Alice and Bob are going on a game show as partners. In this game, there are three doors, which contain a car, the car keys, and a goat, which are arranged randomly and secretly behind the doors. Alice will go onstage first and open two doors and then shut them, while Bob remains backstage in a soundproof booth (he cannot gain any further information about what doors were opened or what they contained). Alice is then led to her own soundproof booth, after which Bob goes onstage and opens two doors. Alice and Bob win the game if Alice finds the car and Bob finds the car keys, while any other outcome results in a loss. The pair are given time to strategize before the game begins.

What is the optimal strategy, and what is the likelihood the pair will win? Without strategizing (i.e., by choosing doors at random), Alice has a 2/3 chance of finding the car, and Bob has a 2/3 chance of finding the keys, for a total win probability of 4/9. But the pair can win with higher likelihood - how?

To clarify a few points: Alice and Bob do not know what is behind any door before opening it, and opening a door reveals only the contents of that door. The prizes behind the doors do not move between rounds. Bob has no way of inferring what doors Alice opened or what she found behind them, other than whatever their previously agreed-upon strategy dictates. And while I've called this a Monty Hall problem due to the doors and prizes, Monty himself does not play any role in this game.

This puzzle adapted from Surprises in Probability- Seventeen Short Stories, by Henk Tijms.

  • $\begingroup$ So if Alice finds the car keys and Bob finds the car, they lose? That's what the question says $\endgroup$
    – Arjun
    Aug 28, 2020 at 19:25
  • $\begingroup$ @BabaYaga That is correct (assuming Bob didn't also find the keys and that Alice didn't also find the car). If Alice does not find the car, or Bob does not find the keys, they lose. The car must be found in the first round by Alice, and the keys in the second round by Bob. $\endgroup$ Aug 28, 2020 at 19:27
  • 2
    $\begingroup$ I get a headache when ever I read the words "alice and bob" lol $\endgroup$
    – Ankit
    Aug 28, 2020 at 22:46

5 Answers 5


They win with probability


with a variant on the 100 Prisoners' Names in Boxes strategy Alice first picks the left door. Bob first picks the middle door. If the first door reveals the car, pick the left door next. If the first door reveals the keys, pick the middle door next. If the first door reveals the goat, pick the right door next.

  • $\begingroup$ Nice. That ensures that they either both fail or both succeed, minimising the number the number of failure cases. $\endgroup$ Aug 28, 2020 at 19:08
  • $\begingroup$ If Alice finds the car in the first door this means she will open that door twice? $\endgroup$
    – Andreas
    Aug 30, 2020 at 7:20
  • $\begingroup$ @Andreas If either Alice or Bob succeed on their first try, their second door pick is entirely irrelevant. $\endgroup$ Apr 5 at 15:20

This solution was derived independently but the result is equivalent to the answer already given by Mark Tilford and the confident tone of a suggestion by Greg Martin led to the realization that this does provide the highest possible probability of winning the car.

Alice and Bob can win with a probability of 2⁄3, or 4 times out of the 6 possible hidden-item arrangements.

The present strategy actually guarantees that Bob will surely find the Key if Alice finds the Car. This surprising result cannot be improved because there is no way to reduce the 1⁄3 probability that the Car remains behind Alice’s unchosen third door.

Imagining a repeating cycle of doors, . . . Left - Middle - Right - Left - Middle - Right - Left - . . . , the hidden items may have two orderings:

  1.  1⁄2 chance that the items are ordered as:  . . . Car - Goat - Key - Car - Goat - Key - Car . . .

  2.  1⁄2 chance that the items are ordered as:  . . . Car - Key - Goat - Car - Key - Goat - Car . . .

The basis of this solution is that Alice and Bob assume that the items are in order # 1.

If # 1 is indeed the order then Alice and Bob will both find their desired items because each will correctly assume which door to open second if their initial choice is wrong. (Which doors are chosen first doesn’t even matter.)

