# Data Management - Probability Game

1. Dec 16, 2012

### billy1024

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?

2. Dec 16, 2012

### JimRoo

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.

3. Dec 17, 2012

### Ray Vickson

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.

4. Dec 17, 2012

### billy1024

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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