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.
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):
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):
Overall odds that the board has a flush in any suit, but none of the five players have a card from that suit:
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:
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:
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.
So, the odds of a split given a flush on the board are:
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:
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.