Probability help, please

[Cerl]

This question is related to a discussion about a game a friend and I were having, and neither of us could figure out how to actually solve this:

You start with 7 boxes. 6 of the boxes contain fluff, the 7th contains the prize.
As each box is opened, it is discarded so that you have a 1 in 7 chance to get the prize with the first box, a 1 in 6 on the 2nd box, and so on.

What is the actual chance that it will take 6 boxes being opened before you get the prize?

Best we can come up with ourselves is a 1 in 7 chance, but is the fact that you have to fail 5 times before getting the prize on attempt #6 going to change that at all?

 

[raquelc]

you can get the probability by multiplying the probability of having fluff on the 1st one chosen (there are 6 with fluff from a total of 7, that is a 6/7 probability) times the probability of having fluff on the 2nd one chosen (now we have 5 with fluff from a total of 6, that is a 5/6 probability) and so on. You will have 6 fractions multiplying

 

[Steven G]

The answer is 1 in 7 or 1/7. The position of the boxes does not matter, unless there is a difference chance that the prize is in some particular place than another place (maybe there is 1/5 chance of it being in position 1, a 2/5 chance of being in position 2, no chance of being in position 3, 4, 5 or 7 and a 1/5 chance of being in position 6). I am assuming that all positions are equally likely.

Look up the Monty Hall Problem. You will like this.

 

[Cerl]

So the answer really is 1/7 after you do all that. Thank you for your help.