What are the odds of the board FLUSHING???

[RedskinRunner325]

Hello everyone, I was just trying to figure out what the odds are of all five of the community cards being the same suit. This doesn't really have any strategic implications that I can really think of, more just me being angry that my Ace pairing in a knockout Heads-up opportunity being foiled by the board deciding make every community card a diamond causing a split pot. Any with any idea of what the odds are when no one is showing any diamonds, I'd be interested to find out.

Thanks for the help,
RedskinRunner

 

[StealthLSU]

obviously depends on if you hold that card, but off the top of my head, assuming you don't have that suit.

13/50*12/49*11/48*10/47*9/46

.0006045 or .06045% That seems right, someone correct me if I'm wrong.

 

[cjatud2012]

I actually saw this happen today at an Sng, the board was JKA9T of hearts. What was even more sick was the dude sitting next to me had the Q of hearts

 

[RedskinRunner325]

obviously depends on if you hold that card, but off the top of my head, assuming you don't have that suit.

13/50*12/49*11/48*10/47*9/46

.0006045 or .06045% That seems right, someone correct me if I'm wrong.

I think your close, but I don't think there should be 5 variables, because the first card has a 100% chance of being the suit that the other four will have to be. And in the situation I described, there are a total of four hole cards that are not diamonds. So if the first card is a diamond, there are then 47 cards in the deck, 12 of witch are diamonds. So would the math be

12/47*11/46*10/45*9/44 ?????

I think that right or close, which would put the odds of the board flushing at around .2773% chance of the board flushing. I think this is close or on the right track, but any other ideas are absolutely welcome.

 

[GDRileyx]

According to Wikipedia, the odds of making 4 to a flush by the turn are .0156. So multiply that by 9/48 to get .002925. Which is about 3 times in 1000 hands, or 333-1.

 

[StealthLSU]

I think your close, but I don't think there should be 5 variables, because the first card has a 100% chance of being the suit that the other four will have to be. And in the situation I described, there are a total of four hole cards that are not diamonds. So if the first card is a diamond, there are then 47 cards in the deck, 12 of witch are diamonds. So would the math be

12/47*11/46*10/45*9/44 ?????

I think that right or close, which would put the odds of the board flushing at around .2773% chance of the board flushing. I think this is close or on the right track, but any other ideas are absolutely welcome.

yup, you are right, knew I missed something

 

[hothandsmgee]

What about the odds of three people flopping a flush. That happened to me recently and of course my q high wasn't good enough. That has to be crazy odds right?

 

[andirayo]

If you have no information about the player's hole cards, the probability of the board flushing (showing 5 cards of the same suit) is


0.198% = 1 * 12/51 * 11/50 * 10/49 * 9/48
If you are a player on the table and you know your own hole cards, there are two options:
Option A) You have suited hole cards.
Then, the probability of the board "flushing" is microscopicly higher:

0.204% = 39/50 * 12/49 * 11/48 * 10/47 * 9/46 + 11/50 * 10/49 * 9/48 * 8/47 * 7/46

Option B) You have offsuit (or double suited) hole cards.
Then, the probability of the board "flushing" is microscopicly smaller:

0.196% = 26/50 * 12/49 * 11/48 * 10/47 * 9/46 + 24/50 * 11/49 * 10/48 * 9/47 * 8/46

 

[TrillUziVert]

I hate it when I have two of the same suit, then there's three of them on the board. I'll be sitting there thinking "I got this". Then there goes a 4th of the same suit, and I just got cheated out my flush.

 

[jfmcd86]

I hate it when I have two of the same suit, then there's three of them on the board. I'll be sitting there thinking "I got this". Then there goes a 4th of the same suit, and I just got cheated out my flush.

All the Time this happens to me