Playing poker with five suits allows for an interesting new lineup of poker hands, including five of a kind, and a “flash.” A flash is the opposite of a flush. In a flush, all five cards are of the same suit. In a flash, each card is of a different suit. Here is the list of hands for five suit poker in order from highest to lowest, including the number of possible ways each hand can be dealt, and the probability of being dealt that hand.

## Five Suit Poker Probabilities

Hand | Number possible | Probability |
---|---|---|

1. Five of a kind | 13 | 0.0001574% |

2. Straight flush | 50 | 0.0006053% |

3. Flash four of a kind | 780 | 0.0094432% |

4. Straight flash | 1,200 | 0.014528% |

5. Flash full house | 1,560 | 0.0188865% |

6. Four of a kind | 3,120 | 0.0377729% |

7. Flush | 6,385 | 0.0773013% |

8. Full house | 14,040 | 0.1699781% |

9. Flash three of a kind | 17,160 | 0.207751% |

10. Flash two pair | 25,740 | 0.3116265% |

11. Straight | 30,000 | 0.3632011% |

12. Flash high card | 153,240 | 1.855231% |

13. Flash pair | 171,600 | 2.07751% |

14. Three of a kind | 197,340 | 2.3891365% |

15. Two pair | 403,260 | 4.8821485% |

16. Pair | 3,403,400 | 41.2039485% |

17. High card | 3,831,000 | 46.3807742% |

All hands | 8,259,888 | 100% |

Some interesting things to note:

- Unlike four suit poker, in the five suit game, a flush beats a full house.
- Although it may seem counterintuitive, a flash without a pair beats a flash with a pair.

You may also play without the “flash” hands, in which case the only changes from four suit poker are the addition of five of a kind, and the fact that a flush beats a full house.

## How Poker Probabilities are Calculated

Calculating the probabilities of poker hands involves factorials and combinations.

Factorials are the number of ways of arranging n objects in an ordered sequence. A factorial is denoted with *n*! and is the number multiplied by every number less than it.

### How many ways the cards in a deck can be arranged

To figure the number of ways the cards in a 65-card deck can be arranged in sequence, we would multiply 65 x 64 x 63 … x 1. The product is 8.247650592 times 10 to the 90th power. It is more than a vigintillion (10 to the 63rd power) but less than a centillion (10 to the 303rd power), and it is more than the number of particles in the universe.

### Total number of possible five-card hands

To calculate the total number of possible five-card hands that could be dealt from a five-suit 65-card deck, we use a combination. The order that the five cards are dealt does not matter, which is why it is a combination and not a permutation.

Combinations may be denoted as C (*n*, *r*) representing the number of ways that *r* objects can be taken from *n*. The formula for C (*n*, *r*) is *n*! / *r*!(*n*–*r*)! so for C (65, 5) we have to divide 65! by 5! times 60! and using a combination calculator, we arrive at 8,259,888.

Now that we know there are 8,259,888 different five-card hands that can be dealt from a 65-card five-suit deck, we must calculate how many of them are represented by each of the poker hands.

### Five of a kind

We can calculate the number of five of a kind hands in our heads. There is only one way to make five of a kind for each of the 13 card values, so there are only 13 possible five of a kind hands. To calculate the probability, we divide 13 by 8,259,888, and our result is 0.000001574 or odds of 0.0001574%.

### Straight flush

The number of straight flush hands is also easy to figure. The ace can be high or low, so for each suit, there are 10 possible straights, from A-2-3-4-5 to 10-J-Q-K-A. Therefore in a five suit deck, there are 50 possible straight flushes. Fifty divided by 8,259,888 is 0.000006053 or 0.0006053%, which is our probability.

### Flash four of a kind

A flash four of a kind has four cards with matching ranks, and all five cards in the hand are of different suits. To determine the number of these hands that is possible, we have to use several combinations. First, we choose one rank out of 13 for the four matching cards. This is C (13, 1), which is 13. Then we select a rank for the non-matching card, using C (12, 1), or 12. For the four matching cards, we need to choose four out of the five suits. That is C (5, 4), or simply 5. Finally, the suit of the non-matching card must be of the only remaining suit, so that is C (1, 1), which is 1. Our final equation is 13 x 12 x 5 x 1 = 780, so there are 780 possible flash four of a kind hands. Dividing this number by 8,259,888 gives us 0.000094432 or odds of 0.0094432%.

