What are the odds?

 Yesterday Poker News published this article about a remarkable hand at the final table of an event in the Lone Star Poker Series. It was all-in pre-flop, Troy Clogston's JsJh against short-stacked Don Iyengar's AdJc. Nothing remarkable about that, or about the way the board fell. What was remarkable was Clogston calling every card of the board before seeing it. Read the article and watch the embedded video there for the details of how it played out. 

Last night I was telling Nina about this. She asked what the probability was. Offhand, I guessed one in a million. This was an actual quick estimate on my part, not just thoughtlessly spouting the cliched "one in a million." Nina thought that was low by an order of magnitude--more like one in ten million. 

So I decided to do the math and see. As I write these words, I still haven't done the calculation, so I don't know the answer. By the time I finish this post, I will. 

We start with the flop. Clogston called for 8-9-10 rainbow. (I don't actually hear "rainbow" in the video, because the announcer talks over it, but I'm assuming it's there, as the text of the article says.) The flop not only comes 8s9h10c, but in that exact order. Now, I don't think the order matters, because surely everybody would give him credit for the call even if the order had been different. So let's ignore the left-to-right order. With 48 unknown cards, and none of the known cards being of the three ranks in question, what is the probability of this guess? 

There are 17,296 different ways of drawing three cards from a deck of 48. (You can find a combination/permutation calculator here.) There are 24 different ways of assembling 8-9-10 rainbow. (Each of four 8s can be matched with each of three 9s and each of two 10s, and 4 x 3 x 2 = 24.) So 24 of the possible 17,296 combinations would fulfill the first guess. Probability: 24/17,296 = 0.001388. This equals about 1 in 720. 

Now we're to the turn. We have 45 unseen cards remaining. So guessing one is the straightforward probability of 1/45, or 0.02222. 

On the river, it's 1/44, or 0.02273. 

For Clogston to get all three components of his guess correctly, we have to multiply those three probabilities: 0.001388 x 0.02222 x 0.02273 = 0.0000007010. This equals 1 in 1,426,000. 

So under the conditions as specified here, my guess was the closer one--and not too far off, for taking only a few seconds to think about it. 

However, perhaps you disagree with my decision about the order of the flop. If you want to make the challenge harder by saying that you would only accept Clogston's guess as correct if he got the order of the 8, 9, and 10 correct, it adds extra difficulty. 

Out of 48 cards, the first card must be an 8; probability 4/48, or 0.08333. The second card must be one of three nines that doesn't match the suit of the 8; probability 3/47, or 0.06383. The third card must be one of the two tens that doesn't match either of the previous suits; probability 2/46, or 0.04348. All three of those must be true simultaneously, so again we have to multiply the independent probabilities together. 0.08333 x 0.06383 x 0.04348 = 0.0002313. This equals about 1 in 4324. (A faster way to get the same result is to realize that you have the same 24 winning ways to make 8-9-10 rainbow, but now we're talking about permutations instead of combination, and a permutation calculator says that there are 103,776 unique orders of three cards that can be drawn from a deck of 48 cards. 24/103,776 works out to 1/4324 exactly.) 

If you now multiply that by the turn and river probabilities calculated previously (they are unchanged for this), you get 0.0000001168, or about 1 in 8,560,000. That is much closer to Nina's guess than to mine. 

So who won our little guessing game? As the classic poker answer goes, "It depends."