**Poker Digest Vol. 4, No. 18, August 24 - September 6, 2001**

After receiving a second e-mail message requesting information on the number of poker hands when a joker is introduced to the deck, and seeing a message on RGP about the number of hands when two jokers are introduced to the deck, I decided to look at counting poker hands with jokers. Let's first look at the number of poker hands with one joker in the deck, where the joker may be used as an ace or as a wild card for straights and flushes.

People sometimes become confused when trying to count poker hands with a joker available because they may omit some possibilities and double count others. I am advocating the approach we are about to use because it is based on a sound mathematical identity, helps keep the bookkeeping straight, and provides a template for the interested reader to handle similar problems on her or his own.

The total number of ways of choosing five cards from fifty-three is
*C*(53,5) = 2,869,685. A well-known mathematical identity for binomial
coefficients tells us that
*C*(53,5) = *C*(52,5) + *C*(52,4). We can interpret
this identity for our problem by noting that five-card hands chosen from a
53-card deck may be partitioned into those that do not contain the joker and
those that do. There are *C*(52,5) five-card hands that do not contain
the joker and *C*(52,4) five-card hands that do contain the joker because
once the joker has been chosen, there are four choices from 52 for the four
cards needed to fill in the hand.

The above identity strongly suggests how to do the counting. First, we use the numbers of types of poker hands for a 52-card deck. Then we use the numbers of four-card poker hands from a 52-card deck. We add the joker to each of the four-card poker hands and tally the kinds of five-card hands we get. Following are tables for the numbers of five-card and four-card poker hands for a standard 52-card deck. These tables are available at my web site: http://www.math.sfu.ca/ alspach.

hand | number |

straight flush | 40 |

four-of-a-kind | 624 |

full house | 3,744 |

flush | 5,108 |

straight | 10,200 |

three-of-a-kind | 54,912 |

two pair | 123,552 |

pair | 1,098,240 |

high card | 1,302,540 |

FIVE-CARD HANDS, 52-CARD DECK

hand | number |

four-of-a-kind | 13 |

straight flush | 44 |

three-of-a-kind | 2,496 |

straight | 2,772 |

two pair | 2,808 |

flush | 2,816 |

pair | 82,368 |

high card | 177,408 |

FOUR-CARD HANDS, 52-CARD DECK

Let's work through the four-card table and determine the effect of adding a joker to each of the hands. There are 13 ways of getting quads. Adding the joker to a hand of four aces produces a new type of 5-card hand, namely, five aces. Adding the joker to any of the other 12 quads produces four-of-a-kind with the joker becoming an ace kicker. So we have a new category of hands produced and we must add 12 to the number of four-of-a-kind hands in the five-card hand table above.

There are 44 four-card straight flushes. Adding the joker to any of them produces a five-card straight flush. So we add 44 to the straight flush entry in the above table.

There are 2,496 three-of-a-kind hands in the 4-card hand table. Each of the 13 ranks occurs the same number of times as trips. So we divide by 13 to obtain 192 four-card hands with trip aces. Adding a joker to these 192 hands yields quad aces. For the remaining 2,304 three-of-a-kind hands, an ace will be the kicker equally often, that is, one-twelfth of the time. Dividing by twelve again produces 192. So these 192 hands become full houses upon adding the joker. The remaining 2,112 hands are still three-of-a-kind after adding the joker which is simply an ace.

There are no more possibilities that can produce quads by adding a joker. We saw above that we get 192 + 12 quad aces upon adding a joker to various 4-card hands. Thus, we add 204 to the appropriate entry of the above table.

There are 2,772 four-card straights. Adding a joker to any of these produces 2,772 five-card straights.

There are 2,808 two-pair, four-card hands. For each pair of distinct ranks, there are 36 two-pair hands. There are 12 pairs of ranks involving an ace. So we have 432 two-pair hands involving an ace. When a joker is added to such a hand, it becomes a full house. When a joker is added to the other 2,376 hands they remain two-pair hands.

We now have completed the possible ways of producing new full houses. There are 432 + 192 such full houses.

There are 2,816 four-card flushes. We certainly obtain a flush upon adding a joker to any of these hands, but some of them produce straight flushes. This happens if there is a gap of only one rank in the four ranks of the flush. It is easy to see there are thirty such sets of ranks. Thus, we get 120 new straight flushes and 2,696 new flushes.

There are 82,368 four-card hands with a pair. Each rank occurs as a pair the same number of times so that there are 6,336 four-card hands with a pair of aces. Adding a joker to such a hand yields trip aces. This is the last way we can get new three-of-a-kind hands so that we have 6,336 + 2,112 of them.

There are 76,032 4-card hands with a pair of rank different
from ace. Each hand has two other ranks giving
*C*(12,2) = 66 choices for
ranks. An ace occurs in 11 of the pairs of ranks so that an ace occurs
as a singleton in 12,672 hands. Adding a joker gives a two-pair hand.
The remaining 63,360 hands remain as a single pair.

There are no more ways to get two-pair hands by adding a joker. So we add 12,672 + 2,376 to the 123,552 above.

There are 177,408 high-card four-card hands. Adding a joker to such a hand may produce a straight, a pair of aces or an ace-high hand. As mentioned earlier, there are 30 rank sets with a single gap. Each rank set allows any of four choices for each rank yielding 256 choices. However, four of the choices result in a four-card flush so these must be removed yielding 252 four-card hands for each rank set. Multiplying 30 by 252 gives 7,560 four-card hands producing a straight when the joker is added.

There are
*C*(12,3) = 220 rank sets of cardinality four containing an ace.
Both {A,2,3,4}
and {J,Q,K,A} are eliminated since they correspond to four-card straights,
and six more are eliminated because they have a single gap and produce
straights when the joker is added. This leaves 212 rank sets with an
ace. Multiplying by 252 produces 53,424 four-card high-card hands with an
ace. Adding a joker gives a pair of aces. The remaining 116,424 hands
become ace-high with a joker.

The next table is the number of five-card hands with one joker in the deck. where the joker may be used as an ace or as a wild card for straights and flushes.

hand | number |

five aces | 1 |

straight flush | 204 |

four-of-a-kind | 828 |

full house | 4,368 |

flush | 7,804 |

straight | 20,532 |

three-of-a-kind | 63,360 |

two pair | 138,600 |

pair | 1,215,024 |

high card | 1,418,964 |

FIVE-CARD HANDS, 52-CARD DECK PLUS ONE JOKER

Now suppose you want the number of poker hands with two jokers added to the deck, where the jokers are restricted as above. We have seen what numbers must be added to the five-card hand, 52-card deck table, for a single joker. For each joker separately, you add the same numbers thereby doubling the increases. Then you consult the table of three-card poker hands and see what effect adding two jokers has on the numbers in the table. This is not difficult, as three-card hands are pretty straightforward.

You can also modify the four-card table to determine the number of poker available when the joker is wild.

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