Friday, December 16, 2011

Hope y'all saw this

I haven't been following Twitter much during this trip, but back in the hotel room for the night, I looked to see if I had any mentions or direct messages, and boy did I! Everybody who knows me, it seems, was trying to make sure I knew about a hand played at the Epic Poker main event today, in which Joe Tehan's Mighty Deuce-Four knocked out both Faraz Jaka with A-A and Vanessa Rousso with Q-Q, three-way all-in pre-flop.


Read the details here (don't miss the Twitter messages from the principals at the end of the story) and here. The three of them talk about the hand on camera here.

My only gripe with the coverage is how some people refer to Tehan's move as a "bluff." How can moving all-in with the most powerful hand in poker be considered a bluff?


Halfway through Albuquerque visit


Day 4 of a week-long trip to Albuquerque to see Cardgrrl and her family. Today was zoo day. We spent a long, long time watching a group of six gorillas. They are endlessly fascinating creatures.
Open the picture in a new window to see it full size. I think it turned out well, for being a cheap point-and-shoot camera on maximal zoom in dodgy light.

Tuesday, December 13, 2011

Math is hard, II

Just minutes after writing the "Math is hard" post last night, I was leafing through the December issue of Ante Up magazine and spotted a column by Antonio Pinzari titled "Going further with math" (page 59). In it he recounts how he has learned the importance of knowing the basic math of poker, and how he drums it into his students.


Then we come to this paragraph:
Again let's go further with two suited cards preflop. You've overcome the 89 percent and flopped two of the same suit, what are the chances of making the flush by the river? Using the Rule of 4 x 2 (if you don't know what that is I suggest you find out fast) you have about a 35 percent chance of making the flush on the turn and an 18 percent chance if you missed the turn card by making a flush on the river.
I'll give him a pass on the awkward sentence construction, but I will not give him a pass on the bad math--not in a column in which he boasts about how important it is to know the numbers and in which he claims to be teaching this math to his poker students. I sure as hell wouldn't hire as a poker coach somebody who claims to know what he's talking about but obviously doesn't.

There are 13 cards in each suit. If I have two of a suit in my hand and there are two more of them on the flop, that's four, leaving nine unaccounted for. There are 47 unseen cards, so the probability that the turn card the dealer puts on the board will complete my flush is 9/47, or about 19.1%. Not 35%, Mr. Pinzari, O Great and Wise Coach of Poker Math. (The second part of his assertion is closer to correct. If I missed on the turn, the probability of spiking a flush on the river is 9/46, or 19.6%. Why is it slightly higher on the river than on the turn? Because the deck is now a little bit more depleted of cards of the other suits.)

I wonder how Mr. Pinzari thinks it is possible that the chance of a flush card hitting on the turn is twice as high as the chance of it hitting on the river. Is there something magical about that particular spot on the board--the one between the flop and the river--that magnetically attracts cards that will complete a flush?

The similarity between Josie's errors and Mr. Pinzari's is striking. Did she read this article, I wonder, and get misled by it?

Let's go on.
Here is a huge number Lee [Childs, in the October issue] didn't cover: 60 percent of all flops contain two suited cards.
BZZZZZZZZZZ! Wrong again.

This is stated without reference to what cards one is holding, so let's run the numbers that way. I.e., we'll just take three cards at random from a full deck of 52, ignoring what might have been dealt to any players. The easiest way to calculate the probability of getting two of one suit and one of a different suit is actually to sneak up on it the back way. It's easier to work out the probability of getting all one suit and of getting three different suits. Then we subtract those values from 100%, and what is left is the probability of getting two suits, since the only possibilities are one, two, or three suits.

We start with figuring the probability of the flop consisting of all one suit. It doesn't matter what the first card is. Let's say it's a club (or "crub," just for my friend Eric, who absolutely loves it when I use that word). The probability that the second card will also be a club is 12/51, because there are 12 clubs left among 51 cards. If that happens, then the probability that the third card will also be a club is 11/50, because there are 11 clubs left among 50 cards. The combined probability is thus 12/51 x 11/50, or 5.1%.

Now we figure out the probability of getting three different suits. Again, it doesn't matter what the first card is. Let's say it's a club again. What is the probability that the next card is something other than a club? Well, there are 51 cards left in the deck, of which 12 are clubs and 39 are non-clubs, so the probability is 39/51. For the river, we have 50 cards left, of which 12 are clubs, and 12 are whatever the suit of the second card was, leaving 26 that can complete our rainbow flop. Thus the combined probability is 39/51 x 26/50, or 39.8%.