### Straight flash

A straight flash has all five cards in order by rank, and each card is of a different suit. We already know that there are 10 possible ranks for the lowest card of a straight: ace through 10. To select this beginning card of the straight, we use C (10, 1). For the suits, we will choose the suit of each card separately, with the number of available suits reduced by one with each succeeding card. So our formula is C (10, 1) x C (5, 1) x C (4, 1) x C (3, 1) x C (2, 1) x C (1, 1). This is simply 10 x 5 x 4 x 3 x 2 x 1 = 1,200. So there are 1,200 possible straight flash hands, and dividing by 8,259,888 total possible hands, we get 0.00014528, or a 0.014528% chance.

### Flash full house

A flash full house is made up of three cards of matching rank and two cards of a different matching rank, with each card being of a different suit. To figure the number of possible hands, we first choose the two ranks out of thirteen that will make up the hand. C (13, 2). Next, from the two ranks we have chosen, we select one for the three-card combination. C (2, 1). Now we choose three suits for the three matching cards, out of five available. C (5, 3). Finally we choose the two suits for the two matching cards, out of the remaining two. C (2, 2). Our formula is C (13, 2) x C (2, 1) x C (5, 3) x C (2, 2). That is 78 x 2 x 10 x 1 = 1,560. That is the number of possible flash full house hands, and we divide it by 8,259,888 to arrive at 0.000188865 or 0.0188865% probability.

### Four of a kind

The formula for all possible four of a kind hands is the same as that for flash four of a kind, except the suit of the non-matching card may be any of the five suits. So the equation is C (13, 1) x C (12, 1) x C (5, 4) x C (5, 1). This is 13 x 12 x 5 x 5 = 3,900. From that we have to subtract the 780 flash four of a kind hands we already counted, which leaves 3,120 regular four of a kind hands. Dividing by 8,259,888, we get 0.000377729, which is a probability of 0.0377729%.

### Flush

To compute the total number of all flush hands possible, we choose five ranks from the 13 available, using C (13, 5). Then we choose one suit from the five available, using C (5, 1). So we use the formula C (13, 5) x C (5, 1). This is 1,287 x 5 = 6,435. From this number we must subtract the 50 straight flushes counted earlier, leaving us with 6,385 ordinary flush hands. Divide by 8,259,888 to arrive at 0.000773013, or a 0.0773013% chance.

### Full house

First, we choose the rank of cards in the hand. A full house has two ranks. So we choose two out of thirteen, which is C (13, 2). Next, we select a rank for the three-card combination. From the two ranks we have selected, we choose one. This is C (2, 1). We choose three different suits for the three-card combination, out of five available. C (5, 3). Finally we choose the two suits for the two-card combination, out of five available. C (5, 2). Our formula is C (13, 2) x C (2, 1) x C (5, 3) x C (5, 2). That is 78 x 2 x 10 x 10 = 15,600. From this number we must subtract the 1,560 flash full house hands we counted earlier, leaving 14,040 regular full house hands. Dividing by the total number of possible hands gives us 0.001699781, a probability of 0.1699781%.

### Flash three of a kind

A flash three of a kind has three cards of the same rank, and all five cards are of a different suit. First, we select the rank for the three matching cards out of 13. This is C (13, 1). Then we choose the rank of the remaining two cards, from the other 12 ranks: C (12, 2). Next we choose the suits for the three matching cards, from the five available. That is C (5, 3). Now the suits of the final two cards, out of the remaining suits: C (2, 1) and C (1, 1). This gives us our equation of C (13, 1) x C (12, 2) x C (5, 3) x C (2, 1) x C (1, 1), which is 13 x 66 x 10 x 2 x 1 = 17,160. Now we divide this number by 8,259,888 to get 0.00207751, which is a 0.207751% chance.

### Flash two pair

To find the probability of two pair with all five cards of different suits, we first choose the ranks of the two pairs, out of 13, using C (13, 2). The rank of the non-matching card is selected with C (11, 1). We choose two suits for the first pair out of five, and two suits for the second pair out of the three remaining, with C (5, 2) and C (3, 2). The suit of the non-matching card must be of the one remaining suit: C (1, 1). So we have C (13, 2) x C (11, 1) x C (5, 2) x C (5, 2) x C (1, 1), which is 78 x 11 x 10 x 3 x 1 = 25,740. So with 25,740 flash two pair hands possible, we use the divisor of 8,259,888 to obtain the quotient of 0.003116265. That is a probability of 0.3116265%.

