# Probability of Poker Hands with Joker

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!

• rank as in like king, queen, etc.? – Chan Hunt Jun 9 '15 at 18:12
• Yes, sorry for confusion! – HCL Jun 9 '15 at 18:14
• I think there is some added complication involving how the Joker's value is chosen. For example, if I have a 10,J,Q,K,JOKER hand of all the same suit, surely I would choose the joker to have a value of A with the same suit. Is this a hand with a joker that does not contain a pair? Or are we counting hands that potentially have a pair? – Jonny Jun 9 '15 at 18:38
• @Jonny; not sure which question in particular you are referring to, but I guess that hand would contain a pair. – HCL Jun 9 '15 at 18:59
• "At least one pair" as in "one or more pairs" or "any hand except high card"? – Hagen von Eitzen Jun 9 '15 at 19:02

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.

• That makes a little bit more sense, however I don't really know how to progress from there. I currently have (1/53) * C(13, 1) * C (12, 3) * C (4, 2) * C (4, 1)^3, which is basically 1/53 * the probability of finding a pair in a poker hand - which doesn't seem correct. – HCL Jun 9 '15 at 18:29
• First you have to figure out the probability of drawing the joker correctly - it's not 1/53. – Sam Clearman Jun 9 '15 at 20:06

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:

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

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:

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

• If you explain the difference between pair and two cards of the same value, then I might be able to answer the second question. – barak manos Jun 9 '15 at 18:50
• They're the same, I think the wording is to accommodate for the fact that a pair can consist of any card and a joker as well. Thank you so much! – HCL Jun 9 '15 at 18:57
• @HCL: In that case, my answer above is wrong. Why does a joker make a pair with any other card? – barak manos Jun 9 '15 at 19:02
• @HCL: Updated the answer according to the definition of the joker as a card that can be used as anything. – barak manos Jun 9 '15 at 19:09

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