one in which $A$ has it (2), so $A$ should want to switch but $B$ should not.

Ilmari Karonen
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I'm not sure Arturo's solution (in comments) is entirely right. It doesn't really matter who's making the random choice, as long as it is indeed random (which we're assuming it to be here).

I would point to the fact that in the game you describe, every player has -- at the time the apples where chosen -- a risk of getting to run from the stage in disgust rather than getting a chance to switch. Since this risk is dependent on whether the apple he was originally assigned contained a worm or not, once a player knows he's not running in disgust, he has information that the player in a classic Monty Hall game doesn't have, and that changes the odds for him.

(From the beginning, the chance that a player has an apple with a worm is 2/3. However, once a player finds himself being given the chance to switch, half of that probability has disappeared into the possibility that he would have been eliminated initially. So the remaining chance of having a worm is now only 1/2 of the initial probability mass, which is the same as the chance of having a good apple.)

This also points to the sometimes underappreciated fact that what is important in Monty Hall-type problems is not simply what has actually happened to you before you make your choice, but also your assumptions about what else might have had happened to you but didn't. If, the original game had been such that Monty has an option not to offer you a switch at all, the entire standard argument for switching collapses. (Imagine, for example, that Monty were a miser and only offered a switch to contestants that had picked the prize door initially).

hmakholm left over Monica
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  • In any case, the point is that the players are not truly playing MH, so they are not each facing "a classic Monty Hall problem". We all agree they are not, we are pointing out different ways in which they are not, I think. – Arturo Magidin Sep 17 '11 at 21:16
  • @Arturo There is clearly a key difference. According to your logic, if in the classical Monty Hall Problem, the initial door was assigned to the contestant, that would have changed the problem, which is not true. – Alex B. Sep 18 '11 at 00:31
  • @Henning: I'm not sure I agree with your reading of what "my logic" is supposed to be. I'm saying that *at least one* of the two contestants left after the first one runs away is not playing the classic MH problem, and so at least one cannot use the classic argument to say he "should" switch; how would that logic apply to a single contestant in the original show who is "given" a door? – Arturo Magidin Sep 18 '11 at 03:51
  • Arturo, I think you're speaking to @Alex, right? – hmakholm left over Monica Sep 18 '11 at 04:09
  • @Henning: Oops. Sorry, yes. I hate it when I do that... – Arturo Magidin Sep 18 '11 at 04:18

The difference is that in the classic problem, the player picked a door. In your problem, the player is having an apple assigned by Monty. The problems are not analogous, and so you can't analyze the second by treating at as instance of the first.

Note that you can't fix this by letting the players pick their apples, because two players cannot pick the same apple.

The original puzzle works because it is an iterative process--the player's choice constrains Monty, so that Monty has to reveal information. In your game Monty is not constrained--he can pick which apple he is going to reveal before the apples are given to the players. His choice reveals nothing useful to them.

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The first 'player' is assumed to always find a worm, and is then excused, so the game is really just two players, and one prize. From the players' perspectives, each has a 50% chance of holding the winning apple. So it doesn't matter whether they swap or not; there's no strategy that can improve the odds of winning. MH is about demonstrating that a consistent strategy of switching improves the odds, so this is quite unlike the MH situation.

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Most of the answers are very good. But I want to give the intuitive crux of the difference between this problem and the classic Monty Haul problem and why that difference matters:

In the classic Monty Haul problem, Monty reveals a door that has no prize. You know Monty can do this, you know Monty will do this. This gives you no new information. If your odds of winning before Monty opened the door were 1/3, they're still 1/3 after.

In this problem, none of the three players knows at the beginning whether or not they'll be the player who is eliminated. Clearly, each of the three players has a 1/3 chance of winning at the start. But once one player is eliminated, each of the two remaining players has new information -- that they will not be the eliminated player. Their odds of winning have now gone up -- from 1/3 to 1/2.

While not as rigorous a proof as some of the other answers, I hope this makes it easier to intuit the distinction and why it matters. Prior to Monty telling one player to eat his apple, no player knew whether or not that person would be them. Monty's choice of player in this example, unlike his choice of door in the classic case, changes things.

David Schwartz
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There is a way to get the Monty Hall problem back. But it requires to wiggle a bit on the assumptions. I.e. the problem does not specify how Monty is choosing the candidate that has to bite into the apple first. He could of course choose this candidate at random among the two candidates with the worm.

But let's assume that in order to boost ratings, Monty always spares the most likeable person from biting into the apple first. Assume further that you are actually the most likeable person in the show. So you know that you are not picked first. Then after one candidate is kicked off the show, you do actually face the original Monty Hall problem: I.e. you initially have a bad apple with probability 2/3, but by switching you can decrease the chance of a bad apple to 1/3.

The fine difference between the situation here and the situation analyzed in the other answers is that here you know (assuming that you are the most likeable person and Monty is the rating-boosting selection strategie) that you don't get kicked off in the first round.

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