### Straight

First we want to figure the total number of straight hands possible. We already know the combination for ranks is C (10, 1). Then we choose a suit for each card, so C (5, 1) five times. That is C (10, 1) x C (5, 1) x C (5, 1) x C (5, 1) x C (5, 1) x C (5, 1), or 10 x 5 x 5 x 5 x 5 x 5 = 31,250. From this total we must subtract the number of straight flushes (50) and straight flashes (1,200) we already counted. That leaves us with 30,000 ordinary straights. We perform our usual division to arrive at 0.003632011, or odds of 0.3632011%.

### Flash high card

One interesting result in five suit poker is that a flash with no pair beats a flash with a pair. To calculate the number of possible hands with a flash and no matching cards, first, we select five ranks among 13: C (13, 5). Then for each card we select among a diminishing number of suits: C (5, 1) x C (4, 1) x C (3, 1) x C (2, 1) x C (1, 1). Our equation is 1,287 x 5 x 4 x 3 x 2 x 1 = 154,440. That is the total number of flashes with all five ranks different. From this number, we must subtract the 1,200 straight flashes counted previously, to arrive at 153,240. We divide by the total number of possible hands as usual, to reach 0.01855231 or 1.855231% odds.

### Flash pair

To determine the number of hands containing one pair with all five cards of different suits, we first choose the rank of the matching pair, with C (13, 1). The three remaining cards are chosen from the 12 remaining ranks, using C (12, 3). We choose the two suits of the pair with C (5, 2). Then, each of the other three cards must be selected from a diminishing number of suits: C (3, 1) x C (2, 1) x C (1, 1). Putting it all together, we have 13 x 220 x 10 x 3 x 2 x 1 = 171,600. We divide by 8,259,888 to get 0.0207751, which is a probability of 2.07751%.

### Three of a kind

Next up, we have the regular three of a kind hand. We can use some of the same combinations as with flash three of a kind. First, we choose the rank for the three matching cards out of 13: C (13, 1). Next we choose the rank of the remaining two cards, from the other 12 ranks, using C (12, 2). Then we choose the suits for the three matching cards, out of five: C (5, 3). The two non-matching cards may be of any suit, so we use C (5, 1) x C (5, 1). Our formula is C (13, 1) x C (12, 2) x C (5, 3) x C (5, 1) x C (5, 1), which is 13 x 66 x 10 x 5 x 5 = 214,500. From this number we must subtract the 17,160 flash three of a kind hands previously accounted for, leaving 197,340. Dividing by 8,259,888 gives us 0.023891365, which is a 2.3891365% chance.

### Two pair

Begin by choosing two ranks from among 13, for the two pairs, using C (13, 2). The rank of the non-matching card must come from among the 11 remaining ranks: C (11, 1). Select two suits for the first pair and two suits for the second pair, using C (5, 2) x C (5, 2). The suit of the non-matching card is chosen with C (5, 1). So our equation is C (13, 2) x C (11, 1) x C (5, 2) x C (5, 2) x C (4, 1), or 78 x 11 x 10 x 10 x 5 = 429,000. From this number we must subtract the 25,740 flash two pair hands we counted before, to arrive at 403,260. We divide by 8,259,888 to get 0.048821485, or 4.8821485% probability.

### One pair

To determine the number of one pair hands, first choose the rank for the pair, using C (13, 1). The ranks of the three non-matching cards are selected from the 12 remaining ranks: C (12, 3). Choose two suits for the pair using C (5, 2). The suits of the three non-matching cards are selected using C (5, 1) x C (5, 1) x C (5, 1). This gives us the formula: C (13, 1) x C (12, 3) x C (5, 2) x C (5, 3), or 13 x 220 x 10 x 5 x 5 x 5 = 3,575,000. Next we subtract the previously calculated 171,600 flash pairs, leaving 3,403,400. Dividing by 8,295,888 gives us 0.412039485, or a 41.2039485% chance.

### High card

Finally, we want to calculate the number of “nothing” hands, where the high card in the hand is the best you’ve got. We select the five ranks out of 13 with C (13, 5). The suit of each card is selected using C (5, 1) x C (5, 1) x C (5, 1) x C (5, 1) x C (5, 1). That gives us 1,287 x 5 x 5 x 5 x 5 x 5 = 4,021,875. From this number we need to subtract the 153,240 flash high card hands and the 6,385 flush hands, as well as all the straights: 50 straight flushes, 1,200 straight flashes, and 30,000 regular straights, leaving 3,831,000. Using the total number of possible hands as a divisor, our quotient is 0.463807742, so the odds of a high card hand are 46.3807742%.

As a final check, you can add up all the hands and probabilities to be sure the right total is reached.