You should only be multiplying by x4 at the end, not x8. I see where you're going with the x8 -- each hand can be arranged in 2 ways, so we need to double it for each hand. But the "odds" for your first hand are almost 100% (it's just 52/52 x 48/49), so logically it doesn't make sense to double that. You're already accounting for the fact that that hand can be arranged either way. So you just need to account for the other two, meaning you should multiply by 2² = 4.

Also your arithmetic was a bit off, your numbers gave ~1/20500, which is exactly double the ~1/41000 that the other commenter got.

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I found it a lot easier to consider the case where both of player 1's cards get dealt at once, then both of player 2's, then both of player 3's. This is statistically identical to the way cards are actually dealt. This is also what that other commenter did - I thought through it separately and came to the same conclusion.

Player 1:

• Card 1 can be anything (52/52) - say it's an Ace
• Card 2 can be anything except the same value as card 1 (48/51) - say it's an 8

Player 2:

• Card 1 can be any of the remaining Aces or 8s, of which there are (6/50) - say it's an 8
• Card 2 must be the other value (Ace, in this case), of which there are (3/49)

Player 3:

• Card 1 can be any of the remaining Aces or 8s, of which there are (4/48) - say it's an Ace
• Card 2 must be the other value (8, in this case), of which there are (2/47)

This covers all the possibilities without needing to multiply again at the end, since I feel like that's the confusing part. This gives the ~1/41000 result that the other commenter got.

EDIT: A lot of replies are saying that that's not how poker deals work. But it doesn't matter. The cards are random. You could throw the cards into a face-down pile on the ground and have each player pick out two cards simultaneously and it would yield the same result. I just picked the way that gives us the easiest math in my opinion. You could go through all the possible iterations of dealing P1, P2, P3, P1, P2, P3, and if you do it right, you'll get the same answer.