# Probability of 3 aces from drawing 7 cards. 1.58 Intro to Probability, 2nd Ed.

This is Problem 1.58 from Tsitsiklis, Bertsekas, Introduction to Probability, 2nd edition.

We draw the top 7 cards from a well-shuffled standard 52-card deck. Find the probability that: a) The 7 cards include exactly 3 aces. b) The 7 cards include exactly 2 kings. c) The probability that the 7 cards include exactly 3 aces, or exactly 2 kings, or both

My solution was

a) $${3! {49 \choose 4} \over {52 \choose 7} }$$

b) $${2! {50 \choose 5} \over {52 \choose 7} }$$

c) $${3! {49 \choose 4} \over {52 \choose 7} } + {2! {50 \choose 5} \over {52 \choose 7} } - {3! 2! {47 \choose 2} \over {52 \choose 7} }$$

The solution book says

book solution

Why do they $${4 \choose 3}$$ instead of (3!)?

We have 7 slots where we can put the cards with each slot accepting 1 card. We want to fill the first 3 with aces and the last 4 with non-aces. There are 3! ways to distribute aces into the first 3 slots.

And then there are $${49 \choose 4}$$ ways to fill the remaining slots.

There is another card problem where they use that same logic to solve the problem: question, answer

a) Divide all cards into two groups: a group of $$4$$ aces and a group of $$48$$ non-ace cards. From the first group you select any $$3$$ aces and from the second group you select any $$4$$ cards.
b) Again divide all cards into two groups: a group of $$4$$ kings and a group of $$48$$ non-king cards. From the first group you select any $$2$$ kings and from the second group you select any $$5$$ cards.
c) Use inclusion-exclusion principle. Adding a) and b) will double count the outcomes of exactly 3 aces and 2 kings, which must be subtracted once. To find the outcomes of exactly 3 aces and 2 kings, you divide all cards into $$3$$ groups: a group of $$4$$ aces, a group of $$4$$ kings and a group of $$44$$ non-ace and non-king cards. From the first group you select any $$3$$ aces, from the second you select any $$2$$ kings and from the last group you select any $$2$$ cards.
• The difference is in permutation/combination. In this problem, again divide all cards into two groups: $4$ aces and $48$ non-ace cards. First distribute the $4$ aces to $4$ players in $4!$ ways (This is ordered permutation. However in previous problem the order of selecting $3$ aces out of $4$ aces was not important). Then divide the rest $48$ non-ace cards to $4$ players, which you can distribute as follows: ${48\choose 12}{36\choose 12}{24\choose 12}{12\choose 12}$. Oct 18, 2018 at 2:51
You're not filling "the first three slots" with the aces, you're just deciding which three aces are among the first seven cards, and which four or the remaining $$48$$ non-aces fill out the first seven. Note that it's $$48$$, not $$49$$ as in your solution, since you can't allow the fourth ace.