I used to think that the biggest problem with the Monty Hall puzzle was that it wasn’t made clear whether Monty always opens another door, but after last week’s Ask Marilyn column, I realized that perhaps an even larger of a problem is that it is not specified that the second time you choose a door, your original door is a choice. Here’s a version of the puzzle that I think makes this clear:
Suppose two players are allowed to choose among three doors. If they both choose the correct door, they each get a car. If only one chooses the correct one, only that player gets the car. Player A chooses door 1. Player B chooses door 2. The host opens door 2 and says to Player A “Well, your door still could be the right one. Do you want to keep it, or switch to door 3?” Marilyn’s advice is that he should switch. But suppose Monty now turns to Player B and says “Sorry, you didn’t choose the right door. But I’ll give you another chance. Do you think it’s door 1, or door 3?” I think that it’s quite clear that both are equally likely. But if it’s truly advantageous for Player A to switch to door 3, it should be advantageous for Player B to do so as well. Whether door 3 is correct or not isn’t going to depend on who chooses it, so if it’s good for one player it’s good for the other.
So going back to the one player problem, suppose that the situation is as follows. First, the player chooses a door. Next, Monty revelas one of the doors and asks the player to choose among the two remaining. The player’s original choice doesn’t affect which door Monty reveals. It should be clear that the player has a fifty fifty chance no matter what he chooses.
GQ seemed like the best place for this, even though I don’t really have a question. Uhh… what’s you favorite color?
In the orthodox version of the Monty Hall problem, Monty always gives you a choice. In that version, it always is a good idea to switch doors.
However as you pointed out, in other scenarios, Monty might not always be on the up and up. He might intentionally be trying to keep you from getting the prize and he might vary his actions depending on whether your original pick was correct or not. In order to analyse the odds in these cases, you need to know more details about what Monty does.
In the scenario you described, the odds favor staying with your first pick if Monty does what you said every time. (You don’t describe what Monty does if both players pick the same door, so I’ll ignore that case except to note it’s identical to the original problem.) When the two players pick different doors, there are three possibilities: either player A picked right, player B picked right, or neither player picked right. Each possibility will occure one third of the time.
If A picked the winning door, then Monty must open B’s door per your rules, and offer A and B the chance to repick. In this case A should stick with his first choice and B should pick the same door. If B picked the winning door, the reverse happens. If neither A or B picked the winning door, Monty can open either player’s door and offer both players the chance to repick. In this case, both players should pick the door that wasn’t originally chosen. If you add up the possibilities, you see that players who stick with the first choice or join with the player who’s staying on his first choice will win 2/3 of the time. Players who switch to the door originally unchosen will only win 1/3 of the time.
I’m not sure if you understand the situation that I was presenting. In my hypothetical, your actions do not affect Monty’s.
I don’t see why people so quickly leap to the conclusion that whatever happened, must always happen. The same strange presumption pops up in the original (the host reveals another door, so he must have a policy of always opening another door). There are two players, one knows he wasn’t right the first time and he has another chance, the other knows one of the incorrect doors, and has to decide whether to switch. That’s all they know, and they have to make their decisions based on that. They don’t get to read the host’s mmind and find out that if they had chosen different doors, he still would have opened one of their doors.
Not to detract from the actual logic puzzle, but actually Monty Hall always knew what was behind all the doors and inside the boxes. The show always had a very thorough run-through before the contestants were let inside the studio. To quote The Ryan “your actions do not affect Monty’s”. Everything was organized beforehand.
I don’t understand the criteria that are used in your example for Monte to select a door. In the “original” Monte was always guaranteed one empty box to open, and he would never open the “prize box”.
Exactly what are the selection criteria in your example(s)? Wholly random? Or, given that A and B have made different selections, does he open the empty box that one or the other must have chosen? And if they’re both empty?
As far as your problem goes, I agree with Little Nemo (again, assuming both players pick different doors, and Monty opens an empty door that one of the players have chosen): sticking with Player A’s door is the best bet.
But in this case, as in the original Monty Hall problem, the player’s choice does affect which door Monty reveals. If the player picks an empty door, Monty is forced to choose the other empty door. Similarly in your problem (if we have read it correctly - we need to know the rules Monty is governed by in order to work out the probablility; a single example isn’t enough), Monty has to open one of the doors that has been chosen by a player. If Player A chooses the prize, he opens Player B’s door, and vice versa.
It is because Monty’s choice is always affected by which door we choose that we can use the knowledge he gives us to gain an advantage. If he just picks a door at random, we lose that advantage.
