# Combinatorics and statistic question (pick all 1s in a hand of 9 cards)

I was going through statistics classes on Khan academy and there was this exercise I was hoping someone could help me with:

1. we have $$36$$ unique cards (normal deck, $$4$$ types - hearts, diamonds etc.). And $$9$$ different groups of cards (from $$1$$ to $$9$$).
2. a hand is a selection of $$9$$ cards which can be sorted any way you want

question: what is the probability of getting all $$4$$ $$1$$s in a hand? (so $$1$$ of diamond, $$1$$ of hearts etc, $$4$$ total) link:https://youtu.be/ccrYD6iX_SY

I'm here because I disagree with the Sal's solution and I was hoping someone could tell me where I could be wrong:

1. we are looking for a probability, and $$\mathrm{(prob\ of\ needed \ result)\ = \frac{\# (needed \ result\ can\ happen) }{ \# (total \ results)}}$$

2. total results are ok and it is the same in the video

(in python)

n_total_outcomes = scipy.math.factorial(36)/(scipy.math.factorial(9)*(scipy.math.factorial(36-9)))

1. but the desired result is where it gets hairy (for me).

2. Next I decided not to use any special formulas and apply logic: we can get Lucky ~ if we have already picked $$\frac 59$$ random cards, we have $$31$$ cards left, and the chance we pick $$1$$s higher ($$\frac 1{31}, \frac 1{30}, \frac 1{29}, \frac 1{28}$$) and unlucky or less probable if we start picking the $$1$$s at the beginning $$\frac {1}{36}, \frac{1}{35}, \frac{1}{34}, \frac{1}{33}$$.

So I just imagined that these are $$2$$ side cases ($$7\times 10^{-7}$$ and $$1 \times 10^{-6}$$), which is still nowhere close to Sal's solution, where he has very high chance. If you watch his video my beef is that he is picking all four 1s at the beginning and then picking up the rest of the hand. It is nowhere guaranteed you will get all four 1s right away.

If anyone could show me the problem with my logic, it would be much appreciated.

• First of all, thank you for posting a well-written question. I will show you how to write fractions by editing the post, but everything else is nice.Please see the edits now. – Teresa Lisbon Jan 27 at 11:50
• But there are so many ways to pick those $4$ cards - it could be first $4$, it could be last $4$, it could be in any order and any of the $4$ out of $9$. So one way to look at it is $\frac{4C4 \times 32C5}{36C9}$. It comes to same as the video you shared. Does that help? – Math Lover Jan 27 at 12:53
• Thank you for your answer. I'm going to go over the topic again. I should have asked differently (your answer is enough): if we deal this 36 cards deck a million times, how many times approximately we will end up with a hand where we have all 4 1s. – Oleg Peregudov Jan 28 at 9:35

it appears that you are thinking he is studying the process of selection. However, this is not the case; he is just considering the final hand consisting of $$4$$ 1s and $$5$$ other cards.
On the other hand you are attempting to consider the process. The way you are doing this would involve you in considering all the different ways the $$4$$ 1s are selected which you have not done.
• The correct probability of $2/935$ is roughly $2$ per $1000$. So it would occur roughly $2000$ times on average. – S. Dolan Jan 28 at 9:47