So which do you think will be harder to get

I messed up my calculation in my earlier post, so I will show the work here to avoid mistakes!

Monte-Carlo simulations are indeed the easiest approach but I think this one was simple enough that it was still doable via a Markov chain at least for the FoT.

For FoT, you must get correct element, crit chance, crit damage, and a socket. You can only get 3 so that the 4th must be obtained by rerolling the movement speed to it. Since all 4 of those stats have the same weights 1000 out of the total 37500, it’s somewhat easy. The difficulty comes from the fact that only 1 element can roll on the amulet and thus changes the total weights when elemental damage rolls on it.

So to compute the probability, you need to consider 4 cases:

Case 1: Correct element rolls first, then two of: ChC/ChD/Socket
Case 2: One of ChC/ChD/Socket rolls first, then element, then ChC/ChD/Socket (what wasn’t rolled)
Case 3: Two of ChC/ChD/Socket rolls first two, then element last
Case 4: All three ChC/ChD/Socket rolls (no element)

*Note: cases 3 and 4 have the same probability since the total weights are only shifted for rolls that occur after the elemental damage.

The probabilities for each case are:

Case 1 (elem 1st): (1000/37500) * (3000/33500) * (2000/32500) = 1.469575e-4
Case 2 (elem 2nd): (3000/37500) * (1000/36500) * (2000/32500) = 1.348788e-4
Case 3 (elem 3rd): (3000/37500) * (2000/36500) * (1000/35500) = 1.234806e-4
Case 4 (no elem): (3000/37500) * (2000/36500) * (1000/35500) = 1.234806e-4
Total = 1.469575e-4 + 1.348788e-4 + 1.234806e-4 + 1.234806e-4 = 5.287975‬e-4

Which can be approximated as roughly 1 in 1891. To get an ancient on top of that would make it roughly 1 in 18911 which is extremely rare!

Can you try your simulation to see if you get a “good” FoT roughly once every 1890 amulets?

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