The extra information can still be used to the contestant’s advantage.
In the original game the “always swap” tactic gives wins 2/3 of the time.
In the modified game where a box is chosen randomly by Monte,
1/3 of the time Monte will open the prize box, (presumably the contestent cannot elect to swap to this prize box)
1/3 of these times (1/9 of the whole) the contestant also has chosen the prise box.
2/3 of the time Monte will open an empty box, (here it is still worth swapping to the remaining unopened box, for exactly the same reasons as the original game)
2/3 of these time (4/9 of the whole) the swap-stategy wins.
Well, it’s that strange presumption that makes the question interesting. Granted, that’s not really the way Monty ran the show, and people usually forget to include that presumption in the problem statement (see Cecil’s answer in the infamous Monty Hall problem column, for example). However, the reason the contestant has an extra edge (in the problem, not on the original show is because he knows that Monty always opens a door, and that door is neither the contestant’s original choice nor the one with the prize behind it. If you eliminate that additional knowledge, then you eliminate the contestant’s edge.
For example, in your proposal (where the player’s original choice doesn’t affect the door Monty chooses), the additional knowledge is eliminated, and, as you say, the chances of guessing the correct door are 50/50.
I guess the next wrinkle is to say you can rechoose any door whatsoever after Monty opens his randomly chosen door.
In which case, cut to the chase and forget your first choice.
1/3 of the time Monty will open the prize door which you take.
2/3 of the time Monty will open a losing door, half the time you choose the winner. 1/3 + (1/2 x 2/3)=2/3, and we are back to the odds of the original Monty problem.
This is not exactly right, some of the time Monte and the contestant will choose the same EMPTY box, in this situation it is clear that the contestant should swap to one of the others, in that situation the contestant will be right 1/2 of those times, giving an expected return in this game of 4/9 (still better than 1/3, not as good as the 1/2 suggested (because of the assumption that if Monte chooses the prize box the contestant can’t swap in))
The Ryan needs to come back and clarify the “rules”
In the game that I suggested name the boxes A, B, C; I believe we can without loss of generality assume the prize is in C
Monte’s and the contestant’s choices can be represented by the ordered pair XY, so for instance, AC means Monte chooses A, contestant chooses C, each ordered pair is equally likely with probability 1/9
Swap pay-out No-swap pay-out
AA 1/2* 0
AB 1 0
AC 0 1
BA 1 0
BB 1/2* 0
BC 0 1
CA 0** 0
CB 0** 0
CC 1 1
==================================================
Total 4 3
That’s expected payouts for each strategy of 4/9 and 3/9
in these cases all that is known is that the prize is in one of the other boxes with equal chance
** using aahala’s version these will have payouts of 1, giving an expected payout of 6/9 as observed.
I’m not leaping to conclusions, I’m describing what the odds are when Monty acts a certain way. If he acts differently, the odds change. I addressed the scenario you described in your original post as best as I could understand it from your description.
Scenario 1 - Three doors, one prize, one player. The player picks a door. Monty always opens a different door which doesn’t have the prize and offers the player a chance to switch doors. Players who switch will win twice as often as players who don’t switch.
Scenario 2 - Three doors, one prize, two players. The two players both pick one door. Monty always opens a different door which doesn’t have the prize and offers the players a chance to switch doors. This scenario is the same as Scenario 1, and the players should switch.
Scenario 3 - Three doors, one prize, two players. The players pick two different doors. Monty always opens one of the player’s doors which doesn’t have a prize behind it and offers both players a chance to pick a new door from the remaining two. In this scenario, players who switch to the door no one originally chose will lose twice as often.
Now you can invent many other scenarios. What if there are more than three doors? What if there is more than one prize? What if Monty only offers a chance to switch some of the time? What if Monty is intentionally trying to make a player win or lose? All of these will affect the odds, so if you want to determine the odds you need to describe precisely what is happening.
Don’t think so, GU. With two players choosing different doors, chances are 2/3 that one of them will have (originally) picked the correct door. Since Monty is forced to reveal a door that one of the players (let’s say player 1) has already picked, there’s a 2/3 chance that the door selected by the player 2 is the winner (and, still, only a 1/3 chance that the remaining unselected door is a winner). So both players should then opt for player 2’s door. It’s like the standard Monty Hall problem, only inverted, since two doors are picked in the first round.
In 300 games, A is opened 150 times. What does this represent?
All the times B picked right(100) and one-half the times #2 is the right answer. Similarly for B, so both need to choose or stick with the unopened player’s door, not door #2.