I was just playing a SNG on Bodog in which the above hand transpired. A king fell on the river, nobody had an ace, and we chopped the pot four ways.
I have seen a board like this on rare occasions before, but this time it occurred to me to wonder how frequently it will occur. Let's find out.
There are 50 unknown cards before the flop is put out. There are 19,600 different flops that can be made from those 50 cards. Obviously, in order to get quads on the board by fourth street, all three flop cards must be of the same rank. Furthermore, we won't be able to get quads on the board on fourth street if the trips on the board on the flop are 10s or 4s, as I have the last one of each in my hand. That means that there are 11 ranks eligible. For each of those, there are four different combinations of three cards that can constitute our all-one-rank flop, making a total of 44 possible flops that meet our criteria. That means that the probability of getting such a flop is 44/19,600, or about 0.0022, or about 1 in 445.
Now that we have an eligible flop, there is only one card left in the deck that will complete our quads. 47 unknown cards remain, so there is a 1 in 47 chance of getting the one we want; probability is 0.021.
The combined probability of an eligible flop and the perfect turn is, therefore, 0.0022 x 0.021, or 4.78 x 10^-5, or 1 out of 20,936. (The answer would be slightly different if I had a pocket pair, but let's not worry about that.)
That's rare enough to get your attention.
(Now the much harder math problem is this: What are the odds that I managed to get all of that right on the first try?)
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Had something similar happen to me once on PokerStars. I was a touch behind (and playing recklessly) with A9, while the other three in the hand had KK, QQ, and TT. The flop was 33x (the x might have even created a set for one of the other players).
Turn and river were 33. As I had the only A I took the pot along with the related flack towards me ("Why were you in with A9?" "Cuz I played foolishly.") and PokerStars ("Figures..." "Only on PokerStars" "FIXED!" - That last one was pretty funny as I had not yet had the chance to pass the dealer the folded up hundred).
No matter what site you play on, you'll see comments from the rigtards swearing that bizarre, suspicious things happen on that site more than on any other.
"Few players recall big pots they have won, strange as it seems. But every player can remember with remarkable accuracy the outstanding tough beats of his career." - Jack King, Confessions of a Winning Poker Player
with no action there you shoulda jammed all in and stolen the pot. :).
Interesting analysis, but I'm pretty sure the math is incorrect. Actually, I think your actual math is fine -- but I think you've started with some wrong assumptions. That said, I don't know the right answer myself, but perhaps we can work together towards the correct solution.
First, you say there are 50 unknown cards, so there are 19,600 possible flops. That's true, but in this case the "unknown" cards are very important. From the screenshot, it looks like you were playing 8 handed, so I'll base my calculations on that.
Just as you excluded 10's and 4's from possible quads because you were holding them, I think you need to exclude others' cards as well. I think of this in a similar way to the Fundamental Theorem of Poker. Your calculation of odds of quads would change if you knew what the other players had.
For example, let's say player 1 has AA, 2 has KJ, 3 has 47, 5 has 99, 6 has J3, 7 has 44, and you have T4. Now we can't make quads with the A, K, J, T, 9, 7, 4, or 3. We can only make quads with the Q, 8, 6, 5, or 2. Of course, the exact odds of which cards can possibly make quads will depend on the deal, and this is the part I can't mathematically figure out.
What is clear is that 8 handed, the range of ranks eligible to make quads ranges from 9 (2 players each with AA, KK, QQ, JJ or any other 4 pockets pairs) to literally impossible (the 8 players together hold at least one card of each rank). So even if we stick with 50 unknown cards and 19,600 flops, only 9 ranks are eligible to make quads (and this is extremely unlikely). Given that, the odds of this particular event occurring are 9 ranks (36 flops / 19,600 possible flops) = 0.0018. Assuming 1 rank available, 4 flops / 19,600 = 0.00020.
Continuing on, the odds of hitting that 4th card would presumably still be 1/47 or .021.
So the odds of this occurrence range from 0.0 to 3.78 x 10^-5. (The case of 1 possible rank would be
4.20 * 10^-6.) In this way of doing the analysis, it's not possible to be as high as you originally stated (4.78 x 10^-5).
But given that I've claimed to "know" all 8 hands, I think we can recalculate based on 36 unknown cards. So 36 choose 3 is 7,140. With 9 ranks eligible, we would have 36 / 7140 = 0.0050. The 4th card would be 1 / 33 = 0.030. In this calculation, 9 eligible ranks would have 0.0050 * 0.030 = 0.00015 or 1.5 x 10^-4. Interestingly, this is more likely than the original odds calculated.
Like you and the original calculations, I'm not sure I did my math correctly, but I did want to provide a different viewpoint for discussion.
A simpler way to come up with the answer:
There are 230,300 4 card (flop + turn) combinations. 11 of those combinations are quads.
11 / 230,300 = 1 in 20,936
If you were holding a pocket pair, then 1 more combinations of quads is possible.
12 / 230,300 = 1 in 19,191
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