# Chances of 5 way chop (flush board)

What is the exact probability that 5 players at a table will be dealt no clubs (or hearts or whatever) ?

Happened to me when all 5 community were all clubs (2,3,5,K,9), thereby a 5 way chop.

• I think intuition tells you/us it's a very very small chance. Maybe 1 in 500 thousand or something like that. I don't think the exact number really matters. – Radu Murzea Dec 3 '13 at 7:31
• you want to have a flush board at the same time? or just no clubs dealt to any person of the 5? thanks i will improve my answer when you tell me what you need. – RayofCommand Dec 3 '13 at 11:33
• @RayofCommand: I believe the question is, how often will the 5 club flush "board play" resulting in a chop, because no one has a club higher than the 2? – Tom Au Dec 6 '13 at 16:53

TL;DR Given a flush on the board, five players will split 19.7% of the time, or about once every 5 hands. Given a fresh deal, the deal will come up with a flush showing and a five-way split 0.0389% of the time, or once every 2,569 hands.

## Some general probabilities

Overall odds that none of the five players have a given suit (say, hearts):

``````comb(39,10) / comb(52,10)
= 39/52 * 38/51 * 37/50 ... 30/43 (equivalent way to do this)
= 0.040186 (about 1 in 25 deals)
``````

Overall odds that none of the five players have a given suit (say, hearts) and the board also has a flush of that suit (i.e., when you don't know whether the board will turn up with a flush):

``````(comb(13,5) / comb(52,5)) * (comb(39,10) / comb(52,10))
= 5/13 * 4/12 ... 1/9 * 39/52 * 38/51 * 37/50 ... 30/43
= 0.0000199 (about one in 50,251 total hands)
``````

Overall odds that the board has a flush in any suit, but none of the five players have a card from that suit:

``````4 * (comb(13,5) / comb(52,5)) * (comb(39,10) / comb(52,10))
= 4 * 5/13 * 4/12 ... 1/9 * 39/52 * 38/51 * 37/50 ... 30/43
= 4 * 0.0000199
= 0.0000796 (about one in 12,562 total hands)
``````

## Some background math on how to calculate the odds of various flush-showing scenarios

Odds that none of the five players have a given suit (say, hearts) when you already know that the board also has a flush of that suit:

``````(comb(39,10) / comb(47,10))
= 39/47 * 38/46 * 37/45 ... 30/48
= 0.122777 (about one in 8 hands when there is a flush on the board)
``````

This seems like it should be our answer, but we have to do some more math in second major section below. For example, it's possible to have a flush on the board and have one or more players have a card in that suit, but still chop because their card(s) are too low to play.

## Okay, so what are the odds of an actual split when you know that the board is flush?

Figuring out the odds of a split pot given a flush on the board is much trickier. For example, if the board is A♥K♥Q♥J♥9♥, then the question is: what are the odds none of the players has the T♥? Obviously, any player with T♥ would win with a straight flush. Likewise, if the board is A♥K♥Q♥J♥5♥, then there's still a chop even if someone has the 4♥, but not if they have the 6♥.

Here's how we calculate this:

``````If the board has a 10-low flush (which would be a straight flush), odds of a split = 1.00000
If the board has a 9-low flush, odds of a split = comb(46,10)/comb(47,10) (46 = number of cards that are not showing and not the 10 of that suit) = .78723
If the board has a 8-low flush, odds of a split = comb(45,10)/comb(47,10) (45 = number of cards that are not showing and not higher than the 8) = .61609
If the board has a 7-low flush, odds of a split = comb(44,10)/comb(47,10) (44 = number of cards that are not showing and not higher than the 7 of that suit) = 0.47919
If the board has a 6-low flush, odds of a split = comb(43,10)/comb(47,10) (43 = number of cards that are not showing and not higher than the 6 of that suit) = 0.37028
If the board has a 5-low flush, odds of a split = comb(42,10)/comb(47,10) (42 = number of cards that are not showing and not higher than the 5 of that suit) = 0.28417
If the board has a 4-low flush, odds of a split = comb(41,10)/comb(47,10) (41 = number of cards that are not showing and not higher than the 4 of that suit) = 0.21651
If the board has a 3-low flush, odds of a split = comb(40,10)/comb(47,10) (40 = number of cards that are not showing and not higher than the 3 of that suit) = 0.16370
If the board has a 2-low flush, odds of a split = comb(39,10)/comb(47,10) (39 = number of cards that are not in that suit) = 0.12278
``````

