jimmyjack
jimmyjack
Joined: Dec 18, 2013
  • Threads: 1
  • Posts: 4
December 18th, 2013 at 1:38:17 PM permalink
I am an odds junkie, but I admit, this is above my head.

Does anyone know what the odds are for 2, 3, 4 and 5 person Texas Hold'em games, where the 5 card river is out, and only ONE player wins with a high card?

Yes, that means, No pairs, No Flushes, and ONE of the 2, 3, 4 or 5 players has a minimum of a 4 or above (the lowest winning High Card can be a 4, as the ace would be a high card and the others would have to be split between 2s and 3s).

What are the smallest and largest odds of this happening?

I can calculate specific hands, I just don't know how to calculate the reverse of the odds, and they have to be odds as in "1 in X", they can't be percentage.

Literally in a situation like this each player has a 7 card chance to get at least a pair, I just don't know how to figure out the odds of no one getting anything except for a singular high card. Can someone help me by showing me the math and process involved? If the Wizard has actually calculated this out, pls show me where and the search terms, because I couldn't find it. Everyone is concerned about winning with an actual hand, I am concerned with the odds of winning with NO hand in a 5 player game. Literally, in this, you're betting the reverse probability in anyone getting a hand - making the odds compounded - and I'm just not sure I am doing it right (I wound up with odds in the millions when I tried to do it myself).
AxiomOfChoice
AxiomOfChoice
Joined: Sep 12, 2012
  • Threads: 32
  • Posts: 5761
December 18th, 2013 at 1:44:03 PM permalink
Quote: jimmyjack


I can calculate specific hands, I just don't know how to calculate the reverse of the odds, and they have to be odds as in "1 in X", they can't be percentage.



I'm not quite sure if this is what you mean, but.... moving between different formats is easy.

To go from "1 in X" to a percentage, just calculate 1/X and multiply by 100. For example, "1 in 2" is 1/2, which is 0.5, which is 50%.

To go from "1 in X" to "Y-to-1" just set Y = X-1 For example, "1 in 3" is "2-to-1" (technically, "2-to-1 against") That's because there are 2 losses for every 1 win (which is the same as 1 win for every 3 tries)

If you meant something else... please explain?
jimmyjack
jimmyjack
Joined: Dec 18, 2013
  • Threads: 1
  • Posts: 4
December 18th, 2013 at 2:05:22 PM permalink
You took that one sentence out of context so there's no way I can respond to your post.

I need to find the odds for NOBODY getting ANYTHING in a 5 player Texas Hold'em game with the river out and only ONE player winning with a high card.

If I can also see the math and how that's calculated I'd appreciate it.
beachbumbabs
Administrator
beachbumbabs
Joined: May 21, 2013
  • Threads: 99
  • Posts: 14232
December 18th, 2013 at 2:24:28 PM permalink
Quote: jimmyjack

You took that one sentence out of context so there's no way I can respond to your post.

I need to find the odds for NOBODY getting ANYTHING in a 5 player Texas Hold'em game with the river out and only ONE player winning with a high card.

If I can also see the math and how that's calculated I'd appreciate it.



jimmyjack,

I'm not so sure you're wrong that it's in the millions, if you're talking about a 4 or 5 player hand. And it also assumes people would stay in all the way to the river with such awful holdings as 2-3 for the others that don't have a 4. So I would have to think it's really a hypothetical question. But mathematically, yeah, I would guess it's very high odds against with 2 players and skyrockets after that. The proof would have to start with all possible combinations of 9 hands (which is in quadrillions, I think; couldn't find the Wiz's ever-useful chart for all possible hands from 2-10 cards) and subtract out all possible combinations of winners from those, then subtract all possible player hands that have any pairs, or have cards higher than a 4. Understand your curiosity, but that's only the start, and when is it ever going to happen? Just a bunch of bluffers having a goof all at the same time? A lot of work in that (for someone else; I don't have the math).
If the House lost every hand, they wouldn't deal the game.
PapaChubby
PapaChubby
Joined: Mar 29, 2010
  • Threads: 11
  • Posts: 494
December 18th, 2013 at 2:31:03 PM permalink
I don't think that there's any math for this.

It's easy to calculate the probability of no pairs on the board. Then its straightforward to calculate no pairs including any number of players' hands. But then you'd have to remove all the possible straight and flush hands, and I don't think there's any direct way to calculate this.

So the only approach that I can think of is to generate every possible combination of boards and players' hands, and count the number that meet your criteria. A computer program could do this fairly readily. It would have to cycle through 4.5e+12 possible combinations (for 5 players).
jimmyjack
jimmyjack
Joined: Dec 18, 2013
  • Threads: 1
  • Posts: 4
December 18th, 2013 at 2:46:35 PM permalink
The process I was using was a variation of the odds found on this siteonly I'd modify it for Texas Hold'em and flip some the values, but the problem is I found myself pushing odds of 1 in 18 million before I thought that I should go and find someone like the Wizard of Vegas to see if there are any actual stats out there.

I'm just not confident enough in my own math when viewing the numbers I came up with.

When you start including 5 players together you wind up with what appear to be exponential increases in the odds.

And yes, this odds calculation is assuming all players are bluffing (2, 3, 4 and 5), and the deck was especially stringy.
AxiomOfChoice
AxiomOfChoice
Joined: Sep 12, 2012
  • Threads: 32
  • Posts: 5761
December 18th, 2013 at 3:24:53 PM permalink
Quote: jimmyjack

You took that one sentence out of context so there's no way I can respond to your post.



Ok, then I was misunderstanding what you were saying.

Quote:

I need to find the odds for NOBODY getting ANYTHING in a 5 player Texas Hold'em game with the river out and only ONE player winning with a high card.

If I can also see the math and how that's calculated I'd appreciate it.



I know how to calculate it but it's a pain in the ass. I'd probably just write some software to iterate through the possibilities and count them.
jimmyjack
jimmyjack
Joined: Dec 18, 2013
  • Threads: 1
  • Posts: 4
December 18th, 2013 at 3:53:48 PM permalink
Well, I'll even settle for the process if it means I have to calculate it. I'm not sure of the process I used, so giving me a process would be good enough from my perspective.

What I do know is that you only need to calculate the odds of NOT getting 3 types of hands ( a straight, a flush and a pair ) in a 2, 3, 4 and 5 player game, making sure that no other player has any hands which would include one of those three types of score making combinations ( since those three types of hands are the progenitors of all score making combinations outside of the High Card ).
AxiomOfChoice
AxiomOfChoice
Joined: Sep 12, 2012
  • Threads: 32
  • Posts: 5761
December 18th, 2013 at 4:20:12 PM permalink
Quote: jimmyjack

Well, I'll even settle for the process if it means I have to calculate it. I'm not sure of the process I used, so giving me a process would be good enough from my perspective.

What I do know is that you only need to calculate the odds of NOT getting 3 types of hands ( a straight, a flush and a pair ) in a 2, 3, 4 and 5 player game, making sure that no other player has any hands which would include one of those three types of score making combinations ( since those three types of hands are the progenitors of all score making combinations outside of the High Card ).



I mean, it's a combinatorics question. Combinatorics is all about finding clever ways to count things so that it's easy (rather than long and brute-force). It's a hassle with hold'em because it's a 7-card game so there is intersection. Eg, you can make both a straight AND a flush, so you can't just add probabilities -- you need to subtract the intersection.

This is why I'd just enumerate through all the possibilities (with a computer). There are a lot of them but there are some short cuts you can take. Although, I dunno, maybe there's a clever way of doing the math that I'm not seeing.

  • Jump to: