# Hypothetically, if Exodia was unlimited (3 copies), what is the probability curve of summoning?

As many of us know, in the game Yu-gi-Oh! summoning all five pieces of Exodia results in an instant win. Currently, the five pieces of Exodia have a "Limited" status which means only one copy of each piece is allowed in a deck.

Now I am curious, what is the probability curve for summoning Exodia if the status of all the pieces were changed to "Unlimited", meaning three copies of each piece would be allowed in a deck? The same win condition exists (must have one of each piece).

Now that triple the number of pieces are present but the win condition requires only one full set, I think order might matter because just drawing five pieces is insufficient, one of each piece must be drawn; multiple pieces of the same kind do not contribute to the win condition.

If I understand correctly, the probability of drawing five pieces for a complete set in the opening draw from a forty-card deck is

(15!(40-5)!)/(40!(15-5)!) ->
(15!35!)/(40!10!) ->
(15/40)(14/39)(13/38)(12/37)(11/36) ~ 0.46%

Through brute force, I see Exodia is guaranteed by card 38. If no pieces are drawn in the opening hand, then the remaining twenty non-piece cards are drawn, then some combination of all three copies of the left legs, left arms, right legs, right arms are drawn before finally drawing Exodia the Forbidden One.

Beyond the end cases, I cannot think of which combinatorial/permutation(s) to apply.

I would like to know what equation(s) to use along with an explanation, especially how/if order matters. A graph of the probability curve would be great as well.