This is a short illustration on the problem of points. Here’s a description of the famous problem of points. Two players play a game of chance with the agreement that each player puts up equal stakes and that the first player who wins a certain number of rounds (or points) will collect the entire stakes. Suppose that the game is interrupted before either player has won (a more colorful way of describing this: the Police had raided the facility). How do the players divide the stakes fairly? We give some simple illustrations to show how to divide the stakes fairly and equitably. See here for a longer discussion.
First Example
Let’s start with a simple example. Two players, A and B, are playing a game of chance in such a way that the first player who wins 3 rounds wins the game. At the time the game is interrupted, Player A had win one round and Player B had won zero rounds. The total amount of stakes put up by the two players was 64 (32 from each player). How can the two players divide the stakes?
The calculation should be prospective and not retrospective. The division of the stake should be done on a prospective basis, i.e., determined by what will happen if the game of chance continues without interruption and not on what had happened in the past. In the simple example here, division of the stakes on a retrospective basis would mean Player gets the total amount since Player A is the sole winner so far. As we will see below, that is not the fair way of splitting the stake.
At the time of interruption, Player A needs to win 2 more rounds in order to win and Player B would need to win 3 rounds to win. This means that to decide a winner, the players need to play 4 (=2 + 3 -1) rounds. Lets’ calculate the probability of each player winning if the game continues.
As a result, Player A’s share is 
Second Example
Same as before, the first player who wins 3 rounds wins the game. Suppose that at the time of interruption, Player A had won 2 games and Player B had won 1 round. How should they divide the stake of 64?
If the game were to continue, Player A needs to win 1 more round to win and Player B needs to win 2 more rounds to win. This means that they need to play 2 (=1 + 2 -1) rounds to determine a winner. Here’s the probability calculation.
As a result, Player A should get three-quarter of the total stake, which is 48, while Player B should get one-quarter of the stake, which is 16.
Remarks
In the above calculation of probabilities of winning, the binomial distribution is used. The fair method to divide the stake should be based on the competitive position of the players. The player in a more competitive position should receive a larger share. Take an extreme situation. Let’s say two players agree that the first player winning 6 rounds wins the game. Suppose that Player A had won 5 rounds and Player B had won just 1 round. Then Player A would be in a stronger competitive position. As a result, Player A should wins almost all of the total stake. In this case, the prospective probability of A winning is 31/32. If the total stake is 64, then Player A’s share is 62.
In conclusion, the division of stakes should be done on a prospective bases, i.e., it should be determined by what will happen in the future had the game continued and not on what had happened. See here for a fuller discussion.
Dan Ma The problem of points
Daniel Ma The problem of points
Dan Ma Probability and statistics
Daniel Ma Probability and statistics
Dan Ma Binomial distribution
Daniel Ma Binomial distribution
Posted: October 3, 2024