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Home/ Questions/Q 481965
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Editorial Team
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Editorial Team
Asked: May 13, 20262026-05-13T01:02:54+00:00 2026-05-13T01:02:54+00:00

I’m coding a board game where there is a bag of possible pieces. Each

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I’m coding a board game where there is a bag of possible pieces. Each turn, players remove randomly selected pieces from the bag according to certain rules.

For my implementation, it may be easier to divide up the bag initially into pools for one or more players. These pools would be randomly selected, but now different players would be picking from different bags. Is this any different?

If one player’s bag ran out, more would be randomly shuffled into it from the general stockpile.

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  1. Editorial Team
    Editorial Team
    2026-05-13T01:02:54+00:00Added an answer on May 13, 2026 at 1:02 am

    So long as:

    • the partition into “pool” bags is random
    • the assignment of players to a given pool bag is random
    • the game is such that items drawn by the players are effectively removed from the bag (never returned to the bag,or any other bag, for the duration of the current game)
    • the players are not cognizant of the content of any of the bags

    The two approaches (“original” with one big common bag, “modified” with one pool bag per player are equivalent with regards to probabilities.

    It only gets a bit tricky towards the end of the game, when some of players’ bags are empty. The fairest to let pick from 100% of the items still in play, hence, they should both pick from which bag they pick and [blindly,of course] pick one item from said bag.

    This problem illustrate an interesting characteristic of probabilities which is that probabilities are relative to the amount of knowledge one has about the situation. For example the game host may well know that the “pool” bag assigned to say player X does not include any say Letter “A” (thinking about scrabble), but so long as none of the players know this (and so long as the partitions into pool bag was fully random), the game remains fair, and player “X” still has to assume that his/her probably of hitting an “A” the next time a letter is drawn, is the same as if all remaining letters were available to him/her.

    Edit:
    Not withstanding the mathematical validity of the assertion that both procedures are fully equivalent, perception is an important factor in games that include a chance component (in particular if the game also includes a pecuniary component). To avoid the ire of players who do not understand this equity, you may stick to the original procedure…

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