All rolled up

Daniel Negreanu tweeted this question last night: "Weird question was raised: what are the odds that at an 8 handed stud table, all 8 players are rolled up? (Dealt 3 of a kind)." He followed that a short time later with this observation: "Bill Chen just told me its 1 followed by about 20 zeros."

This shouldn't be too hard to figure out. For conceptual simplicity, let's assume that the cards are dealt all three to one player, then three to the next, etc. The actual order of dealing doesn't affect the math.

The first player's first card can be anything. His second card must match the first, so must be one of the three remaining of that rank, out of 51 cards left in the deck. The third card he gets must be one of the two remaining of that rank, with 50 cards left in the deck. So the probability that the first player is rolled up is 52/52 x 3/51 x 2/50, or 0.00235, which is 1/425.

The second player can get for his first card any card in the deck except for the fourth of the rank that the first player got, meaning that 48/49 cards will do. Then his next card must be one of the three remaining of the same rank as his first one, with 48 cards left, for a probability of 3/48. His last card being the same rank is a probability of 2/47. The product of these three probabilities is 0.00261, or 1/384.

I'll spare you the repetitive prose for the other players, but you see how the logic goes. Here's what I get for each of the eight players:

1: 0.00234 (1/425)
2: 0.00261 (1/384)
3: 0.00290 (1/345)
4: 0.00324 (1/309)
5: 0.00364 (1/274)
6: 0.00412 (1/243)
7: 0.00468 (1/214)
8: 0.00534 (1/187)

(If it seems strange to you that the probability increases for each subsequent player, it's because the deck is becoming less disordered and more constrained with each set of three ranked cards we remove. Imagine if we took them out four ranked cards at a time--i.e., all the aces out, then all the kings, etc. By the time you got down to the final four cards in the deck, they would necessarily all be of the same rank. The same thing is going on here, though a little less drastically.)

To get the final answer, we just have to combine these probabilities by multiplying them together. The result is 2.16 x 10^20, or 0.0000000000000000000216, or about 1 time out of every 46,300,000,000,000,000,000 deals.

It is safe to conclude that (1) in the whole history of poker, this has never happened without the aid of a mechanic as dealer, and (2) if you ever see it happen, it has been set up. (You might be on "Candid Camera.")

If Negreanu was accurately quoting Bill Chen, and if Chen meant that this would happen one time out of about 100,000,000,000,000,000,000, then he was off only by a factor of about 2. If that was an off-the-cuff guess on his part, I'd call it phenomenally accurate.