So now we know the probability of a flop containing just one suit (5.1%) and of it containing three different suits (39.8%). The only other way a flop can come is with two suits--two cards from one suit and one card from another. That probability must therefore be 100% - 5.1% - 39.8%, which works out to 55.1%, not 60%, as Mr. Math Genius Pinzari claims.

What if we alter things a bit by assuming that we start with two of a suit in my hand? Again I'll use clubs as my example. We use similar logic, but it's made messier because now we have to do the arithmetic separately for each suit, since the probability of clubs hitting the flop is lower than for the other three suits (there being two fewer of them available in the remaining deck). I just filled a sheet of scrap paper with my numbers, and I'll spare you the details, but I work it out to be a probability of 11.0% to flop exactly two clubs, and 14.7% each to flop exactly two diamonds, hearts, or spades. Combining those last three is 44.1%, for a grand total of 11.0 + 44.1 = 54.1% to flop two of any suit, starting with two suited cards in one's hand. In other words, that assumption still doesn't get us to the 60% that Mr. Pinzari asserts. In fact, it's a little bit lower than our first calculation.

Sorry, Mr. Pinzari, but you get an F for a column in which you boast about knowing poker math while showing that you really don't understand it at all.



Monday, December 12, 2011

Math is hard

Lord knows I've made more than any blogger's share of mathematical mistakes in the course of five years of writing about poker. I kind of doubt, however, that I've ever made as many in one post as my pal Very Josie did earlier today.


Here's her first example:
I’m holding a KQ of spades. The flop comes ace of spades, 10 of hearts and 2 of spades. I have a nut flush draw and an inside straight draw. What are the odds that I hit one of these great hands? Hmmm…First thing to do is to count how many cards are out there that will complete my hand. 4 spades are showing, so there are 9 left that will give me a flush; 9. There are 4 jacks in this deck that will give me a straight; 4. Nine plus four is thirteen. Surely you don’t need to be an accountant to figure that out.
It appears that being an accountant is not enough to get it figured out correctly. The jack of spades is being counted twice here--once as a spade and once as a jack. The number of cards that will complete either a flush or a straight is 12, not 13 (as Josie acknowledged after a commenter pointed this out).
There are 13 cards that will give me a big and most likely winning hand. To determine the odds of hitting one of these cards on the turn you take your 13 outs and multiply that by four. 13 times 4 is 52. I have a 52% chance of hitting my winning hand on the turn. If I do not get my card on the turn, it’s time to calculate the odds of hitting it on the river. You take those same 13 outs and this time multiply them by 2. 13 times 2 is 26. I have a 26% chance of hitting my hand on the river.
No, no, no. The "rule of four" gives you the approximate probability that you will hit one of these cards on either the turn or the river. Look at it this way: Each card constitutes about 2% of the deck, so the probability that the next card the dealer shows is, say, the eight of hearts is about 2%. If there are 12 favorable cards whose position is unknown to us (i.e., they might be in the stub of the deck in the dealer's hand, might have been dealt to an opponent, or might be among the burn cards), there is a roughly 2 x 12 = 24% chance that one of them will appear as the turn card on the board. There is another roughly 24% chance that one of them will appear on the river. Combined, that yields about a 48% chance of hitting one on either the turn or river.

A commenter calling himself "Four Hands" correctly made this same point on Josie's blog:
the Multiply by 4 is to calculate the odds of hitting on the turn _OR_ the river. It important for calculating whether or not to go all-in on the flop, but it not accurate if you're calling for a single card, or comparing pot-odds unless you're going to be all-in.

The odds of hitting on the turn are the _same_ as hitting on the river, well, slightly different because you've seen one card, but close enough that the approximation is usually fine.
To which Josie responding, puzzlingly:
NO 4 HANDS! I think you're saying the odds between the turn card and river are pretty much the same, except for that one measly card. i disagree because after the flop you have two chances to hit your hand, yet after the turn you have 1 chance, which is 50% less chance of hitting your hand.

see what i'm saying?
Josie has confused here two different quantities. One is the probability of hitting one of the desired cards on the turn. The other is the combined probability of doing so on either the turn or river. They are both potentially useful numbers, but they are completely different. If your decision is simply whether to call a bet on the flop in order to see the turn card, then the 2 multiplier is your approximation, because you don't yet know if your opponent will bet again on the turn, nor how much he might bet. On the other hand, if one or both of you will be all in on the flop, then you're getting both the turn and river, and you'll be interested in the combined probability that a desired card will hit either spot. "Four Hands" had it exactly right.