We need to know the odds of each of these outcomes. For example, if we know the board has a flush, what are the odds it's a 10-low (straight) flush? A 9-low flush? etc.

``````Odds 10-low = comb(4,4)/comb(13,5) = 0.000777000777000777
Odds 9-low = comb(5,4)/comb(13,5) = 0.00388500388500389
Odds 8-low = comb(6,4)/comb(13,5) = 0.0116550116550117
Odds 7-low = comb(7,4)/comb(13,5) = 0.0271950271950272
Odds 6-low = comb(8,4)/comb(13,5) = 0.0543900543900544
Odds 5-low = comb(9,4)/comb(13,5) = 0.0979020979020979
Odds 4-low = comb(10,4)/comb(13,5) = 0.163170163170163
Odds 3-low = comb(11,4)/comb(13,5) = 0.256410256410256
Odds 2-low = comb(12,4)/comb(13,5) = 0.384615384615385
``````

So, the odds of a split given a flush on the board are:

``````P(10-low) * P(split|10-low) + P(9-low) * P(split|9-low) ... P(2-low) * P(split|2-low)
= comb(4,4)/comb(13,5) * comb(46,10)/comb(47,10) + comb(5,4)/comb(13,5) * comb(45,10)/comb(47,10) + ... + comb(12,4)/comb(13,5) * comb(39,10)/comb(47,10)
= 0.19653
``````

So, the answer is that the probability of a 5-way chop when there is a flush on the board is .19653, or about 1 every 5 hands.

Note: technically, I have oversimplified very slightly here. It's possible for a player to have an in-suit card higher than the lowest card on the board and still get a chop, because the board could be a straight flush. For example, a player holds A♥ and the board is T♥9♥8♥7♥6♥ -- it's still a chop. The math here gets to be a pain, however, and has a very, very tiny impact on the overall calculation, so I've skipped it for simplicity. Some back of the envelope calculations puts the change at around 0.17% (P=0.001750364), or an additional 1 hand per 571, making the total probability 0.19828, or still around 1 in 5 hands.

## Overall odds of a flush board and a five-way chop

We're almost to a general answer of the question, "How likely is it that a hand will have a flush board and split?" The answer is:

``````Number of suits * P(flush board|suit) * P(split|flush board)
= 4 * comb(13,5)/comb(52,5) * 0.19653
= 0.000389
``````

In other words, starting from a fresh deal, approximately one hand in every 2,569 will be a five-way chop with a flush on the board.

I think the real question here is with a 5 card flush on the board what is the chance that out of 5 players none have one of the remaining 8 flush cards? The answer is 12.3%.

The odds of player one’s first card is not a flush suit is 39 of the remaining 47. We can see 5 cards, so 52-5 = 47 available cards. Of those 47 there are eight flush cards remaining so 47 – 8 = 39. That the second also is not a flush card would be 38 out 46. Multiply both fractions for the one player result of 0.685476. 68.5%.

Continue with player #2: 37/45 * 36/44 then multiply player one’s odds for a total of 46.1% for two players.

The third player has 35/43 * 34/42 then multiply the previous values for a total of 30.4%. The fourth player has 33/41 * 32/40 then multiply the previous values for a total of 19.6%. The fifth player has 31/39 * 30/38 then multiply the previous values for a final total of 12.3%.

To continue, if 10 players were still in, it is only a 0.7% chance of no other flushes.