2

Some context about why I'm asking:

I seem to be seeing another player, or myself, having 4 of a kind once every 100 - 150 hands.

This has happened in the region of 20+ times over the course of 2 weeks. I used to see 4 of a kind once every 1000 - 2000 hands.

I'm finding it quite odd, I've had it about 6 times myself in the past 2 weeks. In the period of seeing all of these 4 of a kinds, I've played about 2000 - 2500 hands at most.

How often statistically, should any player at a 9 seat table be getting 4 of a kind? Even if they fold and I don't see it, what's the general odds of 1 in 9 players making 4 of a kind?

As a bonus question, if I may, what are the odds of losing with 4 of a kind to a better 4 of a kind?

Improve this question
1
  • I added a detailed answer for you. I suggest you studying probabilities/statistics, which is a degree-level subject, but will help you a lot even with the very basic items they teach. – Luis Masuelli Jun 30 '16 at 22:06
1

1) According to http://wizardofodds.com/games/poker/, the probability that you can see quads at a 9-player table (assuming no one ever folds) is: 0.013183% or ~ once every 7586 hands

2) For your bonus question, I am not quite sure if that's what you are looking for, but have a look in here: http://forumserver.twoplustwo.com/25/probability/quads-over-quads-odds-866086/ (assuming that 2 players start with 2 pocket pairs,and they're going to see all 5 community cards) it is calculated: ~ 1 in 38,916 hands

++ Quick tip by the way: I would't worry too much if I ever won/lost a big pot with/to quads. Since it happens that rarely, it doesn't make much of a difference in your gameplay, winrate etc. I would focus more on "everyday" stuff, like cbet frequency, opening range, calling down frequencies etc and minimize my leaks in those sections.

Improve this answer
5
  • Hi koita, thanks for answering. I assume the odds for getting 4 of a kind are for 1 player? Do you know enough about statistics to tell me if for 9 players it would be 0.013183% x 9? Thanks for finding the odds on 4oak over 4oak, it's not that I'm worried about losing with 4oak, it's just it happened to me and I wanted to know the odds of it happening, but I couldn't find them anywhere. If I can get a second opinion to confirm the odds in #1 I'll upvote and accept - basically waiting to see if someone comments in agreement. :) – Dom May 24 '16 at 9:10
  • The 0.013183% is for 9 players. In the link I provided, there is a table with all probabilities from 2 to 9 players if you want to get similar data. – koita_pisw_sou May 24 '16 at 9:19
  • It seems a bit small for 9 players, and smaller than I expected for even a single player. Are you sure it's not for a single player in a 9 handed game? The link doesn't have much of a description above the table, and what it does have is ambiguous and could be interpreted in more than one way. – Dom May 24 '16 at 9:30
  • 1
    The thing about probability is that it never said when you'll see those hands. It may be once every 7586 hands mathematically but you could play say 21,000 hands without seeing quads and then you might see quads three times over the next 2,000 hands or so or in a short amount of time. Also that 0.013183% is the percentage chance of it occurring for one player in a 9 handed table. Odds of quads vs quads is even smaller, see here for bad beat odds for quads: wizardofodds.com/games/texas-hold-em/bad-beat-jackpots. I know it seems small but its really not as common as some players believe. – Grinch91♦ May 24 '16 at 9:39
  • *Or rather I should say odds of quads being beaten, not necessarily just quads over quads. – Grinch91♦ May 24 '16 at 9:40
0

How often statistically, should any player at a 9 seat table be getting 4 of a kind? Even if they fold and I don't see it, what's the general odds of 1 in 9 players making 4 of a kind?

The cardinality (whole possible hole + commie cards) is 52 taking 7: 133784560 The desired scenarios for you involve freezing 4 cards being equal (13 scenarios) and 3 out of the remaining 48 cards: 13 * 17296 = 224848.

The odds are like 0.0017 by dividing desired / cardinality, but this will include situations of 4oak being community cards.

If you want to exclude the in-table 4-of-a-kind (you want it only for you), then you will have to calculate like this:

  • cardinality: 674274182400 (52 taking 7 in certain order).
  • scenarios: 13 * 17296 * 5040 (13 pokers, 48 taking 3 free cards, 5040 stands for 7! which contemplates arbitrary sort) - 48 * 47 * 13 * 46 * 120 (scenarios where you don't have one of those cards in your hand, but anyway it is 4oak, while the latter 120 is 5! which contemplates arbitrary sort in the commie cards; arbitrary sort is already contemplated in the hole cards when multiplying 48 * 47).

The result is 0.0014 here.

In this 0.0014, however, another player (only one!) could also have 4oak, if that happens, scenarios are like this:

  • XX vs YY with XXYY? : 13 * 6 * 12 * 6 * 44 * 120 (which stands for: 44 as the ?, 120 as 5! for arbitrary flop order, 6 in both cases for suit combination in hands) = 29652480.
  • ?X vs YY with XXXYY : 13 * 4 * 44 * 12 * 6 * 120 (which stands for: 44 as the ?, 120 as 5! for arbitrary flop order, 4 and 6 for suit combination in hands) = 19768320.
  • XX vs ?Y with XXYYY : 19768320 same situation but reciprocal.

