Probability - Average number to get all toys

So I've been having some massive trouble with this problem and after bashing my head and doing all else possible, I came here.

Question: There are 6 toys included in a happy meal. Each is equally likely to be in the happy meal. Find the # of happy meals that you would need to buy to get all 6 toys **ON AVERAGE**.

Ignore the bold at your peril.

Re: Probability - Average number to get all toys

Quote:

Originally Posted by

**Zystrophys** Question: There are 6 toys included in a happy meal. Each is equally likely to be in the happy meal. Find the # of happy meals that you would need to buy to get all 6 toys **ON AVERAGE**.

This is a famous question. It is known as the coupon problem.

Look at the **Calculating the expectation** section for the answer.

Re: Probability - Average number to get all toys

Thanks.

I don't know what my teacher was thinking, giving a problem like this to high school students. Just a bit too hard for us to do.

Re: Probability - Average number to get all toys

Quote:

Originally Posted by

**Zystrophys** I don't know what my teacher was thinking, giving a problem like this to high school students. Just a bit too hard for us to do.

It is relatively easy for you to calculate the answer:

$\displaystyle 6\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac {1}{5}+\frac{1}{6}\right)=14.7~.$

However, the logic behind the method is quite advanced as you saw.

Re: Probability - Average number to get all toys

The logic was hard, not the arithmetic. Especially if my teacher expected us to somehow recreate it in class.

Thanks Plato.