Probability of creating two same groups

Question

There are 10 different types of stickers. We have a collection of 20 stickers, where each sticker appears twice (each and every sticker has a duplicate). Now we are asked to put 10 stickers in two bags (each bag contains exactly 10 stickers), when there are no same stickers in every bag, so they are distinct. What is the probability of this arrangement?

My attempt was to think about ways to choose an $x$ which will suit the general formula $\cfrac{ x}{{20 \choose 10}}$ but with no success. I also thought about calculating the probability using the intuitive method of choosing the first bag:

$\cfrac{2}{{20}}⋅\cfrac{2}{{19}}⋅\cfrac{2}{{18}}⋅\cfrac{2}{{17}}⋅\cfrac{2}{{16}}⋅\cfrac{2}{{15}}⋅\cfrac{2}{{14}}⋅\cfrac{2}{{13}}⋅\cfrac{2}{{12}}⋅\cfrac{2}{{11}}$

When we have 10 iterations of probabilities of choosing one of 2 different stickers out of what we have at the moment of choosing. But I was wrong. The book says that the probability is just: $$\cfrac{ 2^{10}}{{20 \choose 10}}$$

Which looks like a combination of my two attempts. I still don't get the point. It seems like a relatively easy question, but I find it confusing.

Answer 1

In the book's answer, each of the $10$ distinct types of stickers has $2$ choices for bags, which explains $2^{10}$ of the numerator

If you want to use an intuitive method, it should be

$\cfrac{20}{{20}}⋅\cfrac{18}{{19}}⋅\cfrac{16}{{18}}⋅\cfrac{14}{{17}}⋅\cfrac{12}{{16}}⋅\cfrac{10}{{15}}⋅\cfrac{8}{{14}}⋅\cfrac{6}{{13}}⋅\cfrac{4}{{12}}⋅\cfrac{2}{{11}}$

The numerator skips types of stickers already chosen.

Answer 2

Yes the answer is $ \ \displaystyle \cfrac{2^{10}}{{20 \choose 10}}$

As there are $2$ ways to choose one of the two stamps of each type and there are $10$ types of stamps, which leads to $ \displaystyle 2^{10}$ favorable outcomes.

Coming to your working,

$\cfrac{2}{{20}}⋅\cfrac{2}{{19}}⋅\cfrac{2}{{18}}⋅\cfrac{2}{{17}}⋅\cfrac{2}{{16}}⋅\cfrac{2}{{15}}⋅\cfrac{2}{{14}}⋅\cfrac{2}{{13}}⋅\cfrac{2}{{12}}⋅\cfrac{2}{{11}}$

When you says $\displaystyle \cfrac{2}{20}$, that is choosing $1$ out of $2$ stamps and $1$ out of total $20$ stamps but note there are also $10$ ways to choose the type of stamp for the first stamp. Then after you have chosen the first stamp, you next have to choose a stamp from the remaining $9$ types and so on.

So it should really be,

$\small \cfrac{2 \cdot 10}{{20}}⋅\cfrac{2 \cdot 9}{{19}}⋅\cfrac{2 \cdot 8}{{18}}⋅\cfrac{2 \cdot 7}{{17}}⋅\cfrac{2 \cdot 6}{{16}}⋅\cfrac{2 \cdot 5}{{15}}⋅\cfrac{2 \cdot 4}{{14}}⋅\cfrac{2 \cdot 3}{{13}}⋅\cfrac{2 \cdot 2}{{12}}⋅\cfrac{2 \cdot 1}{{11}}$

Answer 3

It seems to me that the book answer is correct. I will leave it to you to propose a combinatorial argument for that answer.

Below is a simulation in which st.20 contains the specified 20 stickers. Then, at each iteration ten stickers are taken at random without replacement from among the 20 and put into the first bag. The number of uniquely different stickers in that bag is determined.

The goal is for that number to be 10. The proportion of the million iterations with 10 uniquely different stickers is $0.0055 \pm 0.00015,$ which is in agreement with $2^{10}/{20 \choose 10}.$

set.seed(2021)
st.20 = rep(1:10, 2)
x  = replicate(10^6, length(unique(sample(st.20, 10))))
mean(x==10)
[1] 0.005478       # aprx 0.00554
2*sd(x==10)/1000
[1] 0.0001476211   # aprx 95% marg of sim error
2^10/choose(20,10)
[1] 0.005542445
summary(x)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   5.00    7.00    8.00    7.63    8.00   10.00
hist(x, prob=T, br=(4:10)+.5, col="skyblue2")

Answer 4

Think that there are 10 distinct stamps named $A$ to $G$ . We choose one from each of the two copies of each type of stamp. This is the event. Randomly you can choose any 10 out of the 20 stamps. This forms the sample space. $$ P =\dfrac {\binom {2}{1} \times \binom {2}{1} \times \binom {2}{1}… \binom {2}{1}}{\binom {20}{10}}= \dfrac {2^{10}}{\binom {20}{10}} $$^