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Personal Finance (Not Investing) • Actual Roth Conversion

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There is no problem with RetiredAL's equation. It's just the commutative property of multiplication (see that wiki article for more).

The Roth side of the equation is on the left: it costs 12% to put the money into the Roth, after which it grows tax free.
The traditional side of the equation is on the right: it costs nothing to leave it alone, growing tax free, until you withdraw at a 12% cost.
The driving force is not the incidental commutative properties , not really, although your answer treats it as it is which is really misleading. Because of the progressive nature of tax you need to integrate the tax not simply use a phantom 12% bracket that no one is getting consistently. In the Future, that number is more like (Balance)*((1-40.8%)*DD) for starters when applied to large withdrawals, and at this time it is closer to (1-27.8%)*(Balance*DD). Add to that IRMAA, an additional tax,....
Assume whatever you want for a future marginal rate and act accordingly.

If you think that rate will be the same as it is today, regardless of whether that is 12%, 24%, or whatever%, then in the Simplest situation it doesn't matter whether you convert now or withdraw later.

If you use a different assumption, you are likely to reach a different conclusion about your preferred (in)action this year. Whether any assumption turns out to be correct, ay, there's the rub.

Statistics: Posted by FiveK — Wed Dec 10, 2025 2:34 am — Replies 17 — Views 1354



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