Josie's post said that she is twice as likely to hit one of her outs on the turn as on the river. A moment's reflection should reveal that that is an absurd conclusion.

Her next example:
I have J-J, which is definitely okay. The flop is 4-4-8 rainbow (all different suits). I have an over pair and I’ll come out betting here. The question is, what are the odds of improving my great hand. Any jack or four will give me a full house, and there are two jacks and two fours left. I have 4 outs. Four times four is 16. I have a 16% chance of hitting a full house on the turn and since four times two is eight, I have an 8% chance of hitting that full house on the river.
The problem here is more conceptual than mathematical. When I'm in a situation like this, the probability of improving my two pair to a full house is the last thing on my mind. The far more important question is whether I have a better hand than my opponent right now. If he called me pre-flop with Q-Q or K-K, I'm in deep doo-doo--so deep that even making a full house by another 4 hitting the board won't get me out of it. On the other hand, if he called me with 10-10 or 9-9, he's the one that's deep in the doo-doo. The probability of me catching another jack is so low that it's not worth basing any decisions on. It's true that a 4 improves my hand, but it improves the other guy's hand equally, so it's not really a meaningful consideration. If he already has a 4 or 8-8, I'm toast, and only a jack will change that, not another 4.

Another problem with the language here is the use of the term "out." Outs are defined as cards that will improve a currently losing hand to a winning one. If you're already ahead, it's nonsensical to count your outs, or even to speak of having them; it's the other guy that has to be looking for outs. So when Josie says that she has "4 outs," it means that she either thinks she's behind here or doesn't understand the whole concept of outs. And, again, even if it's the former, the number of outs is actually just two, because the remaining two 4s don't move her from being behind to being ahead. Only the jacks can do that. Having a full house is still a losing proposition if the other guy has a bigger one.

But fundamentally this situation does not pose a mathematical problem. It's a hand-reading problem. I have to figure out whether I'm winning or losing. Most of the time it's a favorable flop for me; I was ahead with the jacks before the flop and remain so after the flop. But once in a while an opponent will have smooth-called pre with a bigger pair, or got very lucky with either 8-8 or a 4 in his hand. My primary tasks are (1) to extract the most value from my opponent if he has a pocket pair smaller than mine (which is just about the only thing he could have with which he might pay me off), or (2) spend as little as possible to determine that my hand is second-best on those occasions that he has 8-8 or a 4.

Josie's final example:
We’re playing with the two and three of hearts. The flop is 4 of spades, 5 of hearts and Q of clubs. Right now I have an open ended straight draw and there are 8 cards in the deck that will give me a straight. After the flop I take the number of cards out there that will help me (8) and multiply that by 4 to get 32. There’s a 32% chance I will hit my straight on the turn.
This is wrong, for the reasons given in the first example. It's closer to 16%. To be exact, there are 47 unseen cards, and 8 of them make the straight, so it's 8/47, or 0.170, or 17.0%.
Alas, the turn is a king of hearts. In addition to my open ended straight draw, I also have a flush draw. Now there are 15 cards in the deck that I want. If one of them hits on the river, I’ll have either a straight or a flush. 15x2=30. I now have a 30% chance to hit one of my hands.
That's about right. It usually works out to actually be about 2% more than the rule of thumb predicts. Here, for example, there are now 46 unseen cards. 15/46 = 0.326, or 32.6%. But the difference between the quick estimate of 30% and the actual value of 32.6% will essentially never matter to a poker decision. Either way, you're next going to translate it into odds (about 2:1) for purposes of determining whether a call is worthwhile.


Leaving on a jet plane

Tomorrow I'm leaving for a week in Albuquerque with Cardgrrl and the part of her family that lives there.

Betting stories

I just spent a pleasant hour or so reading this small collection of gambling stories: http://www.mcsweeneys.net/columns/fading-the-vig-a-gamblers-guide-to-life


Each tells the story of a single bet--on a horse race, a boxing match, blackjack, a chess match, along with reflections on life as revealed in the game in question. I like the author's writing style and hope that there will be more in this series.