Suppose I have 6 coupons found in a box of cereal that are randomly distributed throughout the boxes.
If I wanted to find the probability that it takes at most 8 boxes of cereal to find all the coupons, would I be wrong to say that
P(X = at most 8) = sum of binomial distributions for n = 6, 7, 8 and k = 6? Or do I have to venture into more complicated territory (balls and bins for example).
Thanks.
On
The problem can be reduced to a more familiar type of problem (always a good strategy), viz:
Find the probability that all faces of a die thrown $8$ times show up.
Only $2$ favorable patterns showing all die faces will be found.
Favorable ways = [Choose die faces for the pattern] $\times$ [Ways to sequence the pattern]
and divide by $6^8$ to get the probability
$3-1-1-1-1-1:\; \binom61 \times \frac{8!}{3!} = 40,320$
$2-2-1-1-1-1:\; \binom62 \times \frac{8!}{2!2!} = 151,200$
$Pr = \dfrac{40,320+151,200}{6^8} \approx 0.114$
So from the comments I got that there are 6 types of coupons, and each box has a coupon. It can be of any type with equal probability.
Let X = the number of boxes you need open to get all 6 types of coupons.
Let's label the coupons 1,2,3,4,5,6 (like sides of a die :) )
X can be 6 (all 6 get different types) (1,3,4,5,2,6) or 7 (1,3,1,2,4,5,6) or 8 or 9.... onwards
For X to be at most 8, X has to be equal to 6 or 7 or 8, so you got that part right. However it's not a binomial distribution as we have 6 different outcomes instead of 2, and we don't have a specific "success" outcome for each time you open the box.
Let's see for X = 6, you can just arrange all coupon types in any order you like. There are 6! such combinations, and the probability of each is $(\frac{1}{6})^6$
Things are a little different for X = 7 and X = 8
Now if you break it down, the coupon in the last ($X^{th}$) box has to be unique, because it's the point where you have hit all types of coupons. You can't have had the same coupon in a box before that.
We know that only 1 coupon can repeat if X = 7, So we have 6 choices to choose the type of the last coupon, and we have 5 choices to choose the coupon to repeat. Let's say coupon type 5 is the last, and 4 is chosen to repeat. Then we could have:
1 2 4 4 3 6 5
We have 6! ways of arranging the first 5 coupons, but the 4's are identical so we have to divide by 2 as swapping the 4s will give us the same arrangement
So we have $\frac{6!}{2} \times 6 \times 5$ arrangements, each with probability $(\frac{1}{6})^7$
For X = 8, you know that either one type of coupon can repeat 3 times or 2 can repeat once each, e.g.
1 1 2 3 4 1 6 5
or
1 2 5 6 2 5 3 4
For case 1, you have 6 choices for the last coupon, 5 to choose the one that repeats 3 times, and $\frac{7!}{3!}$ ways to arrange the 7 coupons before the last one.
For the second case, you have 6 choices for the last coupon, $\binom{5}{2}$ ways to choose the two types that repeat once once, , and $\frac{7!}{2! \times 2! }$ to arrange them
So you have $6 \times 5 \times \frac{7!}{3!} \times (\frac{1}{6})^8 + 6 \times \binom{5}{2} \times \frac{7!}{2! \times 2!} \times (\frac{1}{6})^8$ for X = 8
You can add the probabilities for X = 6, 7, 8 to get the total.
Your intuition was slightly right, that you might need to go into balls and bins, because in effect you're trying to count "how can I assign n-1 boxes (distinguishable balls) to 5 coupons (distinguishable bins) so that each coupon has a box"? (n-1 because the last box will have a fixed coupon type that you choose). The problem here is that the "balls" or boxes are not identical so their order matters and the problem is harder to solve that way.