# What if two players were playing the Monty Hall Game?

I recently stumbled upon the letter that Monty Hall had sent when he was asked for his permission for his name to be used for the paradox. I thought about this scenario and couldn't figure the answer.

Consider two players playing the game with the host simultaneously. There are three doors,namely A, B and C. Player 1 chooses A and player 2 chooses B. The host,knowing that there is nothing behind door C, opens it. Now, based on the standard solution, both players should change doors. But that doesn't make sense. How can both players claim that two different doors have probabilites of 2/3 of having the prize?

• If there were two players there is no guarantee that there is an empty door to reveal. What if both of the initial choices were empty? So, the logic of the paradox sort of breaks down. – CrockGill Oct 22 '15 at 0:22
• Agreed. But let us talk about this specific scenario, where there is a door to reveal. – statBeginner Oct 22 '15 at 0:43
• Your scenario makes no sense. What if one player wants to switch but the other does not? Do they both end up on the same door and thus have to share the car? – Dilip Sarwate Oct 22 '15 at 3:06
• @statBeginner The Monty Hall game strongly depends on the hosts behavior. So just focusing on the case when there is a door to reveal fails to provide the necesary information. – Taemyr Oct 22 '15 at 4:46

The OP is asking about the probability of winning given that Monty has opened the door that neither player chose. Dilip Sarwate's answer describes the "standard rules" of the two-player game in which this cannot happen: Monty will always throw one player out of the game rather than open the unchosen door.

Why stick to the "standard rules" instead of answering the question? We can easily imagine a game with non-standard rules that will allow for the OP's supposition. For example:

When one player guesses correctly, Monty flips a coin (with bias $b$) to decide whether to open the unchosen door, or throw out the other player. If neither player guesses correctly, Monty flips another coin (fair, we hope!) to decide who to throw out.

In the "standard rules" game, the bias $b$ is set to 1, so Monty will always throw out a player (and you're better off staying put). We could call the other edge case "Monty's a big old softy": if bias $b$ is set to 0, Monty will never throw out a player if he can avoid it. (If he does, you know you'd better switch!)

But we don't need to know $b$ in order to solve the stated problem, because we already supposed that Monty didn't throw out a player. In that case we know that one player must have guessed correctly. If we assume that each player had equal likelihood of guessing correctly, then there is no benefit to switching, exactly as Randy argued in his answer.

We could create other non-standard rules, but the symmetry between Players 1 and 2 remains. The only way to break it is to give one player a greater chance of choosing correctly, or alternatively, allow Monty's choice of whether to throw someone out to depend on who guesses correctly. Both of these options seem unfair.

Note: I'm assuming throughout that the contestants must choose different doors.

Exercise for the reader: find the value of $b$ for which there is no benefit in either switching or staying put, even after Monty throws out the other player.

The two-player game is slightly different.

In the two-player game, Monty can always open one of the chosen doors, tell the hapless $A$ (or $B$) to go back to his or her seat, and then offer the survivor ($B$ or $A$ as the case may be), the choice of sticking with the initial choice or switching to the third, unopened, door. In this game, with the usual assumptions that Monty knows which door conceals the prize, that he can always open one of the doors chosen by the participants, that he will open one such door, and that if he can choose which participant to send back (because the prize is behind the unchosen door) then he will choose at random, the optimum choice for the survivor is to "Stay put". There is a $2$-in-$3$ chance that one of the two doors chosen by the two participants is the winning door, and since the loser has been sent back, the survivor better stay put.

A statement of the above Monty Hall game variant can be found as Problem 4(d) of this problem set and the solution is here

• Thank you for the answer and letting me know about the two-player game. I did not know that a "standard" two player game existed. I was looking for a specific version of the two-player game that I described. – statBeginner Nov 9 '15 at 19:12

In the one person game, the player switches because the odds of having chosen the 'good' curtain the first time is changed by Mr Hall in a way that improves his odds every time the game is played. There is always a 'bad' curtain that can be eliminated, which then changes the odds of success for the unchosen remaining curtain every time. The odds are always changed by Mr Hall from one in three initially to one in two if you switch.

In the two person game, there are 3 possible scenarios:

1) Player X chooses right and player Y chooses wrong and Mr Hall can reveal the third curtain.

2) Player X chooses wrong and player Y chooses right and Mr Hall can reveal the third curtain.

3) Player X chooses wrong and player Y chooses wrong and Mr Hall CANNOT reveal the third curtain.

So if Mr Hall CAN reveal the unchosen curtain, the scenario must be #1 or #2, so the odds for either player to have chosen correctly the first time is 50%. Because choosing differently the second time does not change your odds -- they remain 50% -- there's no reason to switch.

However, if Mr Hall CANNOT open the third curtain, you are in scenario #3, and obviously you should switch curtains.

• "The odds are always changed by Mr Hall from one in three initially to one in two if you switch." What a bunch of nonsense! – Dilip Sarwate Oct 22 '15 at 3:03
• Gently, gently @Dilip. There may be more than one way to understand that quotation. Why not ask for clarification before offering condemnation? – whuber Oct 22 '15 at 5:38
• @whuber The sentence that I quoted and that I denigrated is in the paragraph describing the one-contestant game. One might cavil at (though I choose not to do so) the misuse of odds to describe chances. My complaint is about the change from one-in-three to one-in-two if you switch. There are lots of nuances to the Monty Hall game with respect to Monty's knowledge, whether he always offers a switch, or only when the player has chosen the door with the car, etc. but with the standard rules (stated in the links in my own answer), "to one in two if you switch" is nonsense. – Dilip Sarwate Oct 22 '15 at 11:47
• @Dilip Your explanation is constructive, thank you. – whuber Oct 22 '15 at 15:31