The question marks replace bricks. So you want to discard those scenarios when another one has a different 4oak:

  • cardinality: 674274182400 (52 taking 7 in certain order; this one did not change).
  • scenarios: 13 * 17296 * 5040 - 29652480 - 19768320 - 19768320 = 902154240.

Your odds of getting exclusive 4oak and nobody else getting another 4oak is: 0.0013. In the other 0.0001 case the other one will get another 4oak, so by having an exclusive 4oak there's a 1/14 chance the other one has another exclusive 4oak as well.

Please I need a cross-review in this point! I'd like to check if 0.0013 is a good result or I screwed with the calc application. I admit it looks pretty weird to me this 1/14 difference

As a bonus question, if I may, what are the odds of losing with 4 of a kind to a better 4 of a kind?

Now focusing on the conditional analysis on the game table. If your hand/commie already look like this:

  • XX with XXYYY: One arbitrary opponent has a chance of having 4oak in 44 / 45 * 22 = 0.04444
  • ?X with XXXYY: One arbitrary opponent has a chance of having 4oak in 1 / 45 * 22 = 0.00101
  • XX with XXYY?: One arbitrary opponent has a chance of having 4oak in 1 / 45 * 22 = 0.00101

By having those scenario configurations, you have to map the value of X like this:

  • 2 is worth 0, 10 is worth 8.
  • J Q K A is worth 9 10 11 12.
  • Lets call MAP(X) this mapping that converts the values like I told.
  • 12 will be the maximum value here.

Given those three scenarios which could risk of having another 4oak from only one arbitrary opponent, you require an additional condition: AND the other player beats me with an Y-valued poker, so the values are:

  • 0.04444 * (12 - MAP(X)) / 12 for scenario XX XXYYY.
  • 0.00101 * (12 - MAP(X)) / 12 for scenario XX XXYY?.
  • 0.00101 * (12 - MAP(X)) / 12 for scenario ?X XXXYY.

Disclaimer: These results are probabilistic fractions, which are always in closed interval [0..1]. Since none of these values is zero, you can calculate H = 1 / value and later say "the chances are 1 in H hands".

Improve this answer
-1

Just anecdotal but I was playing live at a casino last week and I flopped quads twice and hit quads on the turn another time in a 5 hour session. That probably represents about 160 - 175 hands. Sometimes the laws of probability can seem to be violated over a small sample size.

On one of these hands, BTW, I came close to hitting the bad beat jackpot ($250K). I had JcJs and the flop was KhJhJd. Of course I didn't know it until the hand was over but one of the other players had Qh10h. That works out to about a 9% chance after the flop of me winning the lion's share of the jackpot. Why can't someone hit a 2-outer on me when I really need it ?!?

Improve this answer
-3

Yes you are ranting. Really you are counting and seeing 4 of a kind once every 100 - 150 hands. If you are counting then you would have a number not a range.

The chance of 4 of a kind in 5 cards is 4,164 : 1

You are not playing with 5 cards you are playing with 7
The chance of 4 or a kind in 7 cards is 594 : 1

9 players is 9 hands so roughly 65 : 1 for a table of nine

It is covered in this wiki that I have linked in several of my answers.
If you google "poker hands probability" it is the top hit.

As a bonus answer stop ranting. If you have 4 of a kind assume it is good and get all you chips in the pot and hope for a call. If there is not another pair on the board and not a straight flush then it is the nuts. It is the right statistical call every time even if the board has another pair or a straight flush.

Out of curiosity could can you even have a quads and a straight flush?

hand 1 AA hand 2 KQ suited

board A A J T x

quads versus quads

hand 1 AA hand 2 KK

board A A K K x

Since there is 50% chance of having a block on the flush you are like 20X more likely to get beat by quads and I know how to run those numbers but at this point I am not interested in helping OP.

Improve this answer
22
  • 2
    Excuse you. Absolutely unnecessary rudeness. No I'm not ranting. I've been playing poker for 12 years and I've never seen 4 of a kind anywhere near as often as I've seen it recently. I've asked for the odds of a player getting it so that I can work out whether it's as unlikely as my gut feeling says it is, as it turns out, it appears to be more common than I thought it was. I assume the reason I'm seeing it more often lately is because I'm playing stakes where people play anything and call everything to showdown. Why would I be sitting there counting every hand and every 4 of a kind? :/ – Dom May 24 '16 at 5:37
  • If you are not counting then just how do you know 100 - 150. I read rant. An answer to your question and -1. You are welcome. – paparazzo May 24 '16 at 5:39
  • Right. So you count to 221 for aces but don't count 4 of a kind. An answer to the stated question and you gave it down vote. Fine, you will never get another answer from me. – paparazzo May 24 '16 at 5:46
  • 1
    Not to get in between you two, but I don't think this is how statistics work :"I know that when I haven't seen aces or kings for over 221 hands they should be along soon" – Chad Mattox May 24 '16 at 21:10
  • 1
    No. I discussed the context of the question. Discussing the user is discussing topics aside of the asked ones, picking aggressive / rude / bully behavior wrt what a user does, picking a fight or flamewar, and starting an answer like you did (without editing it, if the user edited). Accusing a user of bare ranting without the user asking for it, is discussing the user and not the question. That overall behavior is harmful for this community. – Luis Masuelli Jul 1 '16 at 0:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.