Data Management - Probability Game

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Homework Help Overview

The discussion revolves around a probability game designed for a Data Management class, where players retrieve items from locked boxes based on die rolls. The original poster questions whether the placement of items in the boxes affects the chances of winning, particularly comparing uniform distribution versus concentrated placement of items.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the implications of different item placements in relation to winning probabilities. They question whether the average number of rolls needed to win varies based on how items are distributed across the boxes.

Discussion Status

Some participants suggest that the average number of rolls to win may not be the same depending on the distribution of items. There is a mention of the "coupon collector's problem" as a relevant concept, indicating that some understanding is developing regarding the underlying probability principles.

Contextual Notes

Participants are considering the constraints of the game setup, including the maximum number of items that can be retrieved per turn and the potential strategies for item placement. The discussion reflects uncertainty about the optimal packing method and its effect on winning chances.

billy1024
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I am making a probability game for my Data Management class, and have a few things I am unsure about.

Essentially, each player has 6 objects located in locked boxes numbered one through six. Players take turns rolling a die that produces the key to the box that corresponds to the roll. The winner is the one who is able to retrieve all their items first.

My question is, if the players are able to choose where each item is placed (i.e. multiple items in some boxes, no items in some boxes) and each attempt can only recover one item max, would the chances of winning be the same?

The average number of rolls to roll all numbers from 1-6 should be the same as the average number of rolls to roll any combination of 6 numbers from 1-6, right?
 
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I don't think the average number of rolls to win will be the same. Think about the different probabilities between a player who puts 1 object in each locked box versus a player who puts all items in 1 box. Then think about the probability of winning on the first 6 rolls. The player with one object in each box has 6! ways to win; the player will all the objects in one box has only 1 way to win.

After 3 rolls, I'm guessing, the player with 1 object in each box has close to a 50% chance of having 3 objects, the player with all objects in one box has a (1/6)^3 chance of having 3 objects.

I ran a test program 10 times. Only once did I get 6 of a selected number before I got 1 of each number.
 
billy1024 said:
I am making a probability game for my Data Management class, and have a few things I am unsure about.

Essentially, each player has 6 objects located in locked boxes numbered one through six. Players take turns rolling a die that produces the key to the box that corresponds to the roll. The winner is the one who is able to retrieve all their items first.

My question is, if the players are able to choose where each item is placed (i.e. multiple items in some boxes, no items in some boxes) and each attempt can only recover one item max, would the chances of winning be the same?

The average number of rolls to roll all numbers from 1-6 should be the same as the average number of rolls to roll any combination of 6 numbers from 1-6, right?

If you put in one object per box you are guaranteed to retrieve one object on the first toss; with any other packing method you have no such guarantee. After retrieving the first object, the uniform packing method gives you a 5/6 chance of retrieving one object on the second toss. No other way of packing 5 object into the six boxes can do better than that. You can continue like that, and be fairly sure that uniform packing is best, at least in terms of minimizing the expected number of tosses to win.
 
After looking at the extremes and looking up the "coupon collector's problem", I think I understand it a bit more, thanks.
 

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