The trick is to also have some chance of winning if the items are in order # 2. In this case Alice and Bob can win only if both are lucky enough to choose their first doors correctly as they will make incorrect adjustments for their second doors. Therefore:

Bob’s first choice should be one door to the right (cyclically) of Alice’s first choice because the Key is one door to the right of the Car in order # 2.

This strategy can be specified as:

Alice first chooses the Middle door and is content if it reveals the Car being sought.
If it reveals the Goat, however, Alice then chooses the Left door;
had the Key been revealed, Alice chooses the Right door second.

Bob first chooses the Right door and is content if it reveals the Key being sought.
If it reveals the Car, however, Bob then chooses the Middle door;
had the Goat been revealed, Bob chooses the Left door second because that is the next door cyclically to the right of the Right door.

Here is how this works out for the 6 possible item arrangements.

  Ordering # 1:       ... C - G - K - C - G - K - C - G - K - C - G - K - C ...

 (3 arrangements,       Left Mid Right  Left Mid Right  Left Mid Right (Left)
   all winners)           C   G   K       G   K   C       K   C   G   (K)
                        -------------   -------------   -------------

 Alice 1st choice             G               K               C!
  good 2nd choice         C!                      C!        (moot)

  Bob  1st choice                 K!              C               G
  good 2nd choice               (moot)        K!          K!          (K)

  Ordering # 2:       ... C - K - G - C - K - G - C - K - G - C - K - G - C ...

 (3 arrangements,       Left Mid Right  Left Mid Right  Left Mid Right
    1 winner)             C   K   G       K   G   C       G   C   K
                        -------------   -------------   -------------

 Alice 1st choice             K               G               C!
 wrong 2nd choice                 G       K                 (moot)

 Bob   1st choice                 G               C               K!
 wrong 2nd choice         C                   G                 (moot)

SIDE NOTE.  A different strategy is almost as successful while allowing Alice to enjoy making at least one choice randomly.

Alice chooses the first door randomly.
If it is the Middle door, Alice doubles the fun by choosing the second door randomly as well; otherwise Alice next chooses the Middle door.
One of Alice’s choices is sure to be the Middle door.

Bob robotically chooses the Left and Right doors.
The Middle door will not be chosen by both contestants.

As the Car and Key are behind different doors, the prospect of winning improves on pure randomness by eliminating a possible same-door (Middle & Middle) combination of the contestants’ choices and thus increasing the likelihood of a different-door combination. Of the 12 possible outcomes, 1⁄2 are winners:

                        Car is in Middle   Goat is in Middle   Key is in Middle
                          (4 outcomes,        (4 outcomes,       (4 outcomes,
                          all winners)         2 winners)         no winners)

                         Left Mid Right      Left Mid Right     Left Mid Right
                           G   C   K           C   G   K          C   K   G
                         -------------       -------------      -------------

Alice chooses L and M      G   C!              C!  G              C   K
  Bob chooses L and R      G       K!          C       K!         C       G

Alice chooses M and R          C!  K               G   K              K   G
  Bob chooses L and R      G       K!          C       K          C       G

                         Left Mid Right      Left Mid Right     Left Mid Right
                           K   C   G           K   G   C          G   K   C
                         -------------       -------------      -------------

Alice chooses L and M      K   C!              K   G              G   K
  Bob chooses L and R      K!      G           K       C          G       C

Alice chooses M and R          C!  G               G   C!             K   C
  Bob chooses L and R      K!      G           K!      C          G       C
  • 1
    $\begingroup$ It might be worth mentioning why the first probability found is best possible. $\endgroup$ Aug 30, 2020 at 4:27
  • $\begingroup$ Thank you again for the confident tone of your suggestion, @Greg Martin! Made me think afresh about the limit and realize how trivial Bob's role had become. $\endgroup$
    – humn
    Aug 30, 2020 at 5:28

I believe

They can win with probability $\dfrac{1}{2}$, using the following strategy:
Alice opens doors $A$ and $B$, and then Bob opens doors $B$ and $C$. Alice has a $2/3$ chance of getting the car, and so the probability of success cannot exceed this. But Bob may as well work on the assumption that Alice has succeeded, because if she hasn't they've lost anyway. So conditioning on Alice's success means it's more likely the key is in the unopened door, $C$. In a list:
1. c g k.
2. c k g.
3. g c k.
4. g k c.
5. k g c.
6. k c g.
where k is for key, c for car, g for goat, they win in scenarios 1,2,3 - so 50%.

