Unlikely events (non-grumpy content) (WARNING! Math ahead!)

I played at the Venetian again today. Two mathematical long-shots occurred that are worthy of note.

1

During a cash game, I heard a sudden commotion from the next table that could only mean that something really unusual had happened with the cards. I wasn't in a hand at the moment, so I stood up, took a few steps, and saw something I had never seen before: Quad kings beat quad tens, with both players having started with pocket pairs. If the Venetian had a bad-beat jackpot, this certainly would have qualified. I don't remember the order in which the cards came on the board, and I have no idea how the betting went down, but it doesn't really matter.

I wondered about the probability of this happening. Let's take any two pocket pairs, since the probability will be the same for whichever ones we pick. And let's ignore what order the community cards hit the board, because just about any order will get all the players' money in nearly every time (probably the only exception being if the high pair makes quads on the flop with the non-matching card, and the lower pair folds, thus never seeing the turn and river cards).

Here's the ugly math. First, how likely is it that two or more players will simultaneously receive pocket pairs when dealing to a table of ten players? This is something I've never worked out, maybe never even wondered about, but now that I've crunched the numbers, it's kind of interesting. You can find the relevant equations (and a web-based calculator to run them for you) many places, including http://faculty.vassar.edu/lowry/binomialX.html. Here, "p"--the probability of any player being dealt a pocket pair--is 1/17 (because after you get the first card, there are 3 possible pairs to it left and 51 cards left, and 3/51=1/17), or 0.0588. We'll set "n" at 10, assuming 10 players to a table. The probability that nobody gets dealt a pair on a given hand (i.e., k=0) then works out to about 55%, which is quite a bit higher than I would have guessed offhand. The probability of exactly one player being dealt a pair (k=1) is about 34%. The probability of two players being dealt pairs is about 10%, and the sum of the probabilities of all of the other possibilities (i.e., three through ten players being given pairs simultaneously) is the remaining 1% or so.*

(The interesting implication of this: If you're dealt a pocket pair, about 3 out of 4 times you're the only one at the table with a pair. I never knew this, and don't recall seeing it mentioned anywhere else before.**)

So let's settle on 11% as a good estimate of the likelihood that two or more players will, on any given hand, start with pocket pairs.***

We now have 48 unknown cards left, after assigning our two players their pairs (and it doesn't make a bit of difference how many other players are in the hand; their cards are unknown to us, so the math is the same as if only two people were playing). There are five cards to come on the board. There are exactly 1,712,304 different combinations of five cards that we can theoretically pick from 48 cards (as long as we ignore order). (Praise heaven that Microsoft Excel has a built-in function ["COMBIN"] to calculate this for me.) In order to get quads versus quads, four of those five have to be the cards in question, so now eight cards are assigned/accounted for, leaving 44. In other words, 44 of the 1,712,304 possible combinations of five cards we could put out will have the specific four cards we need to see this clash happen, plus one straggler. That makes the probability of quads versus quads (starting with two pocket pairs) 44/1,712,304, or 1 in 38,916. In other words, pocket pairs will only end up in the kind of quads-over-quads outcome that I saw tonight once in about 39,000 times.

Now we combine that with the fact that only about 11% of the time do we even have two or more players starting with pocket pair in order to set this up, and my grand conclusion is that one will see quads over quads only once in about 354,000 hands of Texas Hold'Em. It's even rarer than that, if you account for the fact that the two paired players very often won't both see the flop, since a large pair will usually raise before the flop, causing a small pair to fold. I would guess that one of the pairs folds before the flop roughly half the time, which, if correct, means that one would have to sit through something like 700,000 poker hands before seeing quads over quads hit. No wonder I've never seen it before!

2

I entered the Venetian's 8:00 p.m. tournament. I finished in about 16th place, but only the top five split the money, so I left with nothing to show for it. Oh well. At least I got a story out of the evening.

I had Q-10, and raised with it in late position. I was called by one other player. The flop was Q-7-2. I had top pair, and not many chips left, so I pushed all-in. She called. She had pocket 7s and had flopped a set. I was pretty much dead meat at that point. But then miracle of miracles, the turn card was a 10, giving me two pairs and a ray of hope. Then the river was a lovely third queen, giving me a full house. The only way I could win that hand was with the turn and river coming 10-Q, Q-10, 10-10, or Q-Q, and it happened! An online poker odds calculator says that I had a 1.6% chance of hitting one of these combinations. That makes this an even worse suckout than when I hit a 3% chance full house over two opponents with flopped flushes (see the tale at http://pokergrump.blogspot.com/2007/04/crazy-night-of-poker-non-grumpy-content.html). It's definitely one of my luckiest hands ever.

Incidentally, with experience I'm unquestionably getting better at making quick estimates of poker probabilities. As soon as this happened, I commented to the guy next to me, "That must be in about the 2% range to hit like that." Pat, pat pat. (That was me giving myself a pat on the back for being so smart. Maybe some day I'll be smart enough not to commit all my chips when I'm a 98:2 underdog.)


Editorial comment

If either of these hands occurred online, the chat boxes would fill with conspiratorial notes about how Internet poker is rigged to make improbable events happen more often. But given thousands and thousands of hands being dealt in a large card room like the Venetian every day, extraordinarily unlikely events are bound to occur, even with no puppetmaster manipulating the strings. Randomness is a wondrous thing to behold, given enough chances for rare events to occur. It's no different online, except that nerds sitting in their parents' basement playing poker on a computer all day can get their world view twisted pretty severely, and what are objectively pretty ridiculous theories about how improbable events occur can start to make a weird kind of sense.


