### The Price of Good Parenting

I saw Kinder eggs at the grocery store the other day. I see them every week, actually, but the ones I saw last week stood out because they came in a box of three and had a Hot Wheels label on it (Hot Wheels being one of my favourite childhood toys). The printing on the box explained that there are six different Hot Wheels toys in and each box of three Kinder eggs contains exactly one with a hot wheels toy in it. Kinder obviously wants you to collect all 6, and just in case you weren't sure of that, it also says "collect all 6" on the box.

Suppose you set out to collect them all (or rather, your child convinced you to let him set out to collect all 6). How many boxes would you have to buy, on average, before you collected all 6?

Before we answer this question, let's consider a different question. What is the probability of getting all 6 different cars in the first 6 boxes? To figure this out, we need to know two things, the number of different ways of getting all 6 in 6 boxes and the number of possible outcomes from getting 6 copies of the exact same car to getting all 6 different cars. Dividing the first number by the second gives the probability.

Let's assume that you open the boxes one at a time and that the cars are distributed evenly across the boxes. Suppose that the different cars are numbered 1 to 6 and that as you open the car-containing kinder egg, you write down the number of the car.

If you win all 6 in the first 6, then you have written down the numbers 1 through 6 in some order. How many different ways are there for this to happen? Well, there are 6 possibilities for the first number you write down. Having written down one number already, there are only 5 possibilities for the second number. Similarly, there are only 4 for the third number, 3 for the fourth, 2 for the fifth and 1 for the 6th. If we multiply these numbers together, we get the total number of ways that we could win 6 cars. So the number of ways to win is 6 x 5 x 4 x 3 x 2 x 1 = 720 ways.

Now let's consider the total number of possible outcomes, whether or not you got all 6. There are 6 possibilities for the first box, 6 for the second box, and so forth. To get the total number, then, we just multiply 6 by itself 6 times (or, equivalently, take 6 to the power of 6), to get 6 x 6 x 6 x 6 x 6 x 6 = 46656 possibilities.

The probability of getting all 6 cars in the first 6 boxes is then 720/46656=0.01543=1.5%. Those are better odds than Lotto 6/49, but still not very high. That doesn't help us answer the question of how many boxes you will have to buy, but it does tell us that you most likely will not have all 6 cars from your first six boxes and so you will have to buy more.

Since the cars are distributed evenly across the boxes, the probability of getting any one car is 1 in 6. Thus, if you have already won k cars, then the chance of getting a car you already have in the next box is k/6 and the chance of getting a car you don't have is (6-k)/6.

To determine the average number of boxes needed, we'll start from the end, assuming you only need one more car to complete the set, and work backwards to the case where you only have one car.

If you have already got 5 cars, then there is a 5 in 6 chance that complete the set and 1 in 6 chance that you have to buy another box. This is essentially the same situation is the Roll Up the Rim where there is a 5 in 6 chance of losing and a 1 in 6 chance of winning. There, we concluded that it took an average of 6 tries before the first win. Using the same argument, we conclude that it will take an average of 6 more boxes before getting that last car to complete the set.

If you've got 4, there is a 4 in 6 chance of getting one you already have and 2 in six chance of getting one you don't. By the same reasoning we used above for when you have 5, we can argue that it will take an average of 6/2=3 more boxes before you get the next new car. Then, according to the previous paragraph, it will take 6 more boxes until the set is complete. Thus, if you have 4 cars, it will take an average of 3+6=9 more boxes to complete the set.

Similar arguments can be used if you've got 3, 2, or 1 distinct cars, the only difference being in the probabilities of finding a car you have already or one you don't. If you add up all the numbers from each case, you get 14.7. This is the average number of boxes you need to buy before you get all 6 cars, answering the original question.

Each box costs $3.49. So on average, you are spending 14.7 x $3.49=$51.3. If you are in Ontario, you are paying 13% tax, so the total cost would be $57.97. You are probably better off if you just go out and buy a set of 6 Hot Wheels and a chocolate bar. Both the the cars and the chocolate will be higher quality, and you don't have to worry about dealing with kids who are hopped up on the sugar from 44 Kinder Eggs.

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