Data Management - Probability Game

AI Thread Summary
The probability game involves players retrieving items from locked boxes based on die rolls, with the objective of collecting all items first. Players can choose how to distribute their items across the boxes, affecting their chances of winning. It is determined that uniformly distributing items across boxes maximizes the probability of retrieval, as each roll guarantees at least one object can be retrieved. The discussion references the "coupon collector's problem" to illustrate that the average number of rolls needed varies significantly based on item placement strategy. Ultimately, uniform packing is deemed the most effective method for minimizing the expected number of rolls to win.
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 probabilty 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.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
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