  • $\begingroup$ I am new to this site and i might sound like a doof but can you elaborate more mathematically $\endgroup$
    – Arjun
    Aug 28, 2020 at 17:29
  • $\begingroup$ The reasoning for Bob's choice here nicely relates this puzzle to the original Monty Hall puzzle $\endgroup$
    – humn
    Aug 29, 2020 at 23:57

I believe the answer is:

With a 50/50 shot at winning, they agree to both open one door the same, and the second door different. i.e. Alice opens left and middle, Bob opens middle and right. The order that Bob opens his doors do not matter

Start with all possible permutations:

There are 6 total options for the contents of the doors.

Test the different strategies:

Strategy 1:

If Alice and Bob both open the same two doors then they have a 2/6 or 1/3 chance at winning. This is worse than the random odds, so it's not good.

Strategy 2:

If they instead choose one door to both open, and then split the remaining two doors, this will give them a 3/6 or 1/2 chance at winning.

Strategy 3:

If Bob opens the door that Alice already opened first, he doesn't gain any new info. If it's the car he knows that they aren't guaranteed to lose, but it's still a 50/50 shot for him, so the total odds are still 50%. If it's the Key, then he knows he won, but it's a 50/50 if Alice won, so still 50%. If it's the goat, then he has a 50% chance of guessing right, and Alice only had a 50% chance, so it's only a 25% chance.

Strategy 4:

If Bob opens the unopened door first, if it's the car, he knows they have lost. If it's the Key, he won and he knows Alice won. If it's the goat, then he knows Alice won, and he has a 50/50 shot. Still, each of these comes out to 50% chance, so the order he opens the doors doesn't matter. (2/6 chance for instant loss, 2/6 chance for instant win, and 2/6 chance for a 50/50 win/loss, total of 3/6 wins, 3/6 losses)

  • $\begingroup$ Do the spoilers add something?\ $\endgroup$
    – Arjun
    Aug 28, 2020 at 17:59
  • $\begingroup$ @BabaYaga the spoilers hide something, for those who want to answer without seeing other answers. $\endgroup$ Aug 28, 2020 at 19:50
  • $\begingroup$ @BabaYaga Some like spoilers, others think they’re a nuisance. For complex or multi-stage logic, I think spoilers are unnecessary. $\endgroup$
    – Lawrence
    Aug 29, 2020 at 8:52
  • 1
    $\begingroup$ @Lawrence exactly, you have to click on every single box! $\endgroup$
    – Arjun
    Aug 29, 2020 at 9:08

Alice and Bob start by assuming that the Goat is behind a specific door. If the doors are numbered 1, 2, and 3 then they might, for example, decide to assume the Goat is behind door 3. With this assumption, Alice plans to pick door 1 first and then door 2. Bob plans to pick door 2 first and then door 1. If their assumption is right they will always win. If, however, either one of them gets the Goat on their first pick then they know that their assumption (that door three has the Goat) was wrong.

If Alice picks the Goat on her first pick then she knows that Bob is going to pick the Goat (door 1) on his second pick so the only way they can win now is if Bob picks the key on his first pick (door 2). Thus, Alice knows that the only way they can win now is if she picks door 3 on her second pick instead of door 2.

If Bob picks the Goat on his first pick then he knows that Alice picked the Goat (door 2) on her second pick so the only way they can win now is if Alice picked the car on her first pick (door 1). Thus, Bob knows that the only way they can win now is if he picks door 3 on his second pick instead of door 1.

In short, the general strategy is for them to plan to pick two prearranged doors. Alice in one order and Bob in the reverse order. But, if either picks the Goat on their first pick, they prearrange that they will alter their second pick.

Here are the possible scenarios and the result with the example strategy:


Therefore the probability of a win is 2/3.


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