A reader questioned my approach to calculating the probability of getting quads on the board with the first four cards, in a comment on this post. His objection, in short, is that the cards held by the other players eliminate a bunch of ranks from being possible quads on the board. Since all four of some rank must still be left in the dealer's hand after the pitch, whatever cards are held by other players exclude those ranks from being eligible.
I started to write a reply comment, but it was getting too long. Besides, it's an objection that comes up once in a while when I do a probability calculation, so I thought it would be worth making a separate post about it. It's something that gave me headaches the first several times I pondered poker probability questions. It took me a while to see straight about it, so I'm not surprised that others find themselves confused.
Here's my answer:
Of course our calculation would change if we knew what other cards were. But if you're going to go down that road, then you're going to drive yourself crazy.
From the point of view of an omniscient being, there is no "probability" about this situation at all. The cards have been shuffled, the order of the deck is set, and we will either get quads on the board on the turn or we won't. It is predestined to either happen or not happen, so to speak of probability makes no sense. In fact, that is true from our non-omniscient perspective, too. It will either happen or it won't, and that determination was made when the deck was shuffled and cut. Asking how likely it is that quads will hit the board is sort of a confusing question, because it has already been settled whether it will happen. We just don't know the answer yet.
Think of it this way: Instead of asking, "What is the probability of getting quads on the board after ten hold'em hands have been dealt out?", instead ask it this way: "Of all the times that in a ten-handed game I know only my own two down cards (and they are not a pocket pair) how often, on average, will I see quads hit the board with the first four cards?" That is actually the question that I answered.
If you start adding assumptions or information about what other cards we know, then you're changing the question, so of course you'll get a different answer, all the way up to the ultimate case where you know the location of every one of the 52 cards, at which point you can ask whether quads will come, and the answer is either a definite yes or no.
Here's another way to think about it. Suppose that I'm alone in my apartment. I shuffle the deck, deal myself two cards, but leave the other 50 in the deck. I look down and see that I've dealt myself a 5 and a king. I take the next 18 cards and move them from the top of the deck to the bottom, without looking at any. Now, if I deal out the next four cards face up, what is the probability that they will be all of the same rank?
Of course, it doesn't really matter if I move 18 cards from the top of the deck to the bottom of the deck. Nor would it matter if, instead of doing that, I dealt those 18 cards two to each of nine imaginary players around my table. I could even burn them in my fireplace, and it wouldn't change the math. I could actually have nine friends (well, if I had nine friends, which is hard for a misanthrope) each pick up their two cards and look at them, not telling me what they are. None of it matters. The answer to the question remains the same, if the question is asked the way I suggested in the fifth paragraph above. In fact, I could deal myself two cards, throw away all but the last four cards in the deck, and ask about the probability that they will constitute quads. Or I could deal myself two cards, then pick four cards at random from the 50 remaining and ask the same question. It's all exactly the same, in terms of the math. It doesn't matter which four cards we pick, or in what order, and it doesn't matter what happens to any of the other 46 cards, as long as we don't know what they are. As soon as we know what they are, then we have changed the question we're asking, and the math must change accordingly.
(I'm feeling very tempted to go off on a tangent about Herr Schrodinger and his poor cat, but I think I'll refrain.)
This issue comes up with every probability calculation in poker, though we usually don't think about it. If you have four to a flush on the turn and ask the probability of making your flush on the river, you don't really stop to think that that is, in one way of looking at it, a nonsensical question--because the card that is destined to hit the board last is not going to change its spots; it's either one of your suit or it isn't. The more logical way of framing the question is, "Of all the times that I am in this situation, what percentage of them will, over the long haul, result in making my flush?" That way you're not assigning a "probability" to an outcome that has already been determined concretely (though we don't know which it is yet).
It's also akin to asking, in pre-ultrasound days, whether a baby about to be born is a boy or a girl. It's not 50% boy and 50% girl. It's one or the other. Yet we say, loosely, that the probability of it being a boy is 50%, and the probability of it being a girl is 50%. In reality, it's a very different question than asking, before I flip a fair coin, the probability that it will land tails, because it still could go either way. But the baby is either male or female already. We just don't know which. Our use of the word "probability" for both questions really disguises the fact that the situations are actually different. They sort of look the same superficially, but in one case the outcome has been set already and we just don't know what it is, while in the other the outcome still could be either one.
In spite of those two different realities, the math is identical. If it helps, you can phrase the question analogous to how you might ask about the outcome of the coin flip that has not yet occurred. To set the stage, we specify that after the deck is shuffled and cut, we will deal out ten standard two-card hold'em hands, then discard the top card, put out a flop, discard the next card, and put out fourth street. So the first four cards on the board will be the ones that were at positions 22, 23, 24, and 26 in the cut deck. Now here's our pre-shuffle question: "If I thoroughly shuffle this deck of cards, then do a cut at a random point, what is the probability that the resulting order will have four cards of the same rank at positions 22, 23, 24, and 26?" That's not quite the same question for which I did the math previously, because of the extra condition in the prior example that I knew two ranks that would not work--but it's close. And it gets around the mental stumbling block of asking about the "probability" of an event that has already either occurred or not occurred.
If you want to make the pre-shuffle question truly identical to the one I worked through with the first post, then do this: Remove a 10 and a 4 from a standard deck (representing the two hole cards I had when the situation actually came up). Now ask the question I posed in the previous paragraph. Bingo--the math I worked out gives you the answer.
Whew! That was long-winded. But I hope it clarifies why the answer I gave was correct mathematically, even though it quietly skipped over the problem of speaking of the probability that something will occur, when the real answer is simply either "yes" or "no."