Suppose that after a five-card hand is drawn, the cards in it are put back in the deck and another five-card hand is drawn.

Suppose that after a five-card hand is drawn, the cards in it are put back in the deck and another five-card hand is drawn.

a) What is the probability that the two hands have no card in common?

b) What is the probability that the two hands have at least three cards in common?

I am struggling to get answers for this question since replacement is needed.

• Presumably, the cards are put back in the deck in some fashion and/or the deck is reshuffled? – Thomas Andrews Feb 12 '16 at 22:58
• I assumed shuffled? This is only information I am given besides the presiding information which is:"five-card hands drawn at random, all at once, from the deck." – probability Feb 12 '16 at 23:16

Let our sample space be the set of all possible five-card hands where order is unimportant. These will represent our possible second hands.

Suppose we pick our first hand. Whatever our first hand happens to be, it will use five of the available 52 cards. The probability then that the second time we take a hand of five cards that none of the same cards appear again will be:

$$\frac{\binom{47}{5}}{\binom{52}{5}}$$

This can be seen by the fact that regardless what the first hand was, five of the cards will be considered "bad" or "unwanted", leaving 47 good cards remaining. The number of hands which use only good cards will be $\binom{47}{5}$. Dividing by the size of the sample space gives the probability.

For having at least three cards in common, it is easier to ask the question of "exactly three cards in common or exactly four cards in common or exactly five cards in common."

To do this, count how many hands satisfy this by:

• Pick what the cards common to both in the second hand are
• Pick what the cards in the second hand are which are not ones from the first hand

Apply multiplication principle and addition principle of counting, and then divide by the samplespace size to get the final answer.

The number of second hands with exactly $i$ cards in common with the first with $0\leq i\leq 5$ will be $\binom{5}{i}\binom{47}{5-i}$