Here is a variation of the Monty Hall problem based on the 100 prisoner challenge.
Three objects – a car, a key and a goat – are randomly placed behind three doors. You play the game with a partner. And each of you can open a maximum of two doors, one person at a time. In order to win the car, the first player must find the car, and the second the key.
If the players play at random, the probability of the first player to find the car in two attempts (out of the possible three doors) is 2/3. Similarly, for the second, finding the key is 2/3. The probability of winning is (2/3) x (2/3) = 4/9 = 44.4%.
Strategy
If they do the following strategy, they can maximise the probability to (2/3), which is equal to the chance of any of them reaching their individual goal. Here is how they do it.
Person 1 goes and opens door 1. If she finds the car, she completes her task and over to person 2. If instead, she finds a key, she will try door 2. Whereas, if she finds a goat on the first attempt, she will go for door 3.
The second person might not start 1 out of 3 times, i.e. whenever the first person was unsuccessful. If that is not, she will first open door 2. If it was a key, the team wins. If she finds a goat behind door 2, she will open door 3; if it is a car, she will open door 1. The possibilities are in the table below.
| Sequence | car-key-goat | car-goat-key | key-car-goat |
| Player 1 | door 1: car | door 1: car | door 1: key door 2: car |
| Player 2 | door 2: key (wins) | door 2: goat door 3: key (wins) | door 2: car door 1: key (wins) |
| Sequence | key-goat-car | goat-key-car | goat-car-key |
| Player 1 | door 1: key door 2: goat | door 1: goat door 3: car | door 1: goat door 3: key |
| Player 2 | no game | door 2: key (wins) | no game |
You can see that whenever the first person does her job (wins the car), the second one gets the keys.