*There's a little problem here, in that these occurrences aren't really quite independent, because one person getting a pair reduces the number of pairable cards available to everybody else. But this is close enough for present purposes.

**The exception, of course, is if you're dealt K-K. Then the probability that somebody else has been given A-A is about 99%. Or at least it seems that way....

***Again, this is a bit off, because of the possibility of two players both having the same pair. But I think that will only throw things off a small amount, and since I'm really just interested in ballpark figures, I'll ignore the discrepancy.

Addendum, August 23, 2007

I forgot yet another oddity that occurred during the tournament: 4-of-a-kind appearing on the board. In this case, it was 8-8-8 on the flop, and the final 8 on the turn. If we ignore what order the cards come in again, this will happen only once in 20,825 hands. I have seen this once or twice before, but it's pretty rare. Once it worked out great for me, because an opponent in a tournament flopped a full house. I had just ace-high. But with quads on the board when the river card hit, this idiot tried to bluff me. When there's four of a kind on the board, and I have an ace in my hand, I can't lose--the best he can do is tie me with an ace of his own and split the pot. So I raised, and he folded his cards face up to show his disgust at having his flopped full house go down in flames to an opponent holding just an ace. (He didn't have one.) Hee hee hee!

Addendum, August 25, 2007

If you're at all interested in the first problem I posed above, then you should read the comments. I'm putting this here, rather than in another comment, because I'm now pretty well convinced that I got something wrong, and the corrective information herein is more likely to be seen here than in another comment. (I'm always happy to be given better data or arguments; anytime I can replace bad information or logic with good, I'm better off for it.)

"Froggee" made me rethink how I approached the question of the probability that my pocket pair is the only one dealt on that hand, versus being up against another pair somewhere at the table. I posted the query on the Two Plus Two forum:
http://forumserver.twoplustwo.com/showflat.php?Cat=0&Number=11820546&an=0&page=0#Post11820546

A respondent there pointed me to an incredibly detailed calculation of this (which turns out to be much, much more complicated than I had imagined at first): http://www.math.sfu.ca/~alspach/comp35/ This is part of a larger set of posts on other computations problems related to poker, which, if your brain is wired strangely like mine is, makes for interesting reading: http://www.math.sfu.ca/~alspach/computations.html

Brian Alspach, author of those pages, approaches the problem as not just whether there is some other pair being held by another player, but whether there is a larger pair lurking out there. But that's OK; by using his answer for the case when you're holding pocket deuces, we get what I assume is the answer for whether there is any other pair (since any pair--other than the very rare case of somebody else having the other two deuces--will be larger than deuces). At the very bottom of the page cited above, we see that the probability of there being one larger pair is 32.69%, the probability of there being two larger pairs is 8.19%, and the probability of there being three larger pairs is 1.19%. These sum to 42.07%. The main reason the math is complicated is something I alluded to in a footnote to my original post: Being dealt a pair affects the probabilities of what other players can receive, and thus the standard formulas--which assume completely independent events--are a little off.

Anyway, I conclude from this that Froggee's method of estimation is better than mine was, and that when one has a pocket pair, about 42% of the time somebody else at the table will have another pair.

Incidentally, this doesn't affect the calculation I was aiming at (about quads over quads). I was correct that at a ten-handed table there will be two or more pocket pairs about 11% of the time. I just erred in my side note about what that implies about other players' holdings when I have been dealt a pair.

My thanks to "Froggee" for alerting me to the mistake.

Addendum, August 26, 2007

I thought of another way to verify the above conclusion--sort of a "poor man's Monte Carlo simulation." That is, I don't have an easy way to actually simulate 10,000 random deals, but I can do it mathematically, using the probabilities I derived in the comments:

P(0) = 0.5455
P(1) = 0.3408
P(2) = 0.0958
P(3) = 0.0160
P(4) = 0.0017
P(5) = 0.0001

In 10,000 deals, 3408 of them (ideally) would contain exactly one pair given out somewhere among the ten players. Let's say I'm in seat 1 for the whole time. I would then get 341 of these pairs. There would be 958 deals containing exactly two pairs, of which I would expect to receive 2/10, or 192. There would be 160 deals containing exactly three pairs, of which I would expect to receive 3/10, or 48. There would be 17 deals containing exactly four pairs, of which I would expect to receive 4/10, or 7 (all these are ideals or averages, of course).

Summed up, in 10,000 deals, I would receive a pocket pair 588 times (341+192+48+7) in seat 1. Of these, 341 would be the only pocket pair dealt--that's 58% of the 588 times, meaning that 42% of the time at least one other person would also have been dealt a pair.

OK, that does it, I'm convinced that 42%/58% is the correct answer. That is, 58% of the time that I have a pocket pair, it's the only one at the table, and 42% of the time there's at least one other pair given to an opponent.

Addendum, September 10, 2007

While looking around YouTube at various poker hands from televised shows, I came across this one: http://www.youtube.com/watch?v=KNz-Duyx3Lc

It's from the European Open III, though I've never heard of any of the players and don't recognize the commentators' voices (the American guy sounds very familiar, but I just can't place him). But it's a phenomenally rare occurrence, worth looking at. At a 6-player table, 5 players start with pocket pairs! The probability of this is about 0.0004%, meaning that it will happen about once in every 250,000 hands. Even more bizarre, three of the pocket pairs are aces, kings, and queens. The guy with pocket kings makes a phenomenal--almost impossible--laydown, correctly reading the opponent with aces. But then he gets his heart broken when a third king flops, and he would have taken the huge pot. Even sicker, the guy with queens spikes a third one on the river, meaning that Mr. Aces not only lost the hand, but would have been in third place if the kings had stayed in!

This is the sickest game ever invented!