Probability of Poker Hands with Joker

Question

Need help with a homework question:

If a five card hand from a standard deck of 52 with an added joker (wildcard) is drawn:

• What is the probability that a hand contains at least one pair?
• What is the probability that a hand does not contain at least two cards of the same value?
• What is the probability that a hand contains x number of cards of the same suit?
• What is the probability that a hand contains a full house? (Three cards of one value, and one pair)
• What is the probability that a hand contains a five-of-a-kind? (four of anything and a joker)

I understand how to answer these for a standard deck, however am getting very confused by the addition of a wildcard. I assume the answer to the second question would be the answer to the first question subtracted from 1.

Thanks for any help!

Answer 1

Try breaking the problem into the case where you get the joker, and the case where you don't (this is then identical to the "joker free" problem that you already understand).

For example, if you've drawn the joker, you have a 100% chance of having at least one pair.

Answer 2

Answering the first question:

What is the probability that a hand contains at least one pair?

Calculate $1$ minus the probability of the complementary event:

• The total number of ways to choose $5$ out of $53$ cards is $\binom{53}{5}$
• The number of ways to choose $5$ different cards excluding the joker is $\binom{13}{5}\cdot4^5$

So the probability that a hand contains at least one pair is:

$$1-\frac{\binom{13}{5}\cdot4^5}{\binom{53}{5}}\approx54.07\%$$

---

Answering the second question:

What is the probability that a hand does not contain at least two cards of the same value?

• The total number of ways to choose $5$ out of $53$ cards is $\binom{53}{5}$
• The number of ways to choose $5$ different cards excluding the joker is $\binom{13}{5}\cdot4^5$
• The number of ways to choose $5$ different cards including the joker is $\binom{13}{4}\cdot4^4$

So the probability that a hand does not contain at least two cards of the same value is:

$$\frac{\binom{13}{5}\cdot4^5+\binom{13}{4}\cdot4^4}{\binom{53}{5}}\approx52.30\%$$

Answer 3

One pair
With joker have 5 cards to make a pair

$$\frac{\binom{13}{1}\binom{5}{2}\binom{12}{3}\binom{3}{1}^4}{\binom{53}{3}}\approx63.784\%$$

But I am not sure about that. The joker can act like multiple cards so it is confusing. You also can make a lot of other hands like straight and flush. To do this correctly would be very hard.

Five of a kind

$$\frac{\binom{13}{1}} {\binom{53}{3}} \approx0.000453\%$$

Four of kind

$$\frac{\binom{13}{1}\binom{5}{4}\binom{12}{1}\binom{4}{1}^4}{\binom{53}{3}}\approx0.1087\%$$

I found a table and according to that the last two are correct