Even odds from die with single mark

Question

We want to make a close equivalent of a fair coin throw, by throwing an N-sided fair die T times. The catch: only one side is marked, by a colored disk on that side's center (like the traditional marking for one on a regular 6-sided die). We want something that always yields a decision after T throws.

Update: the position of observers must not matter. Say, the die is thrown on a slowly rotating platter.

Warmup: N=6, T=2.

More difficult: N=4 (a tetrahedron, with rounded edges and vertices). What's the lowest T allowing to reach a probability of 1/2 ± 1/1000?

Bonus: extend to other platonic solids N∈{8,12,20} and common dice N∈{10,14,16,18}.

Answer 1

Given the hint of the oversimplified model, here is another way to do the warm-up question N=6 T=2:

The dot can be on top, on a side, or hidden on the bottom. Take as one outcome that the two dice throws are the same (i.e. both on top, both on side, or both on bottom). The probability of that is $\frac{1}{36}+\frac{16}{36}+\frac{1}{36}=\frac{1}{2}$.

This does not work in the N=4 case, but can be generalised to other dice shapes.

For the other Platonic solids N={8,12,20}, T=1 is enough, as the dot is either in the upper or lower half. There are no face centres directly on the equator.

For the N=4 case the above trick does not work so we have to do it the hard way. Loopywalt came up with an answer for this first, but below is slightly simpler.

The smallest probability case is to land on the dot $T$ times, probability $1/4^T$. This is smaller than $1/1000$ when $T=5$, so let's try that first.
We have the following cases depending on how many times the dot is down (from 5 times face down to 0 times face down):
1* 1/1024
5* 3/1024
10* 9/1024
10* 27/1024
5* 81/1024
1* 243/1024
The first number is the number of different orders in which the throws can occur.
It just so happens that $243+10*27=513$, so a very simple way is to choose one outcome to be that it lands 0 or 2 times on the dot.
With smaller $T$ you would have to find cases with probabilities adding up to exactly $1/2$. You can see it as a knapsack problem in which the weights are multiples of one another. In such a problem the greedy algorithm is optimal, but the greedy algorithm does not give the sum $1/2$ in for each case of $T=2,3,4$, so those values of $T$ do not get us close enough to a fair coin.

Answer 2

Warmup: N=6, T=2.

Actually, this can be done in T=1. Toss the die and see whether the dot is facing you. (Three faces face you, except in the extremely unlikely case that the die lands so that only two faces face you.)

Answer 3

With the clever bits having been handled by @Jaap and assuming nothing of the kind can be done for a 4-sided die, here is an answer to the main question.

Choose T=5.

Group 1: No dot or
exactly one dot but not in the first or last trial or
a dot in exactly the first or last three or four trials or
all dots
Group 2: everything else
p₁ = (243 + 3 x 81 + 2 x (9 + 3) + 1) / 1024 = 511 / 1024

Proof that we cannot do better:

It suffices to show that events cannot be evenly split. In units of 1/N^T there is only one event (all dots) that is not divided by N-1. Therefore one group's probability is divided by N-1 the other's isn't.