# Probability of winning this game

This is adapted from a little game I've played recently:

Suppose that every day you get to select one sealed box out of nine and collect whatever item is in the box. How the $n>9$ items are placed in the various boxes changes every day. (Note that there is only one item in each box, and the $n$ items are unique.) Suppose that of the $n$ items you're really only interested in one of them, so that item is what you hope to pick every day, considered here as "winning the game". Here are two things I'm curious to know:

1) What is the probability of winning on a given day?

2) Over a period of $9<k<n$ consecutive days, what is the probability of winning? More precise perhaps, what is the probability of winning at least once?

I don't really have much to contribute on my own. There are $n!/(9!(n-9!)$ ways that the boxes could be filled, and the event of losing or winning has no bearing on the probability of losing or winning the next day, i.e. independence.

EDIT: The chosen item is replaced.

-
Is the item you get replaced before the next day? The $k \lt n$ in 2) seems to indicate that, but the last sentence says otherwise. – Ross Millikan Oct 1 '13 at 22:31
Yes, it's replaced. – Erik Vesterlund Oct 1 '13 at 22:37

(1) You are right that there are $\binom{n}{9} = \frac{n!}{9!(n-9)!}$ ways of selecting 9 objects out to fill boxes, and $\binom{n-1}{8} = \frac{(n-1)!}{8!(n-9)!}$ of these ways include the object you want. Therefore, the probability that your object is in one of the boxes is $$\frac{\binom{n-1}{8}}{\binom{n}{9}} = \frac{(n-1)!\ 9!\ (n-9)!}{n!\ 8!\ (n-9)!} = \frac{9}{n}$$

Now, after these boxes have their contents fixed, you randomly choose one of the 9 boxes. If your object is in the boxes, the probability that you select that exact box is $\frac{1}{9}$. If your object is not in the boxes, the probability that you find your object is $0$. Therefore, the probability that you find your object is

$$\frac{\binom{n-1}{8}}{\binom{n}{9}}\cdot\frac{1}{9} = \frac{9}{n}\cdot\frac{1}{9} = \frac{1}{n}$$

(2) If games for every day are independent, then this part is just an application of binomial distribution.

-

With your procedure, the boxes don't matter, just the set of items available on any given day. You win with probability $\frac 1n$ each day. Over a series of $k$ days, you lose every day with probability $(1-\frac 1n)^k$, so win at least once with probability $1-(1-\frac 1n)^k$

-
Hm ok, but how can the boxes not matter? Suppose one day the item you want is not inside one of the boxes. Then the probability of winning is zero on that day. Also perhaps I should have been explicit that the items are replaced. – Erik Vesterlund Oct 1 '13 at 22:39
peterwhy goes through the calculation, but the basic point is that each day you choose one item at random, so get each item with probability $\frac 1n$. If you know which items are in the boxes, you are correct that the probability is either $\frac 19$ or $0$, but it all comes out to $\frac 1n$ if you don't. I have updated my answer assuming the items are replaced. – Ross Millikan Oct 1 '13 at 22:41