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.