- #1
SrivastavaHarsh
- 2
- 1
- TL;DR Summary
- A game where n (let n=4; n is always even) players try to find their number inside n boxes by opening at most n/2 boxes.
There are 4 players numbered 1 to 4. There is a room with an entrance door to one side and exit door to opposite side. Inside the room, there are 4 boxes numbered 1 to 4. Inside each box, there is a chit containing a number (with equal probability) from 1 to 4 (inclusive). No two chits can have the same number or more than one number. Players can enter the room one at a time. They can open atmost 2 boxes to find the chit with their number and then exit the room. The players who have exited the room cannot interact with the players who are yet to enter the room. After a player exits the room and before the new player enters, the room is reset to its initial condition. The player who is in the room cannot interact with any other player while inside the room. The winning condition is that all the players find the chit corresponding to their number. Players are allowed to strategize before the start of the game, but cannot change strategy mid-game. What is the probability that the players win the game (also mention the strategy)?
Case 1: The players do not strategize and open 2 boxes at random. Probability that a player finds the chit with his/her number is 1/2. Each players probability of finding the chit is independent of others players. So, Probability of winning = probability that all of the players find their chit = (1/2)^4 = 1/16. This is the lowest possible probability of winning.
Case 2: The players adopt a strategy where they first open the box having the same number as them. If they find the chit with their number inside that box, then they exit the room. Else, they open the box corresponding to the number on the chit that they found inside the first box that they opened. Probability of winning with this strategy is 5/12 which is significantly greater than 1/16 (probability in case 1). Also the probability of individual player to find the chit with his/her number remains 1/2.
In case 1, each player randomly opens boxes with a 50% chance of finding the chit. In case 2, observer sees that each player's starting points is fixed but the chits are placed randomly inside the boxes and so each player still has 50% chance of finding the chit. But unlike case 1 where each player's turn was not only random but also mutually not correlated, in case 2 each player's turn is still random but it follows the distribution of how the chits are placed in the boxes, which is same for all players, making each player's turn mutually correlated. This is the cause of dependence. So, the probability of winning the game in case 2 will depend on how the chits are being distributed (Probability of winning = 1 - Probability of distribution with cycles greater than 2. Example of a cycle greater than 2 : 1st box has 2, 2nd box has 3 and 3rd box has 1).
Can it be said that the increase in probability is due to information being circulated/stored inside the system, here via the structure of the system itself (the system stores the information about how the turns of each player is correlated to one another) which correlates the seemingly independent events? But this behaviour is possible only when the players collude within themselves about how they will interact with the system. If so then should the strategy in case 2 give better results as the number of players increases (Number of players = number of boxes = even. Number of boxes that can be opened by any player =n/2)?
What other strategies are possible for probability of winning higher than 1/16? Is it possible to have a strategy which doesn't fix the starting box for each player making is look random or pseudorandom while still giving better results than 1/16? Also, 1/16 (or 1/2^n in case of n players) seems to be the least possible probability of winning but how to prove it?
Case 1: The players do not strategize and open 2 boxes at random. Probability that a player finds the chit with his/her number is 1/2. Each players probability of finding the chit is independent of others players. So, Probability of winning = probability that all of the players find their chit = (1/2)^4 = 1/16. This is the lowest possible probability of winning.
Case 2: The players adopt a strategy where they first open the box having the same number as them. If they find the chit with their number inside that box, then they exit the room. Else, they open the box corresponding to the number on the chit that they found inside the first box that they opened. Probability of winning with this strategy is 5/12 which is significantly greater than 1/16 (probability in case 1). Also the probability of individual player to find the chit with his/her number remains 1/2.
In case 1, each player randomly opens boxes with a 50% chance of finding the chit. In case 2, observer sees that each player's starting points is fixed but the chits are placed randomly inside the boxes and so each player still has 50% chance of finding the chit. But unlike case 1 where each player's turn was not only random but also mutually not correlated, in case 2 each player's turn is still random but it follows the distribution of how the chits are placed in the boxes, which is same for all players, making each player's turn mutually correlated. This is the cause of dependence. So, the probability of winning the game in case 2 will depend on how the chits are being distributed (Probability of winning = 1 - Probability of distribution with cycles greater than 2. Example of a cycle greater than 2 : 1st box has 2, 2nd box has 3 and 3rd box has 1).
Can it be said that the increase in probability is due to information being circulated/stored inside the system, here via the structure of the system itself (the system stores the information about how the turns of each player is correlated to one another) which correlates the seemingly independent events? But this behaviour is possible only when the players collude within themselves about how they will interact with the system. If so then should the strategy in case 2 give better results as the number of players increases (Number of players = number of boxes = even. Number of boxes that can be opened by any player =n/2)?
What other strategies are possible for probability of winning higher than 1/16? Is it possible to have a strategy which doesn't fix the starting box for each player making is look random or pseudorandom while still giving better results than 1/16? Also, 1/16 (or 1/2^n in case of n players) seems to be the least possible probability of winning but how to prove it?
oh good catch! I saw least